ENGE401 Manual S1 2019 Oneside
User Manual:
Open the PDF directly: View PDF 
.
Page Count: 120
| Download | |
| Open PDF In Browser | View PDF |
School of Engineering, Computer, and Mathematical Sciences
Engineering Mathematics
ENGE 401
2019 Semester 1
Engineering Mathematics - ENGE401
Course Manual
Auckland University Of Technology
This project is on GitHub, find it
and download the source files at:
https://github.com/millecodex/ENGE401
Licenced under MIT General License © 2019
Version 2.1 by Jeff Nijsse, 2019.
Jeff.Nijsse@aut.ac.nz
Version 1.0 by Peter Watson, 2010.
Contents
1 Algebra
1.1 Introductory Algebra .
1.2 Functions . . . . . . .
1.3 Polynomials . . . . . .
1.4 Systems of Equations .
1.5 Chapter Exercises . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
2 Trigonometry
2.1 The Unit Circle . . . . . . . . . .
2.2 Right-Angled Triangles . . . . . .
2.3 Trig Functions of Real Numbers .
2.4 Applications . . . . . . . . . . . .
2.5 Chapter Exercises . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
3 Exponential Functions
3.1 ex . . . . . . . . . . . . . . . . . . . . .
3.2 Logarithmic Functions . . . . . . . . . .
3.3 Exponential and Logarithmic Equations
3.4 Exponential Modelling . . . . . . . . . .
3.5 Chapter Exercises . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
1
3
6
7
11
.
.
.
.
.
15
15
17
23
26
31
.
.
.
.
.
37
39
40
45
48
49
4 Differentiation
4.1 Derivatives from 1st Principles . . . . . .
4.2 Standard Derivatives . . . . . . . . . . . .
4.3 Maximums, Minimums, and Tangents . .
4.4 The Product, Quotient, and Chain Rules
4.5 Parametric Differentiation . . . . . . . . .
4.6 Related Rates . . . . . . . . . . . . . . . .
4.7 Optimisation . . . . . . . . . . . . . . . .
4.8 Chapter Exercises . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
54
56
57
58
60
63
64
66
69
5 Integration
5.1 Standard Integrals . . . . .
5.2 Area . . . . . . . . . . . . .
5.3 Volume . . . . . . . . . . .
5.4 Integration by Substitution
5.5 Integration by Parts . . . .
5.6 Applications of Integration
5.7 Chapter Exercises . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
77
77
79
84
86
89
92
96
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Contents
iii
6 Differential Equations
101
6.1 Basic Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
6.2 Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
6.3 Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
7 Answers
108
1 | Algebra
Engineering Mathematics begins by reviewing foundational algebra. Many of the skills used in
this chapter are foundational mathematical tools that you will need to keep using repeatedly both
in this course and beyond. Refer to the course website blackboard.aut.ac.nz for additional
review material covering the basics of algebra.
1.1
Introductory Algebra
Some of the foundational algebraic properties will be covered here. This section is not comprehensive, and the student should refer to an introductory algebra book if some of these properties are
not clear.
Expanding and Factorising
Multiplying algebraic expressions is usually called expanding and the reverse process is called factorising. We usually use the word factorising in New Zealand, however, most textbooks use the
term factoring. We will use both terms interchangeably in this course. Factor(is)ing and expanding
can be viewed as opposite operations; one undoes the other.
Example
Expand the following algebraic expression (remove the brackets): x(x − 7)
Solution
x(x − 7) = x2 − 7x
Example Expand: (x+3)(x−3). Note there is a mnemonic FOIL that may help you remember
how to expand here: First, Inside, Outside, Last.
Solution
= x2 + 3x − 3x − 9 = x2 − 9
Example
Expand: x(x + 1)(x − 2)
Solution
Begin by expanding the first two terms:
= (x2 + 1)(x − 2)
= x3 − 2x2 + x − 2
Factoring involves removing common terms from expressions and then writing them as products.
Recall that product means multiply and can be shown algebraically by writing terms in brackets.
Example
Factor the following algebraic expression: 2x − 4x2 + 6x3
Solution
Remove a common factor of 2x: 2x − 4x2 + 6x3 = 2x(1 − 2x2 + 3x3 )
Example Factorise: x2 − 5x − 6. Note that this is a quadratic equation and will factor into two
sets of brackets.
Solution For these examples you are required to find a pair of numbers that add together to
1.1. Introductory Algebra
2
give −5 and multiply together to give −6. In this case the numbers are −6 and +1. So the answer
is
x2 − 5x − 6 = (x − 6) (x + 1)
This can easily be verified by expanding the brackets.
Example
Factor: x2 − 4x + 4
Solution
= (x − 2)(x − 2) = (x − 2)2
Solving Equations
An equation is a mathematical expression separated by two lines of equal length (=). There must
be symbols (either numbers or algebraic letters) on both sides of the equality. For example, 5x = 25
is an equation, however, 5(1) + 25 is just an expression. An equation can be solved; in the previous
example, x = 5 is a solution, whereas an expression may be simplified (5(1) + 25 = 30). Conversely,
factoring the quadratic x2 − 4x + 4 = (x − 2)2 is not solving the expression.
When solving an equation the order of operations is important. The acronym BEDMAS is used
for simplifying expressions starting with B=brackets, and ending with S=subtraction. The reverse
is true for solving equations. First you must undo any subtraction or addition to isolate the variable.
Example
Solve the equation for x: x − 11 = 7
Solution To isolate x we will add 11 to both sides: x − 11+11 = 7+11
And simplify: x = 18
Example
Solve the equation: 2x + 5 = 10
Solution To isolate x first we have to subtract 5 from both sides: 2x = 10 − 5
10−5
And then we divide both sides by 2: 2x
2 = 2
5
And simplify: x = 2
A general rule for solving equations is that you can do any mathematical operation to the
equation as long as you do it to both sides. For example, add 5 to both sides, divide both
sides by 2, multiply both side by sin(x), and so on.
Example
Find x: x2 + 1 = 3
Solution Subtract 1 from both sides: x2 = 2 Recall bedmas in reverse
√ order,√now we have an
2
exponent.
√ To solve for a power of 2, take the square root of both sides: x = 2, and simplify
to x = 2
Indices
Indices go by a few different names, sometimes they are called powers or exponents. In the expression x3 the index, exponent, or power is 3. This does not have to be an integer, or even a number:
2
xn has index n; x 3 has a fractional exponent; x−1 has a negative exponent; and xcos x has another
expression for its power.
1.2. Functions
3
The Rules of Exponents
• xn : x is called the base and n the exponent (or power)
• When multiply exponents of the same base, add the exponents: x3 × x4 = x3+4 = x7
• When dividing exponents of the same base, subtract the exponents:
x6
x5
= x6−5 = x1
4
• If an expression is raised to another power, multiply the exponents: x3 = x3×4 = x12
Negative Exponents
One of the most important rules for manipulating mathematics is the exponent of negative one. A
negative power is equivalent to the inverse of the same expression with a positive power. (Inverse
means one divided by the same expression.)
x−1 =
x−4 =
1
x
1
x4
2
7y 3
=
2y −3
7
2−3 =
1
23
=
1
8
Fractional Exponents
Exponents can decimal numbers, integers, expressions, variables, and also fractions. Fractional
1
exponents can written using a root sign, this is called surd form. x 2 is also known as the square
1
1
1
1
root of x. Using rules of exponents you can see that x 2 ×x 2 = x 2 + 2 = x1 = x. Converting between
surd and index form is quite handy, especially when we get to differentiation using the power rule.
Index form:
Surd form:
1.2
1
x2
√
x
3
x4
√
4
x3
1
64 3
√
3
64
a
xb
√
b
xa
Functions
A function is a mathematical relationship between groups. Given an element in one group, the
function says how to get to the other group. For example the function could be a formula that says
if you have x, the output is 2x. This can be written as f (x) = 2x where f (x) is called function
notation and in Cartesian coordinates also means y = f (x). We can depict this function visually
using x and y coordinates. Begin by selecting some inputs (x values) and then calculate the outputs
(f (x) values) from the formula f (x) = 2x.
This method is called making a table of values and in this example any real number for x produces
exactly one output for y (also a real number). The figure below plots the points on an (x, y) grid.
Connecting the points creates the line y = 2x.
1.2. Functions
inputs
x
4
outputs
f (x) = 2x
4 y
−2
f (−2) = 2(−2)
= −4
−1
f (−1) = 2(−1)
= −2
0
f (0) = 2(0)
=0
1
f (1) = 2(1)
=2
2
f (2) = 2(2)
=4
2
f (x) = 2x
x
−4
A table of values for the function f (x) = 2x
−3
−2
−1
1
2
3
4
−2
−4
A plot of the points showing a linear relationship
Lets write a precise definition of a function:
A function f (x) has exactly one output value, y, for any given input value, x.
In the line plotted above, we see that every x value has only one corresponding y value. This means
that y = 2x is a function. Conversely, a relationship that has two or more output values is not a
function.
The circle
values at x = 1: both
√ has two output
√
y = + 3 and y = − 3 are points on the circle.
3 y
The dashed line in the figure represents what
is called the vertical line test. If a vertical line
passes through more than one point on a curve,
then it is not considered a function.
2
1
x
−3
−2
−1
1
−1
−2
−3
2
3
Domain & Range
The domain of a function is the set of all inputs
that are valid; usually these are the x−values.
The range of a function is the set of all outputs
that are valid; usually these are the y−values.
For the line we plotted above, f (x) = 2x, any
value could be substituted into the function,
therefore the domain was all the real numbers.
This is written as: Domain x ∈ R. Similarly the
range was all the y−values, or y ∈ R. We will
return to domain and range later.
1.2. Functions
5
Linear Functions
Linear functions can represented nicely as a straight line on a standard Cartesian (x, y) coordinate
system. The following are all examples of linear functions, and not surprisingly, can be drawn as a
lines.
y
y
y
run
rise
x
(a) zero slope
x
(b) positive slope
x
(c) negative slope
Linear functions have a few characteristics that we will get used to manipulating. The slope of a
line is often represented by the letter m and can be calculated by taking any two points on the line
1
(x1 , y1 ), and (x2 , y2 ) and using the formula: m = xy22 −y
−x1 . This is also known as the gradient, a term
that will be used often in calculus.
slope=m=gradient=
rise
run
=
∆y
∆x
=
y2 −y1
x2 −x1
The standard form for an equation of a line is: y = mx + c where m is the slope described above,
and c is the y−intercept. Alternatively if you know the slope and any given point (x1 , y1 ), the
equation of a line is y − y1 = m(x − x1 ) where x1 and y1 are the coordinates of a point on the line.
Note that a vertical line has an undefined slope. Using the formula above, a vertical line has a
slope of m = ∆y
0 because it has the same x values everywhere. Dividing by zero is undefined (try
on your calculator) and therefore a vertical line is not considered a function.
Question Does a vertical line pass or fail the vertical line test? Why?
Quadratic Functions
A quadratic relationship scales with the square of the input values. The following are all examples
of quadratic functions:
• y = x2 − 5
• s(t) = −4.9t2 − 15t + 3
• b2 + 7b − 1 = 0
1.3. Polynomials
6
Note that they all have a power of 2 in the equation, and that is the highest exponent. This type
of relationship is also referred to as parabolic. When a parabola equation is plotted, the solutions
represent where the function crosses the x−axis. These points are called roots.
Example
Find the roots of the following parabolas.
y
y
y
x
x
x
(a) y = x2
(b) y = 0.3x2 − 1
(c) f (x) = −x2 − 2 ∗ x + 3
Solution The roots are where the function intersects the x−axis. The x−axis is where y = 0,
so we will substitute y = 0 into the functions and solve the equations for x.
(a) Substituting in y = 0 gives the equation 0 = x2 . This solves directly for x = 0. Therefore the
root to y = x2 is 0.
(b) Solve the equation
(c)
2
0 = 0.3x − 1
2
1 = 0.3x
1
= x2
0.3
r
1
=x
0.3
Therefore x = ±1.826
0 = −x2 − 2x + 3
2
0 = x + 2x − 3
here you can divide by − 1
and factor
0 = (x + 3)(x − 1)
Therefore x = −3 and x = 1
The parabola from part (c) above was solved by factoring. Not all quadratic equations can be
solved in this manner. The standard form of a quadratic equation is written ax2 + bx + c = 0. If
we solve this equation for x we get the quadratic formula:
√
−b ± b2 − 4ac
x=
2a
where a, b, and c are coefficients (a 6= 0). Note here there are two possible solutions because of the
plus-minus sign (±).
1.3
Polynomials
A polynomial is a type of function that comes up a lot. The quadratic equations above are all
examples of polynomials. The standard form of a quadratic equation is shown below with some of
the terminology.
1.4. Systems of Equations
7
The prefix poly means many, and polynomials are not limited to three terms. The general form of
a polynomial can be written as:
A1 xn + A2 xn−1 + A3 xn−2 + · · · + An x1 + C = 0
where the terms are written in decreasing powers of x, with
• n ≥ 0, n ∈ Z. This means the exponents must be integers.
• A1 , . . . , An are real numbers.
• C is a constant.
• The order or degree of the polynomial is n (the highest exponent).
• Here, x is the variable. You may have more than one variable in a polynomial, for example
4x2 + y − xy + 4 is a valid polynomial.
1.4
Systems of Equations
A system of equations means having more than one relationship represented within a common
context. Revenue and costs may have different functions but both relate to the same product. We
will study systems composed of two equations and two unknowns. Systems with more equations
and more variables are possible and will be covered in future courses. Three approaches to solve
systems of equations will be covered here:
• graphing
• elimination
• substitution
It helps if you can visualise the shape of the two functions so that the meaning of the solution is
clear in your mind. Later in the course we will find the area between two curves using integration
where the intersection of these two curves represents the solution to a system of two equations. See
section XX.
We know that linear equations in two variables are represented by straight lines. Straight lines will
always intersect unless they are parallel. The coordinates of the point of intersection of the straight
lines is called the solution. you could use a graphical method or one of the two algebraic methods
(substitution or elimination) to find the solution. We will start with a graphical method.
Example
Find the solution to the set of linear relationships given by:
2y = x − 4
y=
x
4
1.4. Systems of Equations
8
y
Solution Here we have two linear equations
(we know they are linear because the highest exponent is 1) and if the lines intersect, that point
of intersection represents a solution.
From the plot we can see the point of intersection
is (8, 2). Therefore the solution to the system of
linear equations is (8, 2). Try plotting the lines
yourself with desmos .
2
y=
x
2
−4
y=
x
4
x
−2
2
4
6
8
10
−2
Its not always convenient to graph a system of
equations to find the solution; you may not have
access to a computer, or the solution may not
be integers. Solving by direct substitution is the
next method.
Solution by Substitution
Example
Consider the system of equations
x2 + y 2 = 25
(1)
3y + x = 15
(2)
Solution Without knowing what the functions look like or plotting them, we can isolate a
variable and substitute it into the other equation. Rearrange equation (2) to isolate x and substitute
into equation (1):
x = 15 − 3y
Equation (1) becomes
(15 − 3y)2 + y 2 = 25
(expand and simplify)
225 − 90y + 9y 2 + y 2 = 25
10y 2 − 90y + 200 = 0
(divide by 10)
2
y − 9y + 20 = 0
(factor the quadratic)
(y − 5) (y − 4) = 0
Either
y−5=0
so
y=5
or
y−4=0
so
y=4
Lastly, back-substitute the y−values into equation (2):
When y = 5, 3y + x = 15
15 + x = 15 ⇐⇒ x = 0. This means (0, 5) is a solution.
When y = 4, 3y + x = 15
12 + x = 15 ⇐⇒ x = 3. This means (3, 4) is a solution. Note there
are two solutions here. Verify by graphing the functions.
Solution by Elimination
The third method is called elimination and works by eliminating one of the variables from the
equation set, then solving for the other variable.
1.4. Systems of Equations
Example
9
Solve the system by the method of elimination:
4x − 3y = 5
(1)
4x + y = 1
(2)
Solution If we subtract equation (2) from equation (1) columnwise then the x term is eliminated because 4x − 4x = 0.
4x − 3y = 5
−
4x + y = 1
0x − 4y = 4
Now there is one equation with one unknown: −4y = 4 so y = −1. Back-substitute into either
previous equation to solve for x = 12 . Therefore the solution is x = 12 , y = −1.
Example
Solve the system by the method of elimination:
3a − 7b = −3
6a − 4
b=
2
(1)
(2)
Solution The first step is to write the equations so the variables line up in columns. Multiply
equation (2) by 2 to get 2b = 6a − 4 and rearrange the terms to match equation (1).
The system now looks like:
Now calculate (4) − (3).
Multiply equation (1) by 2.
3a − 7b = −3
(1)
6a − 14b = −6
(4)
6a − 2b = 4
(3)
6a − 2b = 4
(3)
6a − 14b = −6
−
6a − 2b = 4
0 + 12b = 10
We can’t eliminate any variables because the coefficients
are different.
b =
Back-substitute b into equation (1) to solve for a: 3a − 7( 56 ) = −3. Verify that a =
5
6
17
18 .
Special Cases
Consider the parallel lines shown. What is the
solution to this system? Parallel lines will never
meet by definition and so will have no point of
intersection. In this case the system has no solution – which is a perfectly valid solution! Notice
the slope of both the lines is 13 which means they
are parallel.
Sometimes the two equations will be two different representations of the same line. Imagine two
line plotted on top of one another. In this case
the system has an infinite number of solutions
because every single point on the first function
is also on the second function.
y
2
y=
−4
−2
x
3
+ 0.5 x
2
4
−2
y=
−4
x
3
−2
1.4. Systems of Equations
10
Example If you start with a linear equation such as y − 2x = 3 and multiply each term by a
constant you will get an equivalent equation. If you multiply the equation by another number you
will get a further equivalent equation.
y − 2x = 3
Multiply by 2
2y − 4x =
6
Multiply by −3
−3y + 6x = −9
Solution We know these are both just different ways of writing the original equation y −2x = 3
or y = 2x + 3. To say the system has an infinite number of solutions we are really saying every
point on y = 2x + 3 is a solution.
Guidelines for Solving Systems of Equations
These guideline provide a useful way to tackle any problems where equations are involved.
1. Identify the variables. We often call them x and y, but you may chose any name or
letter you want.
2. Express all unknown quantities in terms of the variables.
3. Set up a system of equations using the facts provided by the problem.
4. Solve the system of equations and use the solution to check it satisfies the conditions
of the problem. Write a sentence describing the answer to the original problem.
Example It takes a boat travelling downstream 1 hour to cover the 20 mile distance. On the
return trip it take the boat 2.5 hours. What is the speed of the boat and the speed of the current?
Solution This is about a boat travelling with the current and against the current and depends
on you knowing that velocities are vectors that can be added and subtracted. Let the speed of the
boat be x mi/h and the speed of the current be y mi/h.
Upstream speed = x − y
Downstream speed = x + y
Total distance
Total time
so Total distance = Speed × Total time
Speed =
20 miles = (x + y) × 1 hour
20 = x + y
Also
20 miles = (x − y) ×
8=x−y
Now we can add equations (1) and (2) and y is eliminated
28 = 2x
x = 14
(1)
5
hours
2
(2)
1.5. Chapter Exercises
11
Back-substitute into either equation (1) or (2) to solve for y = 6.
Check: The boat travels at 14 mi/h and the current travels at 6 mi/h so the effective speed of the
boat is 20 mi/h. At 20 mi/h the 20 mi trip took 1 h. Upstream the speed is 8 mi/h.
Total distance
Speed
20
5
=
= h
8
2
Total time =
Therefore the speed of the boat is 14 mi/h and the speed of the current is 6 mi/h.
1.5
§1.1
Chapter Exercises
Introductory Algebra
1. Remove the brackets and simplify
(a)
(c)
− (x + y)
x2 + 5x − 1 − (2x − 3)
(b) −3(5x − 2y)
(d) (2x − 1) (2x + 1)
2. Calculate the value of
(a)
(15.3)0
(b) 10−2
(c)
ππ
(d)
4
7
(c)
abc
a−2 b2 2c
(d)
1 2
x
2
3
x
(b)
1
1 −5
8
× 0.5
3. Simplify
(a)
2
3a2 b
(b)
x 3 3
x
3
4. Evaluate
(a)
√
4
2.7 accurate to 2 decimal places.
to 3 decimal places.
5. Factorise the expressions by removing the common factors.
(a)
(c)
7y 2 − 14z 2
(a + c)2 − 4(a + c)
(b) x(y − 2) + x2
(d) SA = 2πrh + 2πr2
6. Factorise the quadratics.
(a)
(c)
x2 + 11x + 28
b2 − b − 20
(b) 2x2 − 5x − 12
(d) 3x2 − 7x + 2
7. Solve the equations
(a)
(c)
(e)
7x − 16 = 23 x + 4
x2 − 2x − 8 = 0
x3 − 2x2 − x − 1 = 0
(b) (x − 2)2 = 15
(d) 2x2 + 5x − 4 = 0
(f ) x (x − 1) (x + 2) = 16 x
1.5. Chapter Exercises
§1.2
12
Functions
1. Make t the subject of the equation
(a)
v = u + at
(b) l = l0 (1 + αt)
2. If f (x) = x2 and g(x) = x3 evaluate
(a)
(c)
f (2)
g(3)
(b) f (−2)
(d) f (g(2))
3. The volume of a pipe with length l, inner radius r and outer radius R is V = π R2 − r2 l.
Find the volume when R = 3.1m, r = 2.2m and l = 5.3m.
4. Sketch the following graphs (without using Desmos).
(a)
(c)
(b) 2y = x3 + 1
(d) y = −(x + 1)2
y = 4x − 4
y = x2 − 3
5. Find the x and y intercepts for the graph of y = x2 − 3
6. Use the grid provided to answer the following questions.
(a)
(b)
(c)
(d)
(e)
Show the points A (3, 5) and B (−2, −5) on the graph.
¯
Calculate the slope of the line through A and B. i.e. the slope of AB.
What is the equation of the line AB?
What is the equation of the line parallel to the line AB through the point (−2, 3) ?
Starting from the point B on the graph frame draw a line with a slope of 45 .
6 y
4
2
x
−6
−4
−2
2
−2
−4
−6
4
6
1.5. Chapter Exercises
13
7. Complete the table of values for the function g(x) = x3 − x and sketch the graph
2 y
g(x) = x3 − x
x
−1.5
−1
1
−0.5
x
0
−2
0.5
−1
1
1
2
−1
1.5
−2
8. Graph the three functions on a common screen using desmos . How are the graphs related?
(a) y = x2 ,
(b) y = ex ,
y = −x2 ,
y = −ex ,
y = x2 sin x
y = ex sin 5πx
9. Given f (x) = x and g(x) = sin x, graph f , g, and f + g on a common screen to illustrate
graphical addition.
§1.3
Polynomials
1. What is the degree of the polynomial?
(a)
(c)
y = x2
a99 − 99a2 = −99
(b) 4x3 − 2x2 + 6 = 0
(d) ey 4 − ey 3 + ey 2 − ey 1 + e
2. Are the following considered polynomials?
(a)
(b) π x
π
(c)
Ax2 + Bx + C
3
(d) 4x3 −2x 2 +1 = 0
3. Make up your own polynomials of degree 1,2,3, and 4.
§1.4
Systems of Equations
1. Solve the system of equations graphically. For the non-linear ones you may want to use desmos .
(a)
y= 2x + 6
y= −x + 5
(b)
3y
4
+ 1= x
−x= y
−x + 12 y = −5
2x − y = 10
(d)
x2
9
2
(c)
2. Solve the system using substitution.
+ y18 = 1
y = −x2 + 6x − 2
1.5. Chapter Exercises
14
(a)
x−y =2
2x + 3y = 9
(b)
2x − 3y = 12
−x + 23 y = 4
(c)
x+y =8
y = −8 −
(d)
x + y2 = 0
2x + 5y 2 = 75
x
8
3. Solve the system by eliminating a variable.
(a)
x + 2y = 5
2x + 3y = 8
(b)
x2 − 2y = 1
x2 + 5y = 29
(c)
3x2 − y 2 = 11
x2 + 4y 2 = 8
(d)
12x + 15y = −18
2x + 5y = −3
§1.5
Word Problems
1. A rectangle has an area of 180cm2 and a perimeter of 54cm. What are its dimensions?
2. The admission fee at an amusement park is $1.50 for children and $4.00 for adults. On a certain
day, 2200 people entered the park and the admission fees collected totalled $5050. How many
children and how many adults were admitted?
3. A woman keeps fit by bicycling and running every day. One Monday she spends 12 h at each
activity and covers a total of 12 12 mi. On Tuesday she runs for 12min and cycles for 45min,
covering a total of 16mi. Assuming her running and cycling speeds don’t change from day to
day, find these speeds.
4. A customer in a coffee shop purchases a blend of two coffees: Kenyan, costing $3.50 a pound,
and Sri Lankan, costing $5.60 a pound. He buys 3lb of the blend, which costs him $11.55. How
many pounds of each kind went into the mixture?
2 | Trigonometry
In the first half of this chapter the three trigonometric functions (sine, cosine and tangent) will
be viewed as functions of angles. The second half of the chapter will approach trigonometry as
functions of real numbers.
Pythagoras
Pythagoras is one of the most well known historical figures in mathematics and philosophy, primarily
for his eponymous theorem. Given a right-angled triangle, the square of the hypotenuse equals the
sum of the squares of the other sides.
The Pythagorean Theorem:
a2 + b2 = c2
This diagram will be the basis for a lot of our study of triangles. We will return to the Pythagorean
theorem in section 2.2 when discussing some special trigonometry relationships.
2.1
The Unit Circle
It should come as no surprise that the convention for measuring angles should be the same as for
measuring the distance around the perimeter of the unit circle.. An angle is measured in degrees
and is the amount of rotation between two rays about a vertex. If the vertex is placed at the point
(0, 0) and one ray is placed along the positive x-axis then we let the second ray go through the
point P (x, y) on the unit circle so that the relationship between the angle and the arc length t can
be established. The convention is that the positive direction for measuring angles is anticlockwise
and the negative direction for measuring angles is clockwise. You will be aware that one complete
1
cycle measures 360°. One degree therefore is 360
of one complete cycle. In this subject the word
revolution is often used instead of cycle. Other terms with which you might be familiar are “a
quarter turn” for 90° and “a half turn” for 180°.
If you draw a unit circle (centre (0, 0) radius 1) then you can draw the ray through any terminal
point and show the relationship between t and the angle at the vertex (0, 0).
Definition: If the unit circle is drawn and the distance t is measured around the perimeter from
(1, 0) to the point P (x, y) then we say the angle is measured as t radians.
The abbreviation for radians is rad. This abbreviation will be used in the examples. Some books
2.1. The Unit Circle
16
uses the term “angles in standard position” to describe this situation. We will not define this term
unless it is unavoidable and will stick to rad.
Using the language of geometry, the unit circle
is cut by two rays one through the points (0, 0)
and (1, 0) and the other through the points
(0, 0) and P (x, y).
t is referred to as ”the length of the arc”. The
angle θ between the two rays is referred to as
”the angle subtended at the point (0, 0)”. We
will often leave out the word ”subtended” however as it is implied.
We can say therefore that the angle measured in radians is related to the same angle measured in
degrees. the relationship between these two measures must be understood and must be able to be
derived.
Relationship between Degrees and Radians
One complete revolution is 360° if the angle is measured in degrees and 2π if the angle is measured
in radians. So
2πrad = 360°
or πrad = 180°
You should derive this formula whenever you are asked to convert degrees to radians or radians to
degrees.
Example
(a) Convert 36° to radians (b) Convert
π
3
rad to degrees (c) Convert 1 rad to degrees
Solution
(a)
(b)
180° = πrad
π
1° =
rad
180
π
π
36° =
× 36 = rad
180
5
(c)
πrad = 180°
π
180°
rad =
= 60°
3
3
πrad = 180°
180°
1rad =
π
≈ 57.295 779 51°
≈ 57.3°
Note the similarity between the answers to (b) and (c). This is because π3 = 1.047197551 so you
would expect the values in degrees to be similar. With this terminology we leave out the word
measured when we talk about measuring angles. We say “the angle is 60°” when we mean it has
been measured as 60° or “the angle is π3 ” to mean it has been measured as π3 rad. Notice we always
2.2. Right-Angled Triangles
17
put in the degree symbol and often omit the units when the angle is measured in radians. Should
units be omitted assume the angle is measured in radians. We often use the Greek symbol θ for
the angle subtended at the centre of the unit circle so θ = 60° or θ = π3 are further examples of
terminology that is commonly used.
Arc Length
Let θ be the angle subtended at the centre for the ends of an arc of any circle then the fraction of
θ
θ
the circumference of the circle is 2π
if θ is measured in radians and 360°
if θ is measured in degrees.
The length of the circumference of any circle whose radius is r is 2πr. If θ is measured in radians
Length of arc =
θ
× 2πr = θrorrθ
2π
If θ is measured in degrees
Length of arc =
θ
× 2πr
360°
The simplicity of the first formula shows why working with radians is preferred.
Example Find the length of an arc that subtends an angle of 45° at the centre of a circle whose
radius is 9 cm.
Solution
(1)
Method 1:
(2)
θ
× 2πr
360◦
45
=
× 2π × 9
360
= 2.25πcm
Length of arc =
≈ 7.07cm
If an exact answer is required you should
leave the answer as 2.25π cm. (Or 9π
4 cm.)
2.2
Method 2: Change degrees to radians first
180◦ = πrad
π
1◦ =
180
π
45◦ =
× 45
180
π
=
4
Now, the length of arc = rθ = 9 ×
π
4
=
9π
4 cm.
Right-Angled Triangles
A right angled triangle is uniquely defined, if as well as the 90° angle, you are given one side and
one angle or two sides. To be given two angles does is not enough information as the scale of the
triangle is not known. The sides of the triangle have names with respect to the angle of interest, θ
in the diagram.
Notice that if θ changes, then the sides opposite and adjacent are reversed. These side
names are helpful in defining the trigonometric ratios of sine, cosine and tangent.
2.2. Right-Angled Triangles
18
The phrase SOH-CAH-TOA is useful to remember the ratios.
sine θ =
opposite
;
hypotenuse
cosine θ =
SOH
Example
Find the unknown side x
adjacent
;
hypotenuse
CAH
tangent θ =
opposite
adjacent
TOA
Solution
for x.
Write the sine ratio and solve
x
17
x = 17 sin 14
sin 14 =
x = 4.11
Example
Find the unknown side x
Solution
solve for x.
Write the tangent ratio and
x
15
x = 15 tan 28
tan 48 =
x = 16.7
The calculator gives approximate values of the trigonometric ratios. You must look at your question
to check whether angles are in degrees or radians and ensure the calculator is first set in the right
mode. Questions where degrees are to be used will give angles marked with a ◦ symbol.
When solving for an angle on your calculator you select the appropriate trigonometry ratio and use
the shift button with sin, cos or tan to find sin−1 , cos−1 or tan−1 .
Example
Find the unknown angle x
Solution
for θ.
Write the sine ratio and solve
8
10
sin θ = 0.8
sin θ =
θ = sin−1 (0.8)
θ = 53.1°
2.2. Right-Angled Triangles
Example
Find the unknown angle x
19
Solution
for θ.
Write the cosine ratio and solve
7
cos θ =
11
7
−1
θ = cos
11
θ = 50.5°
Example The height of a steep cliff is to be measured from a point on the opposite side of the
river. The following diagram shows the measurements taken. Estimate the height of the cliff.
Solution
d
= tan 50.0°
58.2
d = 58.2 × tan 50.0°
h
= tan 76.3°
d
h = d × tan 76.3°
= 58.2 × tan 50.0° × tan 76.3°
≈ 284.526397
≈ 284.5m
Example To estimate the height of a mountain above a level plane the angle of elevation of
the top of the mountain is measured to be 30°. 600m closer to the mountain across the plane it is
found that the angle of elevation is 36°. Estimate the height of the mountain.
2.2. Right-Angled Triangles
20
Solution
h
h
= tan 36°and
= tan 30°
x
x + 600
We want h so we eliminate x between these two equations
h
h
x=
andx + 600 =
tan 36°
tan 30°
h
h
=
+ 600
tan 30°
tan 36°
h
h
−
= 600
tan 30° tan 36°
1
1
h
−
= 600
tan 30° tan 36°
tan 36° − tan 30°
= 600
h
tan 30° tan 36°
tan 30° tan 36°
h = 600 ×
tan 36° − tan 30°
≈ 600 × 2.811603815
≈ 1687m
Identities
The unit circle has equation x2 + y 2 = 1 and we define x = cos θ and y = sin θ so
x2 + y 2 = 1
(cos θ)2 + (sin θ)2 = 1
This is always written
sin2 θ + cos2 θ = 1
This is an identity which means it is true for all values of θ. There are many identities in trigonometry, we will only use the Pythagorean identity (above) and one more. Given sin = opp
adj we can
solve for opp = (sin)(hyp). Similarly from cosine: adj = (cos)(hyp). Substituting these into the
tangent relationship:
tan =
opp
(sin)(hyp)
sin
=
=
adj
(cos)(hyp)
cos
The sine and cosine law are two more unique relationships that we will cover in section 2.4.
2.2. Right-Angled Triangles
21
All Students Take Calculus
In the previous section the angles were between 0° and 90°. In this section the angles can take
any value. Initially we consider angles between 0° and 360° and relate these to the radian measure
between 0 and 2π. We remind you that angles are measured anticlockwise from the positive x-axis.
If the point P (x, y) is in the first quadrant, θ is the angle between OP and the positive x-axis and
we complete the right triangle then we have created the following situation.
Let the hypotenuse be r then
p
r = x2 + y 2
Therefore
sin θ =
We now let θ be any angle and define sine, cosine and tangent in the same way. For instance
if P (x, y) is in the second quadrant:
x
is
y
x
y
cos θ = and tan θ =
r
r
x
negative
tan θ
will
be
negative.
sin θ = yr because y is positive sin θ will be
positive. (r is positive by convention.)
cos θ = xr because x is negative cos θ will be
negative.
tan θ
=
y
x
because y is positive and
This pattern can be extended to quadrants 3 and 4. The mnemonic (All Students Take Calculus)
might help you remember which one is positive although you can always work it out if you need to.
The value of a trigonometric function consists of two parts the numerical part and the sign. you
must get both parts correct. In the previous section you related the values of a terminal point
to another point in the first quadrant. A point on the unit circle could be in any one of the four
quadrants.
Quadrant
1
2
3
4
x-coordinate
+
−
−
+
y-coordinate
+
+
−
−
cos
+
−
−
+
sin
+
+
−
−
tan
+
−
+
−
Some people learn this as a mnemonic All sin tan cos. (Meaning all are positive in the first quadrant,
only sine is positive in the second quadrant, only tangent is positive in the third quadrant and only
cosine is positive in the fourth quadrant.)
2.2. Right-Angled Triangles
22
The Area of a Triangle
The fundamental formula for the area of a triangle is Area = 12 × base × height Using the trigonometric functions the height can be replaced and the formula becomes:
Area =
1
× product of two sides × sine of the included angle
2
The formula is particularly easy to remember in symbolic form. Let the triangle have vertices A,
B and C, so the the sides opposite these angles are a, b and c respectively. Then the area can be
expressed symbolically as
1
1
1
Area = ab sin C = bc sin A = ac sin B
2
2
2
h
and h = c sin A
c
1
Area = × base × height
2
1
1
= × b × c sin A = bc sin A
2
2
sin A =
It depends where you draw h and which angle you choose to use as to which formula you finish up
with. The key point to remember is b and c are two sides and A is the angle between them. The
triangle above shows A as an acute angle (between 0° and 90°). If the angle is obtuse (between 90°
and 180°) the formula still holds.
The angle is in the second quadrant and
sin(180 − θ) = sin θ, so sin(180 − θ) = hc
can be written as sin θ = hc or h = c sin θ or
h = c sin A where A is obtuse. So the area is
1
2 bc sin A
Example
its area.
A triangle has two sides of 5cm and 8cm and the angle between them is 150°. Find
Solution
Area =
1
× 5 × 8 × sin 150°
2
It helps to remember that sin 150 = sin 30 =
Area =
1
2
1
1
× 5 × 8 × = 10 cm2
2
2
2.3. Trig Functions of Real Numbers
2.3
23
Trig Functions of Real Numbers
In this next part of the chapter the three main trigonometric functions (sine, cosine and tangent) will
be studied. They will be viewed as functions of real numbers rather than angles. The trigonometric
functions defined in these two ways are identical and there is a simple rule connecting the domains.
Why do we show you the two approaches? Trigonometry will be used to solve a variety of problems
and these can be divided into two groups, dynamic problems and static problems. When dynamic
problems (such as problems involving motion) are being solved real numbers will be used. When
static problems (such as finding distances and angles for triangles) are being solved angles will be
used.
You should be familiar with the fundamental graphs of y = sin x and y = cos x. These graphs are
the basis of this section. Desmos can easily show you the shape of y = sin x and y = cos x so if
you are asked to draw a rough sketch of these curves you should plot a few key points and draw a
smooth curve between them. You will usually be given the required domain however if you are not
you would choose to draw these for one complete cycle (0 − 2π). To sketch y = sin x it is enough
to select as key points x = 0, π2 , π, 3π
2 , 2π.
x
y = sin x
0
0
π
2
1
π
0
3π
2
−1
2π
0
Similarly to sketch y = cos x the same values of x give:
x
y = cos x
0
1
π
2
0
π
−1
3π
2
0
2π
1
You will be aware that these curves repeat this pattern every 2π where x extends in both the
positive and negative directions. 0 to 2π represents one complete cycle. Mathematically we say
sin (x + 2nπ) = sin x for any integer n
cos (x + 2nπ) = cos x for any integer n
Aside: instead of “for any integer n” we can write ∀n ∈ Z. A function that displays this characteristic is described as periodic and for y = sin x and y = cos x the period is 2π.
Transformations
The following six transformations can be applied to any function including sine, cosine, and tangent.
•
Vertical shift
•
Horizontal shift
•
Vertical stretch
•
Horizontal stretch
•
Reflection in the x-axis
•
Reflection in the y-axis
Example Sketch y = sin x − π2 + 3. This may be considered as a sine curve shifted π2 units
to the right. Note for periodic functions like a sine wave a horizontal shift is called a phase shift.
and 3 units upwards.
Solution 1. Beginning with the special points for sin x, a table shows the evolution of the
points.
2.3. Trig Functions of Real Numbers
x
sin x
sin x −
sin x −
π
2
π
2
+3
24
0
π
2
π
3π
2
2π
0
1
0
−1
0
−1
0
1
0
−1
2
3
4
3
2
← x-values, plot these
← y-values, plot these
2. Plot the transformed values with the original special points and connect with a smooth, continuous line. Desmos confirms our transformation. The dashed plot shows y = sin x as reference.
Example
Sketch y = cos(x − π6 ).
Solution You could use a table of values however this is y = cos x with a horizontal shift of
to the right. Sketch the transformation on top of the graph of y = cos x.
π
6
In general y = a sin x represents a vertical stretch of y = sin x by a. If a is negative the transformation can either be described as a negative stretch or (preferably) as a stretch of |a| followed by a
reflection in the x-axis. Recall the reflection of y = f (x) in the x-axis is y = −f (x). The number
|a| is called the amplitude for both y = sin x and y = cos x shown below.
If 0 < a < 1 the fractional stretch causes the curve to shrink vertically. For instance the curve
y = sin x has a maximum value of 1 and a minimum value of −1. the curve y = 21 sin x has a
maximum value of 12 and a minimum value of − 21 .
2.3. Trig Functions of Real Numbers
Example
25
Example
Example
Sketch y = cos(−t)
Sketch y = sin
1
2x
Sketch y = cos(2x)
Tangent
There are other trig functions that we will not be covering here, for example the reciprocal functions
are cosecant= sin1 θ , secant= cos1 θ , and the inverse tangent function, cotangent= tan1 θ . By focussing
on sine, cosine and tangent the majority of problems we encounter can be solved. Tangent is the
odd function out of the trio of common ones.
Previously we learnt that the period for the sine and cosine functions was 2π. Tangent is also a
periodic function and it has a period of π (not 2π). This means that it goes through one complete
sin x
cycle every π. Recall tan x = cos
x . To analyse the behaviour of the tangent function it helps if you
know what you are looking for. Some key values of tangent will show the pattern.
x
− π2
− π4
0
π
4
π
2
sin x
cos x
−1
0
√
−
2
2
0
√
2
2
1
√
2
2
1
√
2
2
0
tan x
−1
0 = −∞
√
√
− 22 ÷ 22 =
0
1 =0
√
√
2
2
÷
2
2 =1
1
0 =∞
−1
As x takes values from − π2 to π2 , tan(x) takes values from −∞ to ∞. This pattern is repeated
every π. A desmos graph can show this relationship. The dashed lines are the asymptotes.
The graph can be seen to have point symmetry. If you rotate the tangent curve through a half
2.4. Applications
26
turn using the origin as axis the curve will lie on top of itself. This is a pictorial representation of
an odd function.
Aside: You must not confuse y = tan x with y = x3 . While they may appear to be similar in shape
the only similarities are that they both pass through the origin and continue towards ∞ in the first
quadrant and −∞ in the third quadrant.
2.4
Applications
The Sine Rule
The Sine Rule is a relationship that allows you to find the sides and angles in triangles without a
right angle. In the next section we use the Cosine Rule to find sides and angles in triangles also, so
as you study these two sections you need to learn which problems require the Sine Rule and which
require the Cosine Rule. Previously, we met
the formula for the area of a triangle given two sides
1
and the included angle area = 2 ab sin C . The Sine Rule and Cosine Rule also require specific
combinations of sides and angles. Using standard side and angle labelling for a triangle → ∆ABC
and let the sides be a, b and c where a is opposite ∠A etc.
The Sine Rule states that in any triangle
sin A
sin B
sin C
=
=
a
b
c
or, inversely as:
a
b
c
=
=
sin A
sin B
sin C
Textbooks may use the term “The Law of Sines” whereas in these notes the term “The Sine Rule
will be used.
Proof of the Sine Rule
The Sine Rule is easy to prove from the formula for the area of a triangle
1
1
1
Area = bc sin A = ac sin B = ab sin C
2
2
2
Multiply right through by 2
bc sin A = ac sin B = ab sin C
Divide right through by abc
sin A
sin B
sin C
=
=
a
b
c
2.4. Applications
Example Find side lengths a and c in the
following triangle.
27
Solution
(a) To find c
c
b
=
sin C
sin B
c
5
=
sin 43
sin 35
5 sin 43
c=
sin 35
≈ 5.95cm(2 dp)
(b) To find a (i) A = 180° − (35° + 43°) =
180°−78° = 102° (ii) It is usually wise to go back
to the original data (i.e. use b and B rather than
c and C).
b
a
=
sin A
sin B
5
a
=
sin 102
sin 35
5 sin 102
a=
sin 35
≈ 8.53cm(2 dp)
These two calculations illustrate the first two cases in which the Sine Rule is used. You will notice
that the triangle has been completely solved in the course of this example. We started with one
side and two angles and we found the other two sides and the other angle.
The third case is not as straight forward. Given two sides and an angle there could be no triangle
formed, one triangle formed or two triangles formed depending on the length of the side opposite
the given angle. You should develop an insight into the reasons why this is so. Imagine the second
side given is the boom of a crane and the angle given is the angle between the boom and the ground.
The side opposite the given angle is represented by the cable. It is clear that for certain lengths of
the cable the hook will not reach the ground. Then as the hook is lowered a point will be reached
when the hook just touches the ground.
It is no surprise that the length a to create this
situation is a = b sin A.
We are after all dealing with the Sine Rule
which
becomes the
fundamental sine formula
opp
a
sin A = hyp = b when the triangle has a
right angle.
2.4. Applications
28
If the cable is held taut and is extended a little more it will touch the ground in two places (provided
it is kept in the same plane). As the cable is extended further the time will come where the cable
is as long as boom. At this point there is only one solution again, the one straight out in front of
the crane). Further extensions of the cable will produce only one solution (straight out in front
of the crane) as the cable will theoretically reach behind the crane boom thus creating a different
triangle altogether.
The two solutions case is often referred to as the ambiguous case. The discussion above shows the
range of values of a that will give two solutions.
b sin A < a < b
Example
Given a = 30°, a = 8 and b = 7 solve the triangle (i.e. find B, C and c).
Because 8 > 7 this is the one solution case.
Solution
(1)
Find B
sin B
sin A
=
b
a
b sin A
sin B =
a
7 sin 30
=
= 0.4375
8
B = sin−1 0.4375 ≈ 25.94°
(2)
Find C
C = 180° − (30° + 25.94°)
= 124.06°
(3)
Find c
c
a
=
sin C
sin A
a sin C
c=
sin a
8 sin 124.06
=
sin 30
≈ 13.3
Example The Ambiguous Case
Given A = 30°, a = 6 and b = 7 solve the triangle.
Solution In this case b sin A = 3.5 and b = 7 so as a lies between 3.5 and 7. This is the
ambiguous case, therefore there are two solutions as the side lengths and angle can product two
different trianges.
It is best to visualise these two solutions on the same diagram so that the isosceles triangle can
2.4. Applications
29
help lead to the two results.
First solution (Proceed as before)
(1)
Find B
sin B
sin A
=
b
a
b sin A
sin B =
a
7 sin 30
= 0.583̇
=
6
B = sin−1 0.583̇ ≈ 35.69°
(2)
Find C
(3)
C = 180° − (30° + 35.69°)
= 114.31°
Find c
c
a
=
sin C
sin A
a sin C
c=
sin A
6 sin 114.31
=
sin 30
≈ 10.9
Second solution
(1) Find the second value of B
B = 180° − 35.69° = 144.31°
(2) Find C
C = 180° − (30° + 144.31°) = 180° − 174.31° =
5.69°
(Or if you remember the rule that the exterior angle of a triangle is the sum of the two interior
opposite angles C + 30° = 35.69° so C = 35.69° − 30° = 5.69°)
(3)
Find c
c
a
=
sin C
sin A
a sin C
c=
sin A
6 sin 5.69
=
sin 30
≈ 1.2
The Cosine Rule
In this section we will state and prove the Cosine Rule (which is called ”The Law of Cosines” in
the textbook). The proof is given here for completeness. You will not be tested on your ability
to reproduce it. The section will give examples where the Cosine Rule is used to solve problems
using the triangle of vectors and we will include revision of bearings and the use of trigonometry
in navigation.
2.4. Applications
30
For any triangle side sides a, b, and c and angle A opposite side a:
a2 = b2 + c2 − 2bc cos A
Proof of The Cosine Rule
To prove: For any triangle ∆ABC, a2 = b2 + c2 − 2bc cos A
By Pythagoras’ Theorem
BC 2 = CD2 + DB 2
a2 = (c − b cos A)2 + (b sin A)2
= c2 − 2bc cos A + b2 cos2 A + b2 sin2 A
= c2 − 2bc cos A + b2 cos2 A + sin2 A
= c2 − 2bc cos A + b2 as cos2 A + sin2 A = 1
This is usually written
a2 = b2 + c2 − 2bc cos A
The diagram has been drawn to simplify the way the proof unfolds. You will see that by placing
the vertex A at the origin the side a is found in terms of b, c, and A. The proof would have been
the same had A and B been as shown and C placed in the second quadrant. (Thus producing a
triangle with an obtuse angle at A.) This rule is symmetrical. You need to be given two sides and
the included angle (b, c and A) and the formula allows you to calculate a. Most textbooks will
therefore show you three equivalent formulae
a2 = b2 + c2 − 2bc cos A
b2 = c2 + a2 − 2ca cos B
c2 = a2 + b2 − 2ab cos C
Example
Given a = 5, b = 6 and C = 50°, find c.
Solution
c2 = a2 + b2 − 2ab cos C
= 52 + 62 − 2 × 5 × 6 × cos 50
≈ 22.43274342
2.5. Chapter Exercises
31
c≈
√
22.43274342
≈ 4.736321718 ≈ 4.7 (1dp)
Given a = 5, b = 6 and C = 130°, find c.
Example
Solution
c2 = a2 + b2 − 2ab cos C
= 52 + 62 − 2 × 5 × 6 × cos 130
≈ 99.56725658
√
c ≈ 99.56725658
≈ 9.97833937 ≈ 10.0 (1dp)
These two examples show that when the two sides and the included angle are given the third side
(opposite the given angle) can be found.
Example
Find the angle given three side lengths; given a = 5, b = 6 and c = 9, find A.
Solution
Because a is the smallest side A will most certainly be an acute angle.
b2 + c2 − a2
2bc
2
6 + 9 2 − 52
=
2×6×9
≈ 0.851851851
cos A =
A ≈ cos−1 (0.851851851)
≈ 31.6°
2.5
§2.1
Chapter Exercises
Unit Circle
1. Find the radian measure of the angle with the given degree measurements.
(a)
(c)
36°
60°
(b) −480°
(d) −135°
2. Find the degree measure of the angle with the given radian measure.
(a)
(c)
3π
4
−1.5
(b) 5π
6
π
(d) − 12
2.5. Chapter Exercises
32
3. Arc length
(a) Find the length of the arc s in the figure. (b) Find the radius r of the circle in the figThe radius is 5.
ure.
4. Find the length of an arc that subtends a central angle of 2rad in a circle of radius 2mi.
5. Find the radius of the circle if an arc of length 6m on the circle subtends a central angle of π/6
rad.
6. Pittsburgh, Pennsylvania and Miami, Florida lie approximately on the same meridian. Pittsburgh has a latitude of 40.5◦ N and Miami is 25.5◦ N. Find the distance between these two
cities. (The radius of the earth is 3960mi.)
7. Find the distance the earth travels in one day in its path around the sun. Assume the year has
365 days and that the path of the earth around the sun is a circle of radius 93 million miles.
§2.2
Right Angled Triangles
1. Find the exact value of sin θ, cos θ and tan θ of the angle θ in the triangle.
2. Find sin θ, cos θ and tan θ in the following triangles.
(a)
(b)
3. Find the side length labelled x for the following triangles.
(a)
(b)
(c)
2.5. Chapter Exercises
33
4. Solve the triangles.
(a)
(b)
5. The angle of elevation to the top of the Empire State Building in New York is found to be 11◦
from the ground at a distance of 1mi from the base of the building. Using this information,
find the height of the Empire State Building.
6. A laser beam is to be directed towards the centre of the moon but the beam strays 0.5◦ from
its intended path.
(a) How far has the beam diverged from its assigned target when it reaches the moon? (The
distance of the earth to the moon is 240 000mi.)
(b) The radius of the moon is about 1000mi. Will the beam strike the moon?
7. A water tower is located 325ft from a building. From a window in the building it is observed
that the angle of elevation to the top of the tower is 39◦ and the angle of depression to the
bottom of the tower is 25◦ . How tall is the tower? How high is the window?
8. Find the area of a triangle with sides of length 7 and 9 and included angle 72◦ .
9. A triangle has an area of 16in2 , and two of the sides of the triangle have lengths 5in and 7in.
Find the angle included by these two sides.
§2.3
Trigonometric Functions
1. Sketch the functions by hand.
(a)
(c)
(e)
y = 1 + sin x
y = −2 sin x
y = |cos x|
(b) y = 1 − cos x
(d) y = 4 − 2 cos x
2. Find the amplitude and period of the function and sketch its graph.
(a)
(c)
(e)
y = cos 4x
y = 10 sin 12 x
y = 3 cos 3πx
(b) y = 3 sin 3x
(d) y = − cos 31 x
3. Find the amplitude, period and phase shift of the function, and plot using desmos .
(a)
(c)
(e)
y = cos x − π2
y = 2 sin 23 x − π6
y = − 12 cos 2x − π3
(b) y = − sin x − π6
(d) y = 3 cos π x + 12
(f ) y = sin (3x + π)
2.5. Chapter Exercises
§2.4
34
Applications
1. Use the Sine Rule to find side x or angle θ
(a)
(b)
2. Solve the triangle using the Sine Rule.
3. Sketch each triangle and then solve using the Sine Rule.
(a)
∠A = 50◦ , ∠B = 68◦ , c = 230
(b) ∠B = 29◦ , ∠C = 51◦ , b = 44
4. To find the distance across a river, a surveyor chooses points A and B, which are
200ft apart on one side of the river. She then
chooses a reference point C on the opposite
side of the river and finds that ∠BAC ≈ 82◦
and ∠ABC ≈ 52◦ . Find the approximate
distance from A to C.
5. The path of a satellite circling the earth causes it to pass directly over two tracking stations A
and B, which are 50mi apart. When the satellite is on one side of thetwo stations, the angle of
elevation at A and B are measured to be 87.0◦ and 84.2◦ , respectively.
(a)
How far is the satellite from station A?
6. A communication tower is located at the top
of a steep hill. The angle of inclination of
the hill is 58◦ . A guy wire is attached to the
top of the tower and to the ground, 100m
downhill from the base of the tower. The
angle between the slope of the hill and the
guy wire is measured as 12◦ . Find AC, the
length of cable required for the guy wire.
(b) How high is the satellite above the ground?
2.5. Chapter Exercises
35
7. Use the Cosine Rule to find side x given AC = 44.3 and θ.
(a)
(b)
8. Use either the Sine Rule or Cosine Rule as appropriate to find x and θ.
(a)
(b)
9. Two straight roads diverge at an angle of 65◦ . Two cars leave the intersection at 2.00 P.M.,
one traveling at 50mi/h and the other at 30mi/h. How far apart are the cars at 2.30 P.M.?
10. A pilot flies in a straight path for 1h 30min. She then makes a course correction, heading 10◦
to the right of her original course, and flies for 2h in the new direction. If she maintains a
constant speed of 625mi/h how far is she from her starting point?
11. A steep mountain is inclined 74◦ to the horizontal and rises 3400ft above the surrounding plain. A cable car is to be installed from
a point 800ft from the base to the top of
the mountain, as shown. Find the shortest
length of cable needed.
12. Three circles of radii 4, 5, and 6cm respectively are mutually tangent. Find the area
enclosed between the circles.
2.5. Chapter Exercises
13. A surveyor wishes to find the distance between two points A and B on the opposite
side of a river. on her side of the river she
chooses two points C and D that are 20m
apart and measures the angles shown. Find
the distance between A and B.
36
3 | Exponential & Logarithmic
Functions
We defined the polynomial ax in chapter 1, however x was restricted to rational numbers. We now
want to explore ax where and x is any real number.
We can show that values exist simply by pressing buttons on the calculator or drawing a graph
using desmos , however an intuitive understanding can be obtained by considering appropriate
values.
Consider 2π . We know π = 3.141592654 . . .. 2π should be between 23 and 24 . That is between 8
and 16. We can evaluate
23.1
23.14
23.141
23.1415
23.14159
23.141592
23.1415926
23.14159265
23.141592654
8.5741877
8.815240927
8.821353305
8.824411082
8.824961595
8.824973829
8.824977499
8.824977805
8.82497783
We could continue this process. If we enter 2π on the calculator the answer obtained is
2π = 8.824977827
The graph of y = 2x , ∀x ∈ R is shown.
Recall y = f (x) is a function and y = f (−x) is the same function with every x replaced with −x,
i.e. y = f (−x) is the reflection of y = f (x) in the y-axis.
x
x
If y = 2x then y = 2−x is the reflection of y = 2x in the y-axis. But y = 2−x = 2−1 = 21 .
Graphing Exercise Use Desmos to verify that y = 2−x and y =
1x
2
are equivalent.
38
The Exponential Function f (x) = ax
Let y = ax where a > 0
When 0 < a < 1
y = ax looks like
When a = 1
y = ax = 1x = 1 looks like
When a > 1
y = ax looks like
f (x) = ax is called an exponential function. a is called the base of the exponential function. The
domain is R. the range is (0, ∞). The x-axis is an asymptote.
Example
Find the equation of the exponential function that passes through (0, 1) and (3, 125).
Solution An exponential function that passes through (0, 1) is of the form f (x) = ax . As
f (3) = 125 we substitute x = 3 and get
a3 = 125
√
3
a = 125 = 5
∴ f (x) = 5x satisfies the conditions.
Transformations of Exponential Functions
Recall the transformations we have met so far
Vertical stretch of a
Horizontal stretch of 1b
Vertical shift of c ↑
Horizontal shift of d −→
Reflection in x-axis
Reflection in y-axis
y
y
y
y
y
y
= f (x)
= f (x)
= f (x)
= f (x)
= f (x)
= f (x)
y
y
y
y
y
y
= af (x)
= f (bx)
= f (x) + c
= f (x − d)
= −f (x)
= f (−x)
Each of these transformations can be applied to an exponential function.
3.1. ex
39
Graphing Exercise Given y = 2x apply the following transformations. Draw a sketch showing
where the graph crosses the y-axis, its shape, its asymptote and one other point it passes through.
1
2
1. Horizontal shift of +2
6. Vertical stretch of
2. Vertical shift of −3
7. Reflection in the x-axis
3. Horizontal stretch of 2
8. Reflection in the y-axis
4. Vertical stretch of 4
9. (Challenge) Reflection in y = −1
5. Horizontal stretch of
3.1
1
3
10. (Challenge) Reflection in x = 1
The Natural Exponential Function
The natural exponential function has wide application in mathematics engineering. It arises naturally and crops up in applications such as finance, population, radioactivity, charge on a capacitor,
and more. We have defined f (x) = ax and √
there is a particular value of a that we denote by the
letter e. It is an irrational number (like π, 2 etc.) and has a button on your calculator. To 10
decimal places it is
e ≈ 2.7182818285 . . .
The natural exponential function f (x) = ex is often simply referred to as the exponential function.
Compound interest can demonstrate an example of how the value above is found. Imagine a bank
that pays 100% interest on your money. Given an initial deposit of $1, at the end of year you
will receive $1 in interest payment and have a total of $2. Compounded interest allows this to
happen at intervals smaller than 1 year. If the interest is compounded twice per year, then after 6
months, you will receive 50% interest and have $1.50. In the second half of the year you now have
an additional $0.50 available to earn the second half interest. Now, $1.50 × 50% = $0.75, so at the
end of the year you have $1.50+$0.75=$2.25.
3.2. Logarithmic Functions
40
1
Lets say the interest is compounded monthly, then after 1 month you will receive 12
×100% = 8.33%
interest for $1+$0.0833=$1.0833. The second month will earn the same rate (8.33%) on $1.0833,
1 12
for a total of $1.1736. After 12 months, your dollar will now be worth $1.00 × 1 + 12
= $2.6130.
compounding periods
interest ($)
total ($)
1 (yearly)
2
3
4 (quarterly)
5
6
1.00
1.25
1.3704
1.4414
1.48832
1.521626
2.00
2.25
2.3704
2.4414
2.48832
2.521626
12 (monthly)
1.613035
2.613035
52 (weekly)
1.692597
2.692597
365 (daily)
1.714567
2.714567
continuous
1.718282
2.718282. . .
As a function, f (x) = ex , is plotted below:
3.2
Logarithmic Functions
The logarithmic function is the inverse of the function f (x) = ax . Recall the inverse of a function
is the reflection of the function in the line y = x. Mathematically this is equivalent to swapping the
x and the y in y = ax . So x = ay is the inverse of y = ax . We have another notation for the inverse
of a function, which is a little more complicated. Let y = f (x) be a function of x then y = f −1 (x)
is the inverse of this function.
Sometimes the inverse of the function is also a function. For example the inverse of y = x2 is
x = y 2 . y = x2 is a function (vertical line test always applies), whereas x = y 2 is not a function
(vertical line test is broken).
3.2. Logarithmic Functions
41
The inverse of y = 10x is x = 10y . y = 10x is a function (vertical line test always applies) and so
is x = 10y .
Another useful fact to remember about inverses concerns the domain and range. The domain of f
is the range of f −1 and the range of f is the domain of f −1 .
We have a notation for x = ay it is y = loga x:
y = loga x ⇔ x = ay
In x = ay substitute y = loga x and we get x = aloga x . This means that given a base of a the power
(or exponent) to which a must be raised to get x is loga x.
Problems involving logarithms will often require us to switch back and forth between y = loga x
and x = ay , however it is also helpful if you can remember to substitute for y and write x = aloga x
so that you can say “the logarithm is the power”.
Example
(a) log10 100 = 2 because 102 = 100
(b) log3 81 = 4 because 34 = 81
(c) log10 0.01 = −2 because 10−2 = 0.01
The graph of y = loga x
The exponential function y = ax where a > 0 is now known and its domain is R and its range is
the positive real numbers. We often write R+ instead of (0, ∞).
The graph of f (x) = ax can be reflected in the line y = x and the result is f −1 (x) = loga x.
Graphing Exercise On the same set of axes draw:
(a)
y = 10x
(b) y = log10 x
(c)
y=x
(d) x = 10y
3.2. Logarithmic Functions
42
Make a comment about each statement below.
1.
Check the graphs in (a), (b) and (c) are you confident that y = log10 x is the reflection of
y = 10x in the line y = x?
2.
When you enter x = 10y describe what takes place.
Graphing Exercise Solution Using desmos we can see the different plots. Plots (b) and (d) are
equivalent, so are on top of each other.
Graphing Exercise On the same set of axes draw:
(a)
y = log2 x
(b) y = log3 x
(c)
y = log4 x
(d) y = log10 x
Graphing Exercise Solution Using desmos we can see the relationship between the different
bases in the log equation:
3.2. Logarithmic Functions
43
Notice the point that is common to all curves and the behaviour of the family of curves for x > 1
and for 0 < x < 1.
Property 1: A property of logarithms is loga 1 = 0 and this can be seen on the graphs where
every graph goes through (1, 0).
The pattern you observe as the base gets bigger might not be evident for values of the base between
0 and 1. On the same set of axes draw the following:
(a)
y = log 1 x
(b) y = log 1 x
2
(c)
y = log 1 x
3
4
(d) y = log 1 x
5
To say the pattern is the same you have to be careful to describe the base. Explain how changing
the base gives the same pattern as for (a) to (d).
Property 2: A second property of logarithms is loga a = 1. That is a1 = a or the power to which
you have to raise a to get a is 1. On your curves above locate a point on each curve that shows
this. You should in each case be looking for the point (a, 1).
Property 3: A third property of logarithms is loga ax = x. This useful property must be understood if logarithm problems are to be mastered. You should understand what loga ax is saying. a
is the base so loga ax = x says “The power to which a must be raised to get ax is x.”
Common Logarithms
When the base is 10 we write y = log10 x = log x. If you see no base you assume it is base 10.
y = log x is called the common logarithm of x. Desmos and other programs with mathematics
incorporated recognise ”log” as ”logarithm to the base 10”. The log key on the calculator gives the
common logarithm of any positive number.
y = log x ⇔ 10y = x
3.2. Logarithmic Functions
44
Natural Logarithms
When the base is e we write y = loge x = ln x. If you see ln x you assume it is loge x. y = ln x
is called the natural logarithm of x. Desmos and other programs with mathematics incorporated
recognise ”ln” as ”logarithm to the base e”. The ln key on the calculator gives the natural logarithm
of any positive number.
y = ln x ⇔ ey = x
Graphing Exercise Use Desmos to sketch the following. Describe in words how each curve is
related to y = ln x.
(a)
y = ln x
(b) y = ln(−x)
(d) y = ln(x − 1)
(e)
y = ln(x) − 1
(c)
y = − ln x
(f )
y = ln(−1 − x)
The Laws of Logarithms
Law 1:
Law 2:
loga (XY ) = loga X + loga Y
loga X
Y = loga X − loga Y
Law 3:
loga X c = c loga X
It is a useful exercise to prove these laws as the proofs show the connection between the laws for
exponents and the laws for logarithms.
Proof of Law 1
Let loga X = u so au = X
and loga Y = v so av = Y
XY = au av = au+v
Take logarithms of both sides to the base a
loga XY = loga au+v = u + v (property 3) = loga X + loga Y
Proof of Law 2
X
au
u−v
Y = av = a
Take logarithms of both sides to the base a
u−v = u − v (property 3) = log X − log Y
loga X
a
a
Y = loga a
Proof of Law 3
X c = (au )c = auc = acu
Take logarithms of both sides to the base a
loga X c = loga acu = cu (property 3) = c loga X //
Example
(a) log
√
Expand using the logarithm laws
1
3 = log 3 2 =
1
2
log 3
3.3. Exponential and Logarithmic Equations
(b) ln
√
a√ b
3c
Example
45
1 1
1
1
= ln ab 2 c− 3 = ln a + ln b 2 + ln c− 3 = ln a + 21 ln b − 13 ln c
Evaluate
4
(a) log2 112 − log2 7 = log2 112
7 = log2 16 = log2 2 = 4 log2 2 = 4
23
(b) log2 823 = log2 23
= log2 269 = 69 log2 2 = 69
√
1
0.001 = log (0.001) 2 =
2
(d) e2 ln 4 = eln 4 = 42 = 16
(c) log
Example
1
2
log 0.001 =
1
2
log 10−3 =
× −3 = − 32 = −1 12
(Let ln 4 = x then ex = 4 so eln x = 4)
Rewrite as a single logarithm term using the logarithm laws
√
√
(a) log 12 + 12 log 5 − log 3 = log 123 5 = log 4 5
2 −1
= log3
(b) log3 x2 − 1 − log3 (x − 1) = log3 xx−1
3.3
1
2
(x+1)(x−1)
x−1
= log3 (x + 1)
Exponential and Logarithmic Equations
The types of problems we meet in this section will be able to be rearranged so that they look like
af (x) = b
Where a and b are real numbers, x is the unknown variable we are trying to find and f (x) is an
expression in x. The technique we will use will be the same for every problem we solve.
Step 1 Our first step is to inspect the problem to see if the unknown variable is in the exponent.
Step 2 Now that we have established that we are solving an exponential equation we rearrange it
until it is in the form af (x) = b
Step 3 Take the logarithm of both sides. In most practical situations we either take logarithms
to the base 10 or logarithms to the base e. there are three situations
Case 1 The problem has reduced to 10f (x) = b. Take logarithms to the base 10.
Case 2 The problem has reduced to ef (x) = b. Take logarithms to the base e.
Case 3 The problem has reduced to af (x) = b where a is neither 10 nor e. You can take logarithms
to the base 10 or e as you wish, either is correct.
Step 4 Solve the equation you obtain.
Example 1 Solve 3x = 5. This is an example of case 3 above.
Solution
Two methods will be explored below:
3.3. Exponential and Logarithmic Equations
Method 1
Take logarithms to the base 10
46
Method 2
Take logarithms to the base e
log 3x = log 5
ln 3x = ln 5
x log 3 = log 5
log 5
x=
log 3
≈ 1.464973521 ≈ 1.46(2 dp)
x ln 3 = ln 5
ln 5
x=
ln 3
≈ 1.464973521 ≈ 1.46(2 dp)
This shows it is immaterial whether you take logarithms to base 10 or logarithms to base e.
Example 2 Solve 32x+1 = 5
Solution Take logarithms to the base 10:
log 32x+1 = log 5
10
1+e−x
9
= 2.25
4
= 2.25 − 1 = 1.25
1 + 105x =
(2x + 1) log 3 = log 5
log 5
2x + 1 =
log 3
log 5
2x =
−1
log 3
1 log 5
x=
−1
2 log 3
≈ 0.23248676 ≈ 0.23(2 dp)
Example 4 Solve
Example 3 Solve 4 1 + 105x = 9
Solution This must first be rearranged:
105x
10 log 105x = log 1.25
5x = log 1.25
1
x = log 1.25
5
≈ 0.019382002 ≈ 0.019(3 dp)
= 3.
Solution
10
3
10
−x
−1
e =
3
10 − 3
7
=
=
3
3
−x = ln 7 − ln 3
1 + e−x =
x ≈ −0.85(2 dp)
Solving Logarithmic Equations
Whereas exponential equations have the unknown variable in an exponent, logarithmic equations
are equations containing the logarithm of an unknown variable.
Step 1 You recognise you are dealing with a logarithmic equation by inspecting the problem to
see if you have a logarithm of a term containing the unknown variable.
3.3. Exponential and Logarithmic Equations
47
Step 2 Rearrange the equation until it is in the form
loga (term containing unknown) = b
where b is a number.
Step 3 Take “antilogarithms ”. That is if loga X = b then ab = X
Step 4 Solve the equation. (a and b are known and X is an expression containing the unknown
variable.)
Example 1 Solve ln x = 5 .
Solution
Take antilogarithms with base e:
e5 = x
x ≈ 148.4131591 ≈ 148.4(1 dp)
Example 2 Solve 5 + 4 log (5x) = 17.
Solution
Isolate the log and solve:
4 log (5x) = 17 − 5 = 12
log 5x = 3
103 = 5x
1000
x=
= 200
5
Check: Substitute x = 200
LHS = 5 + 4 log (5 × 200)
= 5 + 4 log 1000 = 5 + 4 log 103
= 5 + 4 × 3 = 5 + 12 = 17 = RHS
Example 3
Solve log x + log (x − 1) = log 4x
Solution
log x + log (x − 1) − log 4x = 0
x (x − 1)
log
=0
4x
1
log (x − 1) = 0
4
1
100 = (x − 1) = 1
4
x − 1 = 4; x = 5
Check:
LHS = log 5 + log 4 = log (5 × 4) = log 20
RHS = log 4 × 5 = log 20 = LHS
Example 4 The velocity of a sky diver t
seconds after jumping is given by
v (t) = 80 1 − e−0.2t
After how many seconds is the velocity 70 ft/s?
Solution
70 = 80 1 − e−0.2t
70
= 1 − e−0.2t
80
70
e−0.2t = 1 −
80
7
1
= 1 − = = 0.125
8
8
− 0.2t = ln 0.125
ln 0.125
t=
≈ 10.39s
−0.2
3.4. Exponential Modelling
3.4
48
Exponential Modelling
The natural exponential function, ex , is used in a variety of situations where there is exponential
growth or decay.
Example The exponential function can be used to model the way populations grow and diseases
spread. The following example is about the spread of an infectious disease in a small city whose
population is 10, 000. After t days the number of people who have caught the disease is modelled
by the function
10000
f (t) =
5 + 2495e−0.84t
(a) How many people had the disease initially? Initially in this sense means right at the beginning,
or when the clock is still at zero. Substitute t = 0 into the function:
f (0) =
10000
10000
=
= 4 people
0
5 + 2495e
2500
(b) How many people have the disease after 1 day? Substitute t = 1 into the function:
f (1) =
(c)
10000
10000
10000
=
=
≈ 9.24; 9 people
5 + 2495e−0.84
5 + 2495(0.4317)
1082.1
How many people have the disease after 5 days? Following from part (b), find f (5):
f (5) =
10000
10000
≈ 235.77; 236 people
=
42.41
5 + 2495e−0.84(5)
(d) Use desmos to graph the function and describe its behaviour.
The graph has distinctive characteristics. It starts at a particular non-zero value (when t = 0) and
increases slowly at first then more rapidly. It slows down after a time and levels off because the
exponential function in the denominator
0 when t
∞. Graphs with these characteristics are
called logistic curves. The particular model is called a logistic growth model.
3.5. Chapter Exercises
3.5
§3.1
49
Chapter Exercises
ex functions
1. Use your calculator to evaluate to 5dp
(a)
e4
(b) 2e−0.7
(c)
e3.1
(d) ee
2. Use Desmos to sketch:
(a)
f (x) = e−x
(b) g(x) = 2e0.1x
3. Find the exponential function f (x) = ax whose graph is given.
(a)
(b)
(c)
(d)
(c)
h(x) = −2.1e−0.12x
3.5. Chapter Exercises
50
4. Sketch the transformed graph
(a) The graph is y = 3x .
Draw y = −3x .
(c) The graph is y =
x
Draw y = 4 + 12 .
1x
2 .
(b) The graph is y = 2x .
Draw y = 2x − 3.
(d) The graph is y = 10x .
Draw y = 10x+3 .
3.5. Chapter Exercises
51
(f ) The graph is y = e−x .
Draw y = e−x − 1.
(e) The graph is y = ex .
Draw y = −ex .
§3.2
Logarithmic Functions
1. Express the equation in exponential form.
(a) log5 25 = 2
(d) log2 81 = −3
(b) log5 1 = 0
(e) ln 5 = x
1
3
(c)
(f )
log8 2 =
ln y = 5
(b) 10−4 = 0.0001
(e) ex = 2
(c)
(f )
8−1 = 18
e3 = y
(b) log3 1
(e) log9 81
√
(h) log10 10
(c)
(f )
log3 32
log7 710
(i)
log5 0.2
(l)
(o)
eln 5
ln (1/e)
(c)
(f )
log3 243 = x
logx 8 = 23
2. Express the equation in logarithmic form
(a) 53 = 125
(d) 2−3 = 81
3. Evaluate the expression
(a) log3 3
(d) log6 36
(g)
1
27
2log2 37
log3
3log3 8
(j)
(m) log8 0.25
(k)
(n) ln e4
√
4. Use the definition of the logarithmic function to find x.
(a) log2 x = 5
(d) log10 x = 2
(b) log2 16 = x
(e) logx 16 = 4
5. Use the Laws of Logarithms to rewrite the expression in a form with no logarithms of products,
quotients roots or powers.
(a)
log2 (2x)
(d) log2 AB
√
(g) ln ab
2
(b) log2 (x (x − 1))
√
(e) log3 x y
q
(h) ln x yz
(c)
(f )
(i)
log 610
√
log5 3 x2 + 1
p
log 4 x2 + y 2
3.5. Chapter Exercises
52
6. Evaluate the expressions
√
(a) log5 125
(d) ln 6 − ln 15 + ln 20
(b) log 2 + log 5
(e) 102 log 4
(c)
(f )
log4 192 − log4 3
log log 100010,000
7. Rewrite the expression as a single logarithm
(a)
§3.3
log3 5 + 5 log3 2
(b) log2 A + log2 B − 2 log2 C (c)
ln 5+2 ln x+3 ln x2 + 5
Logarithmic Equations
1. Find the solution of the exponential equation, correct to four decimal places.
(a) ex = 16
(d) e1−4x = 2
(g)
5x = 4x+1
(b) 102x = 5
(e) 80.4x = 5
(c)
(f )
3ex = 10
5−x/100 = 2
(h) 23x+1 = 3x−2
(i)
100 (1.04)2t = 300
2. Solve the equations for x
(a)
(c)
(e)
(g)
§3.4
x2 2x − 2x = 0
e4x + 4e2x − 21 = 0
log (3x + 5) = 2
log x + log (x − 1) = log (4x)
(b)
(d)
(f )
(h)
4x3 e−3x − 3x4 e−3x = 0
ln x = 10
2 − ln (3 − x) = 0
log5 (x + 1) − log5 (x − 1) = 2
Modelling
1. A radioactive substance decays in such a way that the amount of mass remaining after t days
is given by the function
m(t) = 13e−0.015t
Where m(t) is measured in kilograms.
(a) Find the mass at time t = 0.
(b) How much of the mass remains after 45 days?
2. A sky diver jumps from a reasonable height above the ground. The air resistance she experiences
is proportional to her velocity, and the constant of proportionality is 0.2. It can be shown that
the downward velocity of the sky diver at time t is given by
v(t) = 80 1 − e−0.2t
where t is measured in seconds and v(t) is measured in feet per second (ft/s).
(a) Find the initial velocity of the sky diver.
(b) Find the velocity after 5s and after 10s.
(c) Draw a graph of the velocity function v(t).
(d) The maximum velocity of a falling object with wind resistance is called the terminal
velocity. From the graph in part (c) find the terminal velocity of the sky diver.
3.5. Chapter Exercises
53
3. The population of a certain species of bird is limited by the type of habitat required for nesting.
The population behaves according to the logistic growth model
n(t) =
5600
0.5 + 27.5e−0.044t
where t is measured in years.
(a) Find the initial bird population.
(b) Draw a graph of the function n(t).
(c) What size does the population approach as time goes on?
4. A 15g sample of radioactive iodine decays in such a way that the mass remaining after t days
is given by m (t) = 15e−0.087t where m(t) is measured in grams. After how many days is there
only 5g remaining?
5. A small lake is stocked with a certain species of fish. The fish population is modelled by the
function
10
P =
1 + 4e−0.8t
where P is the number of fish in thousands and t is measured in years since the lake was stocked.
(a) Find the fish population after 3 years.
(b) After how many years will the fish population reach 5000 fish?
4 | Differentiation
Rate of Change
We are all familiar with the concept of average speed. If you travel a distance of 120 km in 2 hours
then your average speed is 60 kph. There is a formula that you will have met
Average speed =
distance travelled
time elapsed
A distance/time graph can be drawn. The average speed can be expressed using function notation
Average speed =
s(b)−s(a)
b−a
Finding the average rate of change is important in many contexts and in fact the average rate of
change can be defined for any function.
The average rate of change of the function y = f (x) is
change in y
change in x
or
f (b)−f (a)
b−a
(1)
The average rate of change is the slope of the secant line between x = a and x = b on the graph
of f , that is the slope of the line that passes through (a, f (a)) and (b, f (b)).
Example Calculate the average rate of change for the function f (x) = x2 + 4 between the
following points:
(1) x = 2 and x = 6
Solution
Using the function notation
in (1) above,
(2) x = 5 and x = 10
Solution
f (5) − f (10)
29 − 104
=
= 15
5 − 10
−5
(3) x = a and x = a + h
(h 6= 0)
Solution
f (a) − f (a + h)
a − (a + h)
f (2) − f (6)
8 − 40
=
2−6
−4
=
−32
=8
−4
=
a2 + 4 − ([a + h]2 − 4)
a+h
Can this be simplified further?
Tangents
We now investigate the process of changing the value of (b − a) in formula (1). As (b − a) is made
smaller and smaller the slope of the secant approaches the slope of the tangent at x = a. The
notation for this process is as follows.
Definition: The tangent line to the curve y = f (x) at the point P (a, f (a)) is the line through
lim f (x)−f (a) provided that the limit exists. This means that as the value of x
P with slope m = x→a
x−a
55
gets close to a the function remains smooth. Imagine zooming in on a function, from far away it
may appear smooth, but up close it could have some gaps or discontinuities. Limits do not exist
at sharp transitions in a graph, or where the function does not exist (think of piecewise functions).
We sometimes refer to the slope of the tangent line to a curve at a point as the slope of the curve
at that point. The idea is that if we zoom in far enough towards the point then the curve looks
almost like a straight line. The more we zoom in the more the parabola looks like a straight line.
Using function notation for the tangent line is usually easier to use and is often preferred. The
f (a + h) − f (a)
slope of the secant line between x = a and x = a + h is
. This looks familiar from
h
example 3 above. We can now use this as a definition:
lim
slope = m = h→0
f (a + h) − f (a)
h
lim , and we would say ‘the limit as h approaches 0’. Note that if h = 0 the
This is limit notation, h→0
function is now undefined (math error). So h is allowed to get close to zero, but not actually equal
zero.
Velocities
When this process is applied to the average velocity the result of computing the average velocity
over shorter and shorter time intervals produces the instantaneous velocity. Let s = f (t) be the
displacement at time t then the instantaneous velocity is
lim
v(a) = h→0
f (a + h) − f (a)
h
This is often referred to as the velocity at t = a.
Suppose y is a quantity that depend on another quantity x. Thus y is a function of x and we write
y = f (x). If we change from x1 to x2 then the change in x is called an increment in x and is
denoted by ∆x = x2 − x1 and the corresponding increment in y is denoted by ∆y = y2 − y1 .
The average rate of change of y with respect to x over the interval [x1 , x2 ] is denoted by
∆y
f (x2 ) − f (x1 )
=
∆x
x2 − x1
and can be interpreted geometrically as the slope of the secant line. As we did with velocities we
can compute the average rate of change over smaller and smaller intervals by letting x2 approach
x1 and therefore letting ∆x approach 0. The limit of this process is called the instantaneous rate
of change of y with respect to x at x = x1 . This is interpreted geometrically as the slope of the
tangent to the curve y = f (x) at (x1 , f (x1 )).
lim
Instantaneous rate of change = ∆x→0
∆y
f (x2 ) − f (x1 )
= lim
∆x x2 →x1
x2 − x1
4.1. Derivatives from 1st Principles
4.1
56
Derivatives from 1st Principles
Because the expression
lim f (a+h)−f (a)
h→0
h
occurs so widely it is given a special name and notation.
The derivative of a function at a number a, denoted by f 0 (a) is
lim
f 0 (a) = h→0
Definition of the derivative:
f (a + h) − f (a)
h
The process of finding the derivative using the above definition is called finding the derivative from
first principles. So far we have found a derivative of a function f at a fixed number a. If we replace
a in this equation with a variable x we obtain
lim
f 0 (x) = h→0
f (x + h) − f (x)
h
*Note the difference from above
given any number x for which this limit exists. We assign to x the number f 0 (x). So we can regard
f 0 as a new function which we call the derived function or the derivative of x.
Alternate Notation for the Derived Function
Given the notation y = f (x) for the function the following alternative notations for f 0 (x) are
common:
dy
df
d
f 0 (x) = y 0 =
=
=
f (x), also = Df (x) = Dx f (x)
dx
dx
dx
The symbols D and
d
dx
are called differential operators.
Derivatives of Polynomial Functions
The constant function f (x) = c is considered a polynomial of degree zero. Using the method of
first principles we can find the derivative as follows:
lim
f 0 (x) = h→0
f (x + h) − f (x)
c−c
lim
lim 0 = 0
= h→0
= h→0
h
h
Higher Power Polynomials
When f (x) = x it can be shown from first principles that f 0 (x) = 1. Similarly when f (x) = x2 it
can be shown that f 0 (x) = 2x and when f (x) = x3 it can be shown that f 0 (x) = 3x2 .
Example
Find f 0 (x) for f (x) = x4 from first principles.
Solution
lim
f 0 (x) = h→0
lim
= h→0
lim
= h→0
f (x + h) − f (x)
h
4
(x + h) − x4
h
4
x + 4x3 h + 6x2 h2 + 4xh3 + h4 − x4
h
4.2. Standard Derivatives
57
4x3 h + 6x2 h2 + 4xh3 + h4
h
lim (4x3 + 6x2 h + 4xh2 + h3 ) here, substitute h = 0
= h→0
lim
= h→0
f 0 (x) = 4x3
This pattern will follow for any similar polynomial: If f (x) = xn then f 0 (x) = nxn−1 . Or alternatively
The Power Rule for Differentiation:
d n
(x ) = nxn−1
dx
This pattern implies
n must be a positive integer. It can be shown that from the definition of
that
dy
d 1
1
a derivative dx
=
−
or
y = x−1 then dx
= −1 × x−2 , which proves the power rule for n = −1.
2
x
x
√
Similarly if the exponent is a fraction it can be shown that the power rule holds e.g. if f (x) = x
1
1
then f 0 (x) = 2√1 x or f (x) = x 2 then f 0 (x) = 21 x− 2 . It can be shown that the power rule holds for
any real number n.
4.2
Standard Derivatives
The basic functions have easily repeatable patterns to find their derivatives. The common ones are
summarized in the table below:
Function
Derivative
f (x)
f 0 (x)
Notes
notation
xn
nxn−1
‘power rule’
ex
ex
exponential
ln(x)
1
x
logarithmic
sin(x)
cos(x)
cos(x)
− sin(x)
tan(x)
sec2 (x)
trigonometric
The Natural Exponential
Recall the natural exponential function from section 3.1. Here we can see why precisely it is so
h
lim e −1 = 1. The
special. Using limit notation, we can say that e is the number such that h→0
h
derivative is:
The Derivative of the Natural Exponential Function
d x
(e ) = ex
dx
4.3. Maximums, Minimums, and Tangents
58
The natural exponential function is unique because it has its own derivative! Geometrically
this means that the slope of the tangent at any point is the same as the y-coordinate, f (x), of that
point.
4.3
Maximums, Minimums, and Tangents
Turning points occur at the boundary between regions in a function. A function that transitions
from increasing to decreasing must have a turning point. In the function below, f (−1) = 2 means
the point (−1, 2) is a turning point. More specifically this is a local maximum of our function.
It is the highest point in its local neighbourhood. Notice that the other boundary has different
properties: it is the lowest point. So f (1) = −2 means the point (1, −2) is a local minimum.
The boundary means that the slope of the
tangent has gone from increasing to decreasing. Or from positive to negative. During this
transition it had to go through zero, and so a
turning point can be found anywhere the slope
is equal to zero: f 0 (x) = 0. Remember that
the first derivative represents the slope of the
function.
y
local maximum
f 0 (−1) = 0
2
1
x
-2
-1
1
2
-1
-2
Example
Find the turning points for
f (x) = x3 − 3x
f 0 (1) = 0
local minimum
Solution
Set the derivative equal to zero and solve for
values of x.
dy
= 3x2 − 3
dx
0 = 3x2 − 3
3 = 3x2
1 = x2
x = 1, and x = −1
Therefore the x−values of the turning points
are 1 and −1. Note there are two turning
points in our function so there should be two
corresponding solutions.
f (1) = (1)3 − 3(1)
= 1 − 3 = −2
Therefore (1, −2) is the first turning point.
Similarly, (−1, 2) is found as the other turning point. These are both shown in the figure
above.
Second Derivative Test
We know from the graph that these turning points are maximum and minimum. The graph is not
always provided to identify maxima and minima, so it helps to have a test to determine if a point
is a max or min.
We know first derivative represents the slope of the function. Looking at the parabola y = x2 and
following the function from left to right we can plot some values of the first derivative. The graph
4.3. Maximums, Minimums, and Tangents
59
tells us that the turning point is at (0, 0) and is a minimum. The value of the 1st derivative is
-2,-1,0,1,2 : these numbers are increasing. This trend is always true for a local minimum.
How do we test for this trend of increasing slopes? The rate of change of the slopes is the second
derivative, and if this number (at the point (0, 0)) is positive, then the point is a minimum.
This is called the second derivative test.
Second Derivative Test
• If f 0 (x) = 0, and f 00 (x) > 0, then the point (x, f (x)) is a local minimum. The graph
in this neighbourhood is concave up.
• If f 0 (x) = 0, and f 00 (x) < 0, then the point (x, f (x)) is a local maximum. The graph
in this neighbourhood is concave down.
Example Use the second derivative test to determine if the turning point (−1, 2) is a maximum
or minimum on the function f (x) = x3 − 3x
Solution Find the second derivative and substitute the turning point.
f 0 (x) = 3x2 − 3
f 00 (x) = 6x
f 00 (−1) = 6(−1) = −6
−6 < 0
therefore the turning point (−1, 2) is a local maximum.
Points of Inflection
There is one last feature of the initial function that can be found. There is a point in between the
maximum and minimum points where the change in slopes of the function change from decreasing
to increasing. This is called a point of inflection. It is where the concavity of the function changes
between down and up.
y
point of inflection
x
concave down
concave up
4.4. The Product, Quotient, and Chain Rules
60
The second derivative test can be used to determine if a function is concave-up (> 0) or concave
down (< 0). Note there is no = in these inequalities. In between is where the point of inflection
can be found:
Second Derivative Test for Concavity
• If f 00 (x) = 0, then the point (x, f (x)) is a point of inflection.
Example
Find where the function changes from concave-up to concave-down.
f (x) = 3x3 − 12x2 + 7
Solution
Find the second derivative and set equal to zero to find the point of inflection.
f 0 (x) = 9x2 − 24x
f
00
f (x) = 18x − 24
0 = 18x − 24
4
=x
3
Sub back into f (x) to find the y−value:
4.4
3
2
4
4
4
=3
− 12
+7
3
3
3
= −7 29
Therefore, the point
flection.
4
2
3 , −7 9
is a point of in-
The Product, Quotient, and Chain Rules
The Product Rule
Let f (x) = x and g(x) = x2 . What is the derivative of f (x) × g(x)? The question helps to show
that the answer is NOT f 0 (x) × g 0 (x)
f (x) × g(x) = x × x2 = x3 and we know the derivative of x3 is 3x2 . Also we know that f 0 (x) = 1
and g 0 (x) = 2x so f 0 (x) × g 0 (x) = 1 × 2x = 2x not 3x2 .
So the derivative of the product of two functions is not the product of the derivatives of each
function. In symbols this can be written
(f g)0 6= f 0 g 0
Theorem If f and g are both differentiable then
d
d
d
[f (x)g(x)] = f (x) [g(x)] + g(x) [f (x)]
dx
dx
dx
The Product rule is often seen in an abbreviated form as (uv)0 = uv 0 + vu0 .
The Quotient Rule
Let u = f (x) and v = g(x) be differentiable functions of x then we can show that
4.4. The Product, Quotient, and Chain Rules
61
d
d
g(x) dx
[f (x)] − f (x) dx
[g(x)]
d f (x)
=
2
dx g(x)
[g(x)]
or in abbreviated form as
Example
0
u
vu0 − uv 0
=
v
v2
See blackboard course notes for examples of the product and quotient rules.
Chain Rule
When functions are combined with other functions, they are often called composite functions.
These require special treatment when differentiating.
Let f (x) = x2 and g(x) = 2x+1 then (f ◦ g) This means f ‘composed of’ g is f (g (x)) = f (2x+1) =
(2x + 1)2 .
Also, g ‘composed of’ f would be: (g ◦ f ) (x) = g x2 = 2 x2 + 1 = 2x2 + 1.
The differentiation rules we have met so far allow us to differentiate pairs of functions that have
been added, subtracted, multiplied or divided. They do not allow us to differentiate an expression
that is made from a function that is within another function.
The following are all examples of composite functions.
1. We can differentiate x2 but we can’t use the same procedure to differentiate (1 − x)2 . Here
we can imagine if f (x) = x2 and g(x) = 1 − x then (f ◦ g) (x) = f (1 − x) = (1 − x)2 .
2. We can differentiate
1
x2
but we can’t use the same procedure to differentiate
1
.
x2 +1
2
3. We can differentiate ex but we can’t use the same procedure to differentiate ex .
A name often used for functions of this type is function of a function.
Once we recognise we are dealing with a composite function we need a procedure to differentiate
it. You will find that you are far more likely to be required to differentiate a composite function
in a practical situation than a simple one. It can be proved that the derivative of the composite
function f ◦ g is the product of the derivatives of f and g. This important rule is given the name
the Chain Rule. A substitution method is often used to add clarity to the differentiation process.
Let y = u2 and let u = 1 − x. Then
−2(1 − x) = 2(x − 1).
dy
du
= 2u and
du
dx
= −1. Now
dy
dx
=
dy
du
· du
dx = 2u × (−1) = −2u =
dy
dy
du
=
is what gives the rule its name. Because of the
dx
dx
du
apparent cancelling it is particularly easy to learn in this form.
The Leibniz form of the Chain Rule
As an aside let us verify the rule for this example. Given y = (1 − x)2 . We will expand the right
hand side of the equation. It becomes y = x2 − 2x + 1. So y 0 = 2x − 2 = 2(x − 1) as before.
Using function notation the Chain Rule states: If f and g are both differentiable and F = f ◦ g is
the composite function F (x) = f (g(x)), then F is differentiable and F 0 = f 0 (g(x))g 0 (x).
4.4. The Product, Quotient, and Chain Rules
62
A Comment on the Leibniz form of the Chain Rule
dy du
= du
· dx gives the impression that the du could cancel but remember we have not defined du.
dy
We have defined du
as the rate of change of y with respect to u and du
dx as the rate of change of u
with respect to x. However the apparent cancelling helps us to remember the way the differentials
are arranged. it also helps us to accept the extension of the Chain Rule to cover a function of a
function of a function etc. e.g.
dy
dx
Lety = f (u),u = g(v)andv = h(x)
dy
dy du dv
Then
=
·
·
dx
du dv dx
Example
Solution
where
Find F 0 (x) when F (x) =
1
.
x2 +1
Using function notation F (x) = (f ◦ g) (x) = f (g(x))
f (u) = u−1 and g(x) = x2 + 1
f 0 (u) = −u−2 andg 0 (x) = 2x
and
F 0 (x) = f 0 (g(x))g 0 (x)
−1
= 2
· 2x
(x + 1)2
−2x
=
2
(x + 1)2
Using the Leibniz notation let u = x2 + 1 and y = u−1 then
dy du
= −u−2 (2x)
du dx
−1
−2x
=
(2x) =
2
(x2 + 1)
(x2 + 1)2
F 0 (x) =
To use the method we need to bring a new variable into the problem we are trying to solve. It
is recommended that you use the variable u wherever possible so that you follow through using a
pattern you are familiar with.
In summary: if g is differentiable at x and f is differentiable at g(x), then the composite function
F = f ◦ g defined by F (x) = f (g(x)) is differentiable at x and F 0 is given by the product
F 0 (x) = f 0 (g(x)) · g 0 (x)
In Leibniz notation, if y = f (u) and u = g(x) are both differentiable functions, then
dy
dy du
=
·
dx
du dx
The Chain Rule will be found in many situations where functions are added, subtracted, multiplied
or divided. As an example we will focus on combining the Chain Rule with the Product Rule,
however any combination of these rules could be found in a problem.
4.5. Parametric Differentiation
63
2
Example
Differentiate xe−x .
Solution We can see that there is a product of two functions present in this example, i.e.
2
f (x) = x and g(x) = e−x . Also g(x) is a composite function.
We have from the Product Rule
(f g)0 = f g 0 + gf 0
By inspection we can see that of the four expressions on the right side of this equation f , g and f 0
can be put into the equation immediately and only g 0 requires some effort to be worked out.
g(x) is a composite function so g 0 (x) is computed using the Chain Rule.
Let u = −x2 then
du
dx
= −2x. Also g(u) = eu so
dg
du
= eu .
dg du
dg
=
dx
du dx
= eu · −2x
2
= e−x · −2x
= −2x e−x
2
So g 0 (x) = −2x e−x
2
Putting this all together
(f g)0 = f g 0 + gf 0
2
2
= x · −2x e−x + e−x · 1
2
= e−x 1 − 2x2
4.5
Parametric Differentiation
Parametric Curves x and y are both given as functions of a third variable t (called the parameter ).
Let the equations be
x = f (t) and y = g(t)
Each value of t gives a point (x, y). As t varies the point (x, y) = (f (t), g(t)) traces out a curve in
the coordinate plane called a parametric curve.
Using the Chain Rule to find the Derivative
dy
dx
Given the parametric equations x = f (t) and y = g(t) define a parametric curve. If f and g are
both differentiable the Chain Rule gives
dy
dx
dy
·
=
dt
dt
dx
provided y is also a differentiable function of x. So provided
dy
dy dx
=
÷
dx
dt
dt
dx
dt
6= 0
4.6. Related Rates
64
=
Example
4.6
dy
dt
×
dt
dx
See blackboard course notes for examples of parametric differentiation.
Related Rates
The concept of related rates is best understood by exploring some examples.
Example 1. Air is being pumped into a spherical balloon so that its volume is increasing at a
rate of 100 cm3 /s. How fast is the radius of the balloon increasing when the diameter is 50 cm?
You have to find the related rates in the question. Let V be the volume and r be the radius at
time t. The volume is increasing at the rate of 100 cm3 /s so dV
dt = 100. The question asks how fast
is the radius increasing. In other words what is dr
?
dt
Solution
and dr
dt .
We need a formula that connects V and r if we are to find the relationship between
dV
dt
The formula
4
V = πr3
3
is of the form V = f (r) so we can find
dV
dr
From the Chain Rule we can write
Substituting
dV
dt
= 100 and
dV
dr
dV
= 4πr2
dr
(1)
dV
dV dr
=
·
dt
dr dt
(2)
= 4πr2 in equation (2) we get
100 = 4πr2 ·
dr
dt
dr
100
=
dt
4πr2
25
= 2
πr
Now we substitute r = 25. (Diameter = 50 so radius = 25)
dr
25
=
dt r=25 π × 252
1
=
25π
So the radius is increasing at the rate of
1
25π cm/s
Example 2. A ladder 5 m long rests against a vertical wall. If the bottom of the ladder slides
away from the wall at the rate of 0.5 m/s how fast is the top of the ladder sliding down the wall
when the bottom of the ladder is 3 m from the wall?
4.6. Related Rates
65
Solution Let the origin be placed at the corner where the wall meets the floor, let x be the
distance of the foot of the ladder from the wall and let y be the distance of the top of the ladder
from the corner. The ladder forms a right angled triangle whose sides are x, y and with hypotenuse
dy
5. We are given that dx
dt = 0.5 and are asked to find dt when x = 3.
Pythagoras theorem gives
x2 + y 2 = 52
(1)
Differentiate equation (1) with respect to t
2x
Solve for
dy
dt
dy
dx
+ 2y
=0
dt
dt
dy
x dx
=− ·
dt
y dt
Using Pythagoras theorem when x = 3 and the hypotenuse = 5, y = 4
Substitute
dx
dt
= 0.5, x = 3 and y = 4
dy
3
= − · 0.5
dt
4
= −0.375
The top of the ladder is moving vertically downwards at the rate of 0.375 m/s
Example 3. A water tank has the shape of an inverted circular cone with a base radius of 2 m
and height of 4 m. If water is being pumped into the tank at a rate of 2 m3 /min find the rate at
which the water level is rising when the water is 3 m deep.
Solution Let V , r and h be the volume of water the radius of the surface and the height at
time t. We are given
dV
= 2m3 /min
dt
We are asked to find
dh
dt
when h = 3.
Draw a diagram to show that the relationship between r and h can be found by similar triangles.
r
2
=
h
4
h
r=
2
The formula for the volume is
1
V = πr2 h
3
Substituting equation (1) in equation (2)
1 h 2
V = π
h
3 2
(1)
(2)
4.7. Optimisation
66
=
π 3
h
12
Differentiate with respect to t
dV
π
dh
=
· 3h2 ·
dt
12
dt
π 2 dh
= h ·
4
dt
So
dh
4 dV
·
=
dt
πh2 dt
Substitute h = 3 and
dV
dt
=2
4
dh
·2
=
dt
π (3)3
8
=
9π
The water level is rising at the rate of
4.7
8
9π
m/min.
Optimisation
Optimisation is the process of using calculus to find the best result for a situation involving a
changing quantity (variable). Examples include maximizing profit, minimizing cost, maximizing
volume, minimizing amount materials used, and so on. As long as the quantities in question can
be represented by a function, calculus can used to find the special points of the function and their
nature.
A general method to approach an optimisation problem
1. Write down the known variables and draw a diagram
2. Form an equation of the situation by placing the unknown variable in terms of the
known variables
3. Use the facts of the problem to reduce the expression until it becomes a relationship
between the unknown quantity and one of the known quantities
4. Optimize by finding a maximum or minimum value
Example A farmer wishes to fence a paddock using an existing wall as one side of the paddock.
She has 100 meters of fencing and wants to know the dimensions of the paddock to enclose the
maximum area.
4.7. Optimisation
67
fence length 2
area of the
paddock
fence length 1
existing wall
Solution Following the steps outlined above, draw a diagram and label the variables. Let
T represent the total fence length which cannot be more than 100m. Write this as an equation:
T = 100. There are 3 individual lengths that make up the total, so let l represent the long side of
the paddock, and w represent the short side.
T = l + 2w
100 = l + 2w
The quantity to be optimised is area (A) of the paddock (remember it is a rectangle):
A = lw
The next step is to represent the quantity to be optimised (area) as a function of one other variable.
We can rearrange our earlier equation to solve for length: l = 100−2w. This can now be substituted
into the area equation:
A = lw
= (100 − 2w) w
= 100w − 2w2
This is now an equation that can optimised using calculus.
dA
= 100 − 4w
dw
0 = 100 − 4w
w = 25
Therefore the paddock width is 25 meters to maximize the total area. Going back to the original
constraint of 100m total length means that the paddock length is 100 − 2(25) = 50 meters.
Example A cylindrical can is to be made to hold 1 litre of oil. Find the dimensions that will
minimise the cost of the aluminium to manufacture the can. Note that 1 L = 1000 cm3 .
Solution The minimum cost of the aluminium will be the minimum surface area of the cylinder.
The can can be deconstructed into two circles and a rectangle. We will label the variables required
to calculate area.
Let SA represent surface area of the can. The total surface area is:
SA = πr2 + πr2 + (height ∗ circumference)
68
r
circumference
4.7. Optimisation
height of can
= 2πr2 + 2πrh
Note that we have 2 variables that are unknown, r, and h. We need to express one variable in
terms of the other in order to proceed. Use the additional information in the problem. The volume
of the can must be 1000 cm3 .
V = πr2 h
1000 = πr2 h
1000
h=
πr2
Substitute this form for h into the surface area function:
SA = 2πr2 + 2πrh
1000
2
= 2πr + 2πr
πr2
2000
= 2πr2 +
r
This function is now the surface area in terms of a single variable, r, and can be optimised:
dSA
2000
= 4πr − 2
dr
r
2000
0 = 4πr − 2
r
= 4πr3 − 2000
2000
r3 =
4π
r = 5.419 cm
Therefore the final dimensions of the can optimised for minimum cost are radius= 5.42 cm, and
height= 10.8 cm.
4.8. Chapter Exercises
4.8
§4.1
69
Chapter Exercises
Differentiation from First Principles
1. Find the derivative of the function at the given value using the method of first principles.
(a) f (x) = 5x2 + 3x − 1 at the number 2
(b) f (x) = 1 − 3x2 at the number 2
(c) f (x) = x4 at the number 1, given (x + h)4 = x4 + 4x3 h + 6x2 h2 + 4xh3 + h4
Graphing Exercises
2. The graph of f (x) = (x − 1)(x + 2)(x − 3) is drawn below. On the same set of axes, sketch
the graph of the derived function, f 0 (x). What is the shape of the graph of the derived function?
4.8. Chapter Exercises
70
3. The graph of f (x) = (x + 1)(x − 1)(x − 2)(x + 3) is drawn below. On the same set of axes,
sketch the graph of the derived function, f 0 (x)
What is the order of f (x)? What is the order of f 0 (x)?
4. The graph of f (x) = ex is drawn below. On the same set of axes, sketch the graph of the
derived function, f 0 (x)
What is the shape of the graph of the derived function?
4.8. Chapter Exercises
71
5. The graph of f (x) = 10x2 ex (x + 1) is drawn below. On the same set of axes, sketch the graph
of the derived function, f 0 (x)
6. Differentiate
(a) f (x) =
(d) y = x13
(g) y = √1x
§4.2
1. Find
(a)
(c)
(e)
1
x2
√
3
(b) y = x2
(e) y = x−4
(c)
(f )
y = x5
y = x3/4
Standard Derivatives
dy
dx
y = 25 x5
y = −3x4
y = 3x13
(b) y = −10
(d) y = x24
2. Differentiate
(a)
(c)
√
t
√
3 5
f (z) = z
f (t) =
(b) f (t) =
√
t3
(d) f (x) = 2x3.2
3. Find the derivative
(a)
(c)
f (x) = 2x3 − 3x2 + 4x − 1
x8 + 12x5 − 4x4 + 10x3 − 6x + 5
(b) f (x) = x2 + x + 1 +
1
x
4.8. Chapter Exercises
4. (a)
(c)
ds
dt
Given s = 4t2 − 7t + 5 find
d(3u4 )
Find du
(e) Given f (x) =
the graph.
(g)
72
ex
− x find
Differentiate g(x) = x3
(b) Find
d(3x)
dx
(d) Given f (x) = 2x − 3 find Df (x)
f 0 (x).
Sketch (f )
5
Differentiate f (x) = (3x)3
(h) Find the derivative of f (x) = ex − xe
5. Find the derived function
(a)
f (x) =
(c)
g(x) =
1
√
4 3
x
2
ex + 2ex
1
x3
√
−
(b) f (x) =
2
+ xe2 + xe
(d) f (x) =
√
x+ 3 x
√
4x
√
3
x+
√
5
2
6. Sketch the graphs and label your axes in radians
(a)
(c)
f 0 (x) where f (x) = sin x
h0 (x) where g(x) = tan x
(b) g 0 (x) where g(x) = cos x
7. Differentiate the trigonometric functions
(a)
(c)
(e)
√
(b) f (x) = x sin x
(d) y = cosx x
y = x2 sin x
h(x) = tan(5x)
g(t) = cos(ωt + δ)
8. Find the tangent line to the curve y = ex cos x at the point (0, 1).
9. A ladder 10m long rests against a vertical wall. Let θ be the angle between the top of the
ladder and the wall and let x be the distance between the bottom of the ladder and the wall.
If the bottom of the ladder slides away from the wall, how fast is x changing with respect to θ
when θ = π3 ?
10. Find the derivative
(a) sin 4x
(d) tan 3x
(b) π2 sin πx
(e) 3 tan (x + 2)
(c)
(f )
5 cos 3x
sin3 x
sin2 3x
(h) sin3 (x − 1)2
(i)
tan2 2x
(g)
§4.3
Maximums, Minimums, and Tangents
1. The function y = 7x − 3 has no turning point. Why?
2. Turning points are sometimes referred to as stationary points. Find the stationary points of
the following functions:
(a)
y=
x3
3
− 4x
(b) y = x2 +
16
x
(c)
1
f (x) = −x4 +4x3 −4x2 +
3. Find the turning points and use the second derivative test to show if they are local maxima or
local minima:
4.8. Chapter Exercises
(a)
(c)
73
y = 4 − (3 − x)2
f (x) = (3 − x)(x + 2)(x + 5)
(b) y =
x3
3
−x
4. Consider f (x) = xex . Do you think this function has a local max or min? Think of the graph
of y = ex . Plot the function using Desmos and find the turning point.
√
5. Find the equations of the tangent line and the normal line to the curve y = x x at the point
(1, 1).
6. Find the points on the curve y = x4 − 6x2 + 4 where the tangent line is horizontal.
7. At what point on the curve y = ex is the tangent line parallel to the line y = 2x?
8. Find a point a on the curve f (x) = x3 + 2x2 + 3x + 4 where f 0 (a) = 2
9. Find where the tangent line to the function f (x) = x3 − x + 1 is parallel to the line y = x.
Graphing Exercise
10. If f (x) = 2x2 − x3 find f 0 , f 00 , f 000 , f (4) . Use Desmos to graph f , f 0 , f 00 , and f 000 on a common
screen. Describe whether these graphs are consistent with a geometric interpretation of these
derivatives.
11. If f (x) = x1 find f 0 (x) and f 00 (x) then graph f , f 0 and f 00 on a common screen. Are your answers
reasonable?
§4.4
Product, Quotient, & Chain Rules
1. Use the product rule to find the derivative
(a)
f (x) = xex
(b) g(x) = x2 ex
2. Differentiate with respect to x
(a)
(c)
(b) x−3 ex
(d) (x + 2) (x − 2) ex
x3 e x
(x + 1) ex
3. Use the quotient rule to find the derivative
(a)
y=
(c)
f (t)
3x+1
2x−1
2t
= 1+t
2
(b) y =
find
df
dt
ex
x+1
(d) f (x) =
A
B+Cex
4. Differentiate
(a)
(c)
3x2
1−x
√
x
x+2
(b)
(d)
x
x+2
2x2 +3x+2
ex
Exercises involving Tangent and Normals
5. Find the equation of the tangent line and normal line to the curve y = xex at the point (0, 0).
6. The curve y = 1/(1 + x2 ) has the name witch of Maria Agnesi.
(a) Find the equation of the tangent line to this curve at the point 1, 12 .
(b) Use Desmos to draw the graph of the curve and the tangent line on the same grid.
4.8. Chapter Exercises
74
7. The curve y = x/(1 + x2 ) is called a serpentine.
(a) Find the equation of the tangent line at the point 1, 12 .
(b) Use Desmos to draw the graph of the curve and the tangent line on the same grid.
Composite Functions
√
8. Let f (x) = x and g(x) = x3 find (f ◦ g) (x) and (g ◦ f ) (x).
9. Given h (x) = ex and j(x) =
x2
2
find (h ◦ j) (x) and (j ◦ h) (x).
10. The chain rule
Find F 0 (x) when F (x) =
5
(b) Given y = 1 − x2 find
(a)
√
1 + x2 .
dy
dx .
2
(c) Find the derivative of ex .
x
(d) Find the derived function for ee .
11. Differentiate with respect to x
(a)
(c)
(e)
(g)
(3x + 2)3
1
2x+1
√
2x − 5
√1
x+2
3
(b) (5x + 3) 5
3
(d) (4−x)
3
√
3
(f )
5 − x2
(h) e2x
3
12. You may have to combine rules to differentiate the following
(a)
y = x2 e−2x
(c)
√p+1
(e)
(x − 3)3 (x + 2)
2√
ex x + 1
(g)
§4.5
p2 +1
(b) (1 − 2x)2 e−x
2
(d) x2 + 3 (x − 4)
√
(f )
x + 1 (x − 1)2
(h)
x2 +x+2
(x+1)2
Parametric Differentiation
1. Use desmos to sketch the parametric curves
(a)
(c)
(e)
x = t2 − 2t and y = t + 1 where 0 6 t 6 4 (b) x = cos t and y = sin t where 0 6 t 6 2π
x = 2 cos t and y = sin t for 0 6 t 6 2π
(d) x = cos t and y = sin 2t where 0 6 t 6 2π
x = 1.5 cos t − cos 40t and y = 1.5 sin t − sin 40t where 0 6 t 6 2π
2. Given y = 2t and x = t2
dy
(a) Find dx
.
(b) By finding the appropriate value of t show that (1, 2) lies on the parametric curve.
(c) Find the equation of the tangent line to the parametric curve at (1, 2).
4.8. Chapter Exercises
75
3. Given x = cos t and y = sin t find the equation of the tangent line at the point
Where does this curve have horizontal or vertical tangent lines?
4. Given x = cos t and y = sin 2t find the equation of the tangent line at the point
Where does this curve have horizontal or vertical tangent lines?
5. Find
(a)
(c)
(e)
§4.6
dy
dx
√
√
2
2
,
2
2 .
√
3
3
,
2
2 .
√
in terms of the parameter
x = t2 + t
y = t3 − t2
x = at2
y = 2at (a is constant)
x = 2 sin θ
y = cos 2θ
(b) x = et cos t
y = et sin t
(d) x = a sin θ
y = b cos θ
(f ) x = a (θ − sin θ)
y = a (1 − cos θ)
Related Rates
1. If V is the volume of a cube with edge length x and the cube is expanding as time passes, find
dV
dx
dt in terms of dt .
2. Each side of a square is increasing at a rate of 6cm/s. At what rate is the area of the square
increasing when the area of the square is 16cm2 ?
3. (a) If A is the area of a circle with radius r and the circle expands as time passes find
terms of dr
dt .
dA
dt
in
(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius
of the oil spill increases at a constant rate of 1m/s how fast is the area of the spill increasing
when the radius is 30m?
4. If a snowball melts so that its surface area decreases at a rate of 1cm2 /min, find the rate at
which the diameter decreases when the diameter is 10 cm. The following steps may help you
to solve this problem.
(i)
(ii)
(iii)
(iv)
(v)
What quantities are given in the problem?
What is the unknown?
Draw a picture of the situation for any time t.
Write an equation that relates the quantities.
Solve for the unknown quantities.
5. At noon ship A is 150km west of a ship B. Ship A is sailing east at 35km/h and ship B is sailing
north at 25km/h. How fast is the distance between the ships changing at 4:00 P.M.?
6. A plane flying horizontally at an altitude of 2km and a speed of 800km/h passes directly over
a radar station. Find the rate at which the distance from the plane to the station is increasing
when it is 3km away from the station.
§4.7
Optimisation
1. Divide 50 into two parts such that the product of the two parts is a maximum.
2. Find the number that exceeds its square by the greatest amount.
4.8. Chapter Exercises
76
3. A farmer has 2400m of fencing and wants to fence off a rectangular field that borders a straight
river. He needs no fence along the river. What are the dimensions of the field that has the
greatest area?
4. A farmer wishes to fence off a corner of a field where there is an existing hedge on two sides.
The hedge is to be used to fence the two sides. If he has 300m of fencing available, find the
dimensions a and b so that he encloses the maximum area.
5. The volume of the cone is given by the formula V (r) =
at its maximum volume.
10πr2
3
3
− πr3 . Find the radius of the cone
6. The total cost of holding a large event is composed of the venue hire, V , times the venue tax,
T , plus the entertainment, E. These values can all be estimated based on the number of people
attending the event, p, such that: V (p) = 50p+80, T (p) = 0.002p−0.4, and E(p) = 36000−60p.
Find the minimum cost.
7. Find the area of the greatest rectangle that can be inscribed in a semicircle of radius r.
8. If 1200cm2 of material is available to make a box with a square base and an open top. Find
the largest possible volume of the box. (Hint: There is no wasted material.)
5 | Integration <
R
>
In mathematics we are often given a function f and asked to find a function F whose derivative
is f . F in this situation is called the integral of f . It is usual to develop a list of integrals by
differentiating a range of functions then using those to work backwards. The terms integration and
anti-differentiation are synonymous. Generally we will use the term integration, however, both are
acceptable.
Using derivatives to infer the integral
1. Because y = x2 gives y 0 = 2x we can say that an integral of 2x is x2 .
2. Because y = x2 + 1 gives y 0 = 2x we can say that another integral of 2x is x2 + 1.
3. Because y = x2 + c where c is a constant gives y 0 = 2x we can say that an integral of 2x is
x2 + c.
4. Because y = ex gives y 0 = ex we can say that an integral of ex is ex . In fact we say that the
integrals of ex are ex + c where c is a constant.
5. Because y = ln x + c where c is a constant gives y 0 = x1 we can say that a family of integrals
of x1 is ln x + c. This is often written ln kx where c = ln k is the constant
6. This can be generalised. y = ln |x| + c also gives y 0 =
ln |x| + c.
1
x
so in general the integrals of
1
x
are
The process of ‘reversing’ or ‘undoing’ a derivate has its own symbol, the integrand.
Z
5.1
f 0 (x)dx = f (x)
Standard Integrals
The examples shown above begin the process of establishing the start of a list of integration rules.
It is not the intention here to list all of the rules that are required however at this stage let us
explore the Power Rule to establish a rule for antedifferentiating an expression of the form y = xn
f
x
x2
x3
x4
f0
1
2x or
3x2
4x3
f
x
1 2
2x
1 3
3x
1 4
4x
f0
1
x so we can say
x2
x3
f0
1
x
x2
x3
f
x+c
1 2
2 x + c or
1 3
3x + c
1 4
4x + c
f
1
x
x2
x3
F
x+c
1 2
2x + c
1 3
3x + c
1 4
4x + c
5.1. Standard Integrals
78
This establishes the pattern and if you think about the rule for differentiating y = xn you can soon
establish the rule for integrating xn .
The Power Rule for Integrating
Z
xn+1
xn dx =
+ c, where n 6= −1
n+1
If f (x) = x−1 then the integral of f is ln |x| + c or ln |kx|
All of the differentiation rules we have met so far lead to integration rules. For instance we can
establish standards for sin x, cos x, and sec2 x. The standard integrals are summarized in the table
below.
function
f (x)
Rintegral
f (x)dx
xn , n 6= −1
ex
xn+1
n+1 + C
ex + C
1
x
ln |x| + C
sin(x)
− cos(x) + C
cos(x)
sin(x) + C
sec2 (x)
tan(x) + C
To allow us to combine these integrals and thus extend the range of questions we can tackle we use
two important rules for integrals
Sum Rule The integral of the sum of two functions is the sum of the integrals of the
functions.
Z
Z
Z
[f (x) + g(x)] dx = f (x) dx + g(x) dx
This is easily extended to the sum or difference of a number of functions.
Constant Multiple Rule The integral of a constant times a function is the constant times
the integral of the function.
Z
Z
cf (x) dx = c
f (x) dx
Z
clearpage Example
Integrate the following functions; find
f (x).
5.2. Area
(a)
79
f (x) = 3x2
Applying the power rule:
Solution
Solution
3x2+1
+C
2+1
= x3 + C
Z
(c)
(b) f (x) = 7
Here we are integrating a constant:
Z
7 dx = 7x + C
3x2 dx =
2
(d) f (x) =
f (x) = x 3
1
√
2 x
+
√1
2
Solution The power rule still applies to frac- Solution Here we need to combine the sum
tional indices:
rule and the power rule:
Z
2
x 3 +1
x dx = 2
+C
3 +1
Z
2
3
=
Z
Z
1
1
( √ ) dx + ( √ ) dx
2 x
2
Z
1
1
1
=
x− 2 + √ x + C
2
2
√
x
= x+ √ +C
2
f (x) dx =
3x5/3
+C
5
Find f (x) given f 00 (x) = 6
Example
Here we have a second derivative, indicated by the double-prime symbol, f 00 . Knowing
Solution
R 0
that f (x) = f (x) we can safely assume that
Z
So f 0 (x) =
R
f 00 (x) =
R
f 00 (x) dx = f 0 (x).
6 dx = 6x + C. Now we need to integrate a second time to get f .
Z
(6x + C) dx = 6x2 + Cx + D
We end up with two unknown values, C and D as opposed to just a single value.
The previous two examples of equations involving derivatives. Any equation involving derivatives
of a function is called a differential equation. We will look into this subject in the next chapter.
5.2
Area
In this section we attempt to find the area under a curve. That is the area that lies between a
curve and the x-axis from x = a to x = b. The area is bounded by the x-axis, a continuous curve
y = f (x) and the two vertical lines x = a and x = b. This is shown in the figure below with the
area shaded in.
5.2. Area
80
Figure: A parabola showing the area
under the curve between x = 0, and
x = 1.5. Note the area stops at the
axis. Area as calculated by integration is always in reference to the axis.
Previously, when we wanted to find the slope of a tangent line we found the slope of a secant line
lim . A similar procedure will be used to find the area. We first
and applied the limiting process h→0
approximate the area with rectangular strips then we take the limit of the areas of these rectangular
strips by making the strips narrower and narrower and thus the number of strips between x = a
and x = b greater and greater.
Example
0 and 1.
Consider the curve y = x2 . Use rectangles to find the area under this curve between
Solution Consider 4 strips by constructing vertical lines at x = 14 , x = 12 , x = 43 and x = 1.
Define the left sum as the sum of the rectangles whose left side is formed using x = 14 , x = 12 , x = 34
and x = 1. Define the right sum as the sum of the rectangles whose right side is formed by x = 14 ,
x = 21 , x = 34 and x = 1.
1 2 1 1 2
·0 +
+
4
4 4
1
9
1
+
+
=0+
64 16 64
Left sum =
1 1 2 1 3 2
+
4 2
4 4
1+4+9
7
=
=
= 0.21875
64
32
1 1 2 1 1 2 1 3 2 1
+
+
+ (1)2
Right sum =
4 4
4 2
4 4
4
1
1
9
1
1 + 4 + 9 + 16
15
=
+
+
+ =
=
= 0.46875
64 16 64 4
64
32
It is clear from a diagram that the actual area is larger that the left sum and smaller than the right
sum.
0.21875 < A < 0.46875
We could repeat this process with a larger number of strips but it is clear the process would quickly
become tedious. We can however simulate this process using a spreadsheet like Excel.
For the curve y = x2 set up a spreadsheet to show 10 strips or 20 strips or 30 strips etc. The table
gives the results you should obtain for a selection of numbers of strips
5.2. Area
81
n
10
20
30
100
1000
Left sum
0.2850000
0.3087500
0.3168519
0.3283500
0.3328335
Right sum
0.3850000
0.3587500
0.3501852
0.3383500
0.3338335
It can be seen that a very accurate approximation to the area can be obtained as the number
of rectangles increases. It should be clear that as n → ∞ both the left sum and the right sum
approach the area under the curve we write
lim Left Sum = lim Right Sum
A = n→∞
n→∞
This process can be generalised by selecting any height within each rectangular strip and finding
the area of each strip using this height. Let there be n strips and consider the ith strip. Select
a value of x in the ith strip call it xi . The height of this rectangle will be f (xi ). Consider the
situation described above where the area is bounded by the x-axis, a continuous curve y = f (x)
and the two vertical lines x = a and x = b.
With n rectangles the length of the base of each rectangle is ∆x =
b−a
n
The area of the ith rectangle is f (xi )∆x
The sum of all the rectangles is
n
X
f (xi )∆x = f (x1 ) ∆x + f (x2 ) ∆x + . . . + f (xn ) ∆x
i=1
And
"
A = lim
n→∞
n
X
f (xi )∆x
#
i=1
If fPis a continuous function defined on the interval [a, b] then as n → ∞ the number represented
by n f (xi )∆x → A the area under the curve y = f (x). This number is called the definite integral
i=1
Rb
Rb
of f from a to b and is denoted by a f or a f (x)dx
" n
#
Z b
X
f (x)dx = lim
f (xi )∆x
a
n→∞
i=1
This process is called a Riemann sum after the GermanR mathematician Bernard Riemann (18261866) who defined the integral in this way. The symbol was introduced by Leibniz and is called
the integral sign.
Definite Integrals
The method of computing Riemann sums is often long and to achieve a result that is accurate
enough requires a computer. Both Sir Isaac Newton and Leibniz discovered a much simpler way
based on the integral. This discovery is called The Evaluation Theorem.
5.2. Area
82
Given F is an integral of f i.e. F 0 = f , provided f is continuous on the interval [a, b] then
Evaluating definite integrals
Z
b
f (x)dx = F (b) − F (a)
a
This is an amazing result in view of the fact that it replaces such a complex procedure as finding
Riemann sums over greater and greater numbers of elementary rectangles.
R1
Example Evaluate 0 x2 dx.
Solution Because we know a particular integral of f (x) = x2 is F (x) = 31 x3 We have from the
Evaluation Theorem
Z 1
1
1
1
x2 dx = F (1) − F (0) = · 13 − · 03 =
3
3
3
0
Looking back at the calculation of left sum and right sum above we can now see that the actual
area that we were endeavouring to calculate was in fact 1/3 or 0.3̇.
These are some of the different notations for using the Evaluation Theorem
F (b) − F (a) = F (x)|ba = [F (x)]ba = F (x)]ba
Definite Integrals and Areas
Areas above the x-axis have positive definite integrals and areas below the x-axis have negative
definite integrals. You need to be careful therefore to answer the question that is being asked. If
the question is about definite integrals then they are evaluated by following the definition. If the
question is about areas you must find the parts of the question that have areas above the x-axis and
those parts that have areas below the x-axis and evaluate them separately. The definite integral
calculates the result as the net sum of the positive and negative areas. To find the total area you
drop the signs (|Ai |) and treat each area as if it is positive before adding them together
X
Total area =
|Ai |
i
To illustrate this point consider y = x3 − x.
Example
Find the area under the curve y = x3 − x between x = −1 and x = 1.
Solution This can be factorised to give y = x(x + 1)(x − 1). This cubic curve crosses the x-axis
at −1, 0, and 1. Here, this area must be found in two parts. (Refer to the figure on the next page.)
5.2. Area
83
Figure: A cubic showing how area
‘under’ the curve is evaluated. The
area for −1 ≤ x ≤ 0 is positive (above
the axis), and the area for 0 ≤ x ≤ 1
is negative.
R1
R0
(x3 − x)dx
4
2
= x4 − x2 |10
= [ 41 − 12 ] − 0
= − 41
Therefore the total area is
| − 41 | + 14 = 12 .
3
−1 (x − x)dx
2
4
= x4 − x2 |0−1
= 0 − [ 14 − 12 ]
= + 14
0
We will compare with a single integral from −1 to 1.
Z 1
1
1
x4 x2 1
− |−1 = − − (− ) = 0
(x3 − x)dx =
4
2
4
4
−1
Area must be non-negative, and so this result is nonsensical given the context.
Example
The energy, or electrical charge, that a capacitor can discharge is found by taking the
integral of the voltage-time function. This can
neatly be represented as the area under the
voltage-time curve. Find the total discharge
from the capacitor after 5 seconds. The units
for charge are coulombs.
Solution Integrate the V (t) function to
find the area:
voltage(V )
8
6
V (t) = 8e−t
4
5
V (t)dt = 8
e−t dt
a
0
5
= 8 −1e−t
0
−5
= 8 −e − (−e0 ) = 8.054 coulombs
Z
b
Z
area =
2
time(s)
−1
1
2
3
4
5
Example A logo is formed by the shaded area between the cubic function f (x) = 4x − x3 and
a parabola g(x) = 2x − x2 . The two curves intersects at x = 0 and x = 2. Find the shaded area.
6
5.3. Volume
84
Solution
4
Z
2
b
(f (x) − g(x))dx
Shaded area =
a
Z
x
0.5
1
1.5
2
2.5
2
(4x − x3 ) − (2x − x2 ) dx
=
0
3
Z
2
(2x − x3 + x2 ) dx
0
x2 x4 x3 2
=
2 −
+
2
4
3
0
16 8
− (0)
=
4−
+
4
3
8
= units2
3
=
−2
−4
Indefinite Integrals
R
Rb
Although f (x)dx looks very similar toR a f (x) dx they are quite different and must not be
confused or used in place of each other. f (x) dx is a function of x or a family of functions of x
Rb
and a f (x) dx is a number. They are connected of course, provided f (x) is a continuous function
of x on [a, b]. In this case the Evaluation Theorem gives the connection between them.
Z
b
b
Z
f (x) dx =
f (x) dx
a
a
The indefinite integral represents either a particular integral or a family of integrals. These will use
a constant C where C takes a different value for each member of the family. C is called the constant
of integration. Our ability to find indefinite integrals depends on the number of derivatives we have
met so far and as we meet more and more different functions and learn how to differentiate them
we add to our list.
5.3
Volume
If a function is revolved around an axis it creates a volume between the axis and the function.
Similar to how if we integrate a function, it results in an area — if we integrate and area it results
in a volume.
Volume rotated around the x-axis
Z b
Z b
2
Volume =
πy dx =
π(f (x))2 dx
a
a
Volume rotated around the y-axis
Z
Volume =
d
2
Z
πx dy =
c
c
d
π(f (y))2 dy
5.3. Volume
85
Example A connector was obtained by revolving the function f (x) = 3x + 1 around the x−axis
for 1 ≤ x ≤ 35 . Calculate the volume of the connector.
Solution
Rb
volume = a π[f (x)]2 dx
=π
R
5
3
1
(3x + 1)2 dx
3
= π (3x+1)
3·3
=
π
9
5
3
1
63 − 43 =
152π
9
≈ 53.058 units3
Example Fluorescent and incandescent light bulbs are often filled with the inert gas krypton.
Find the volume of krypton gas required to fill the bulb shown.
You have estimated the function to be:
x2
5
You will have to calculate 1 −
for − 1.5 ≤ x ≤ 2 cm
f (x) = 1 −
Solution
x2
5
2
before integrating
b
Z
π[f (x)]2 dx
volume =
a
2
x2
dx
=π
1−
5
−1.5
Z 2 4
x
2x2
=π
−
+ 1 dx
5
−1.5 25
5
2
x
2x3
=π
−
+x
125
15
−1.5
32
16
=π
−
+ 2 − (−0.06075 − −0.45 − 1.5) = 2.3π ≈ 7.23 cm3
125 15
Z
2
5.4. Integration by Substitution
86
Example Calculate the volume of the √
container found by rotating the curve y = x3
around the y−axis for 0 ≤ y ≤ 5.
Here the volume is created by rotation about
the y-axis, and therefore we need to adjust
our formula.
First we will solve the equation
√
y = x3 for x, and then integrate.
Solution
Volume =
Rb
a
πf (y)2 dy
Rearrange the function to isolate x:
√
y = x3
y 2 = x3
p
3
y2 = x
Integrate to find volume:
Z
V =π
5
5
Z
=π
4
y 3 dy
0
7
=π
y3
7
3
5
0
3π 7
=
53 − 0
7
≈ 57.6 units3
2
(y 3 )2 dy
0
5.4
Integration by Substitution
Earlier we stated that every differentiation rule leads to an integration rule. The chain rule for
differentiation leads to the substitution rule for integration, i.e. the substitution rule for integration.
A series of examples will illustrate how integration by substitution works.
R √
Example Find the integral: 2x 1 + x2 dx.
Solution This cannot be integrated using the techniques discussed so far. This is an example
of a function that can be integrated using the substitution rule.√ They are often recognised by
noting the presence of a composite function. In this example the 1 + x2 is a composite function.
We introduce a new variable which simplifies the composite function.
Let u = 1 + x2 . We then differentiate du
probably noticed that we have throughout
dx = 2x. You have
R √
the chapter on integration written expressions like 2x 1 + x2 dx where the dx has been placed
at the right of the expression we are integrating to denote we are integrating with respect to the
variable x. We now write the expression du
dx = 2x as du = 2x dx. In this course we accept that this
can be done and leave the proof to a later course.
If we now look back at the original question we are ready to substitute expressions containing u
5.4. Integration by Substitution
87
for expressions containing x.
Z
Z p
p
2
1 + x2 · 2x dx
2x 1 + x dx =
Z
√
=
u du
Z
1
= u 2 du
1
=
u 2 +1
+C
1
2 +1
3
=
u2
3
2
+C
√
2 u3
=
+C
3
q
2
=
or =
(1 + x2 )3
3
+C
3/2
2
1 + x2
+C
3
Notice the last few steps are tidy-up steps. It is usual to begin and end with an expression in x so
at the end we back-substitute the expression we substituted with u.
The Substitution Rule can be stated formally as follows. If Ru = g(x) is a differentiable
function
R
0
whose range is an interval I and f is continuous on I, then f (g (x)) g (x) dx = f (u) du. dx
and du are known as differentials. The Substitution Rule permits us to replace g 0 (x) dx with du.
R
Example Find x sin x2 dx
Solution
Using the procedure from the previous example try u = x2 . This gives du = 2x dx.
Z
Z
2
x sin x dx = sin x2 · x dx
Z
1
= sin u · du
2
Z
1
=
sin u du
2
1
= − cos u + C
2
1
= − cos x2 + C
2
It is clear the challenge of the Substitution Rule is to come up with a suitable substitution. Earlier
examples often have fairly obvious substitutions but they can quickly become quite complicated
and result in many Rfalse starts.
x
Example Find √1−x
dx
2
Solution Let u = 1 − x2 . Then du = −2x dx. So x dx = − 21 du. These are now substituted
into the original expression
Z
Z
Z
p
x
1
1
1
1
√
√ du = −
dx = −
u−1/2 du = − 2u1/2 + C = − 1 − x2 + C
2
2
2
u
1 − x2
5.4. Integration by Substitution
88
In each of the above 3 examples you could use desmos to compare the original function and the
x
integral to see if the result is reasonable. For example 3 try graphing the original function y = √1−x
2
√
and the integral y = − 1 − x2 on the same√axes and check the original function represents the
slope of the tangent lines to the curve y = − 1 − x2 .
Evaluating Definite Integrals by Substitution
There are two ways to evaluate a definite integral when you have used the Substitution Rule for
the integration.
Method 1 Find the indefinite integral then use the Evaluation Theorem.
Method 2 Make the substitution to the integrand and differential (as before) and also use the
same substitution to change the limits to those for the new variable (in our questions we usually
use u).
R8√
Example Evaluate 0 x + 1 dx. Graphing provides us with the following screen
Figure: The curve is clearly
continuous. If we let u =
x + 1 then u0 = 1, this is
also continuous.
Solution Method 1 Substitute u = x + 1 then du = dx. Substituting these values in the
indefinite integral we get
Z
Z
√
√
2
2
x + 1 dx =
u du = u3/2 + C = (x + 1)3/2 + C
3
3
So
Z
8√
Z
x + 1 dx =
√
8
x + 1 dx
0
0
8
2
(x + 1)3/2
3
0
2
2
= (9)3/2 − (1)3/2
3
3
2
2
= 27 − 1
3
3
2
2
52
1
= 17
= (27 − 1) = 26 =
3
3
3
3
=
Method 2 Again we let u = x + 1 so du = dx. Also we calculate the new limits for u using
u = x + 1.
x = 0 gives u = 0 + 1 = 1
5.5. Integration by Parts
89
x = 8 gives u = 8 + 1 = 9
Now the definite integral is transformed into a definite integral in u.
Z
8√
Z
x + 1 dx =
0
1
9√
u du
9
2 3/2
u
3
1
2 3/2 2 3/2
= 9 − 1
3
3
2
1
= (27 − 1) = 17
3
3
=
A check on the graph will show that an area of 17 13 appears to be reasonable.
Method 2 is usually preferred as the step where the indefinite integral is first calculated has been
neatly incorporated into the method. The difficulty with method 2 is that once the values for u
have been calculated the integral is completely transformed and we never return to the original
question. We answer a different question that, because of the transformation, has the same answer.
This concept might cause confusion. However a graph of the situation shows what has happened.
It is clear that the area under the first graph between 0 and 8 is the same as the area under the
second graph between 1 and 9.
5.5
Integration by Parts
Recall the rule for the differentiation of a product
d
(f (x) · g(x)) = f (x) · g 0 (x) + g(x) · f 0 (x)
dx
We can antidifferentiate each side and write the process as follows
Z
f (x) · g(x) =
f (x) · g 0 (x) + g(x) · f 0 (x) dx
5.5. Integration by Parts
90
Z
=
Z
0
f (x) · g (x) dx +
g(x) · f 0 (x) dx
This is rewritten in a particular way to become the formula for integration by parts.
Z
Z
0
f (x) · g (x) dx = f (x) · g(x) − g(x) · f 0 (x) dx
Z
Z
0
or f (x) · g (x) dx = f (x) · g(x) − f 0 (x) · g(x) dx
(1)
(2)
This formula is written in a number of different ways in textbooks. Here are two ways
Let u = f (x) and v = g(x) then du = f 0 (x) dx and dv = g 0 (x) dx so using the substitution rule
equation (1) becomes
Z
Z
u dv = uv − v du
(3)
R
R
Let f (x) · g 0 (x) dx be regarded as the the integral of the product of two functions then g = g 0 .
We can write equation (2) as
Z
f g0 = f g −
Z
f 0g
(4)
Equations (3) and (4) are the common forms that are best to remember.
The success of this method depends on the discovery that a simpler integral results from this
process. Sometimes the process results in a worse situation than you started with so should be
abandoned. Sometimes you produce a pattern which leads to a solution after 2 or more applications
of the integration by parts rule. The patterns that produce solvable problems can be discovered as
different questions are tried.
R
Example Find x cos x dx. This can be seen as
R the integral of a product. The two functions
0
are f (x) = x and g(x) = cos x. So f (x) = 1 and g(x) = sin x are easy to find. Notice though
that f 0 (x) = 1 gives us a clue that integration by parts is going to be a fruitful method.
Solution Using equation (2)
Z
Z
x cos x dx = f (x) · g(x) − f 0 (x) · g(x) dx
Z
= x · sin x − 1 · sin x dx
Z
= x · sin x − sin x dx
= x · sin x − (− cos x) + C
= x sin x + cos x + C
R
Example Use integration by parts to find ln x dx. This is a function that has arisen in the
course as the integral of x1 , however we are now able to use this fact to help with this question.
Solution Notice there is only one function here so we have to create two functions by stating
5.5. Integration by Parts
91
ln x = 1 · ln x. The two functions are therefore f (x) = ln x and g(x) = 1. Can we find f 0 and
Yes!
Z
Z
ln x dx = ln x · 1 dx
Z
1
= ln x · x −
· x dx
x
Z
= x ln x − 1 dx
R
g?
= x ln x − x + C
R
Example Find ex cos x dx This is an example where a pattern is established and perseverance
leads to the solution.
R
Solution Let f (x) = cos x and g(x) = ex . Can we find f 0 and g? Yes!
Z
Z
x
x
e cos x dx = cos x · e − − sin x · ex dx
Z
= ex cos x + ex sin x dx
This is an integral that is very similar in appearance to the original question with the cos x transformed
into sin x. We continue by repeating the integration by parts process with the new integral
R x
e sin x dx
R
Z
ex sin x dx
z
x
x
Z}|
x
{
x
e cos x dx = e cos x + sin x · e − cos x · e dx
Z
= ex cos x + ex sin x − ex cos x dx
Notice the original integral has
R xnow appeared on the right hand side of the equation. We now
simply solve the equation for e cos x dx and add the constant of integration.
Z
2 ex cos x dx = ex cos x + ex sin x
Z
ex
ex cos x dx =
(cos x + sin x) + C
2
This can easily be verified by differentiating.
Definite Integrals
Integration by parts can be combined with the Evaluation Theorem to evaluate definite integrals.
If we assume f 0 and g 0 are continuous then we can use the Evaluation Theorem and write equation
(1) as follows
Z b
Z b
b
0
f (x) · g (x) dx = f (x) · g(x)]a −
g(x) · f 0 (x) dx
(5)
a
a
R1
Example Evaluate 0 xex dx. A graph of y = xex shows the area required and gives us an idea
of the answer to expect.
5.6. Applications of Integration
Solution
grated
92
Notice x becomes simpler when differentiated and ex is unchanged when it is inte1
Z
x
xe dx =
0
xex ]10
Z
1
−
1 · ex dx
0
= xex ]10 − ex ]10
= 1e1 − 0e0 − e1 − e0
= e1 − e1 + e0
= e0 = 1
5.6
Applications of Integration
Rectilinear Motion
We will use integration to analyse the motion of an object moving in a straight line. Let the
position function for the object be s = f (t) where t is the time. The velocity function is v(t) = s0 (t).
Therefore the position function is the integral of the velocity function. Also the acceleration function
is a(t) = v 0 (t) so the velocity function is the integral of the acceleration function. We can obtain
the position function from the acceleration function by antidifferentiating twice. This process will
generate two constants of integration so we need two additional pieces of information to find the
particular solution. Usually s(0) and v(0) are given.
Example A particle moves in a straight line with an acceleration of a(t) = 4t + 2. If the initial
velocity is −4cm/s and the initial displacement is 5cm, find the position function.
Solution
Because v 0 (t) = a(t) = 4t + 2 we can antidifferentiate a(t) to obtain v(t)
v(t) = 4
t2
+ 2t + C1 = 2t2 + 2t + C1
2
Substitute t = 0 because we are given the initial velocity i.e. v(0) = −4
v(0) = 2 · 02 + 2 · 0 + C1 = −4
SoC1 = −4and therefore
5.6. Applications of Integration
93
v(t) = 2t2 + 2t − 4
Because s0 (t) = v(t) we can antidifferentiate v(t) to obtain s(t).
t3
t2
+ 2 − 4t + C2
3
2
2 3
2
= t + t − 4t + C2
3
s(t) = 2
Substitute t = 0 because we are given the initial displacement i.e. s(0) = 5
2 3
· 0 + 02 − 4 · 0 + C2 = 5
3
SoC2 = 5and therefore
2
s(t) = t3 + t2 − 4t + 5
3
s(0) =
You should know about the gravitational force near the earth that produces a downwards acceleration given the symbol g. We assume g is constant and about 9.8m/s2
Example A ball is thrown vertically upwards with a speed of 24.5m/s from the edge of a cliff
that is 147m above the ground.
(a)
Find the position function.
(b) Find the time when the ball reaches its
maximum height.
(c)
Find the maximum height above the ground. (d) When does the ball hit the ground?
Solution (a) It is usual to choose the positive direction to be upwards this means that the
acceleration due to gravity must be −9.8m/s2 .
Soa(t) = v 0 (t) = −9.8therefore
v(t) = −9.8t + C1
Substituting v(0) = 24.5 we get that C1 = 24.5 so
v(t) = −9.8t + 24.5
Because s0 (t) = v(t) we antidifferentiate v(t)
s(t) = −9.8
t2
+ 24.5t + C2
2
Substitute s(0) = 147. We get C2 = 147.
Sos(t) = −4.9t2 + 24.5t + 147
This expression will hold true until the ball hits the ground provided you assume the ball continues
on its trajectory without striking the cliff.
5.6. Applications of Integration
94
(b) The maximum height is reached when v(t) = 0.
−9.8t + 24.5 = 0
24.5 = 9.8t
24.5
t=
9.8
= 2.5
So the maximum height is reached after 2.5s.
(c) The maximum height is found by substituting t = 2.5 in s(t)
s(2.5) = −4.9 · 2.52 + 24.5 · 2.5 + 147
= −30.625 + 61.25 + 147 = 177.625m
(d) The ball hits the ground when s(t) = 0
So − 4.9t2 + 24.5t + 147 = 0
t2 − 5t − 30 = 0
Use the quadratic formula t =
√
−b± b2 −4ac
2a
√
25 − 4 × 1 × −30
2
√
5 ± 145
=
2
≈ 8.5s or − 3.5s
t=
5±
You will be aware that the position function is a parabola so it is not surprising that the theoretical
result shows a solution about 3.5 s before the ball is thrown. The answer is that the ball hits the
ground about 8.5 s after it is thrown. You will also be aware that we did not test whether the
value of t when v(t) = 0 gives a maximum or minimum. This is also because we are aware of the
behaviour of a parabola with a negative t2 coefficient.
Work
The strategy we use to allow us to apply calculus to a problem in engineering is the same as we
used to evaluate areas. The physical quantity is divided up into a large number of small parts, each
one approximately equal to the theoretical quantity it represents. These are then summed and a
limit taken as the number of small parts, n → ∞. This process allows us to evaluate the resulting
integral.
You will recall that from Newton’s Second Law of Motion
F = ma = m
d2 s
dt2
(1)
Where s(t) is the position function, m is the mass of an object and F is the force required to
2
produce an acceleration of a. (Where a = ddt2s ).
5.6. Applications of Integration
95
We usually measure mass in kilograms (kg), distance in metres (m) and force in Newtons (N)
(N = kg · m/s2 ). If the acceleration is constant then the force to produce that acceleration will
also be constant.
Work = force × distance or
W = Fd
(2)
If F is in Newtons and d is in metres then W is in Newton-metres. One Newton-metre is called a
Joule (J).
Example A mass of 3.5kg lifted 0.5m requires a force of F = ma = 3.5 × 9.8 = 34.3N. This is
the force required to counter the force exerted by gravity. Calculate the work done.
Solution The work done is calculated using equation 2
W = F d = 34.3 × 0.5 = 17.15J
If the force is variable this formula can no longer be applied. Let the force acting on an object
as it moves along the x-axis in a positive direction from a to b be f (x), where f is a continuous
function of x. Divide the interval from a to b into n subintervals of width ∆x where ∆x = b−a
n .
For simplicity we let the end points of these subintervals be x0 , x1 , x2 , ..., xn We select the ith
subinterval and select a representative x-value in this interval x∗i . The work done when we move
the object from xi−1 to xi is
Wi ≈ f (x∗i )∆x
The total work done is
W =
n
X
f (x∗i )∆x
(3)
i=1
As we did with area we find the limit as n → ∞ of this sum. As this is a Riemann sum the limit
is a definite integral
Z b
n
X
W =
f (x∗i )∆x =
f (x) dx
(4)
a
i=1
Example If the force on a particle is given by the equation f (x) = 3x2 − 2xN, how much work
is done moving the particle from x = 2 to x = 3?
Solution
x-axis.
The graph of f (x) = 3x2 − 2x shows that for the interval [2, 3] the area is above the
Z
b
W =
f (x) dx
a
Z
=
3
3x2 − 2x dx
2
3
x2
x3
= 3 −2
3
2 2
3
2 3
= x −x 2
= 33 − 32 − 23 − 22
= 18 − 4 = 14 joules
Example
Hooke’s Law states that the force required to maintain a spring stretched x units
5.7. Chapter Exercises
96
beyond its natural length is proportional to x, the spring displacement, such that f (x) = kx where
k is the spring constant. This law holds provided x does not get too large. A force of 20N is
required to stretch a spring from its natural length of 5cm to a length of 7cm. How much work is
required to stretch the spring from a length of 12cm to 13cm?
Solution As the force is measured in Newtons we convert all lengths to metres. Our first step
is to use Hooke’s Law to evaluate the spring constant
f (x) = kx
So20 = k · 0.02
k = 1000
Sof (x) = 1000x
The work done in stretching the spring from 12cm to 13cm is
Z 0.08
W =
1000x dx
0.07
= 500x2
0.08
0.07
= 500 0.082 − 0.072
= 500 (0.0064 − 0.0049) = 500 × 0.0015
= 0.75J
5.7
Chapter Exercises
§5.1
Standard Integrals
1. Find the general integral in each case.
(a) 4x7
(d) 3.2ex
(g) x1.2
(j)
2
x− 3
(b) 10
1
(e) 2x
(h) xπ
(c)
(f )
(i)
(k) (2x + 1)2
(l)
8x3
√3
x
4
√
3 4
x
ex +
xe
2. Find the indefinite integrals.
1
x2
(a)
R
(d)
R
(1 − x) (2 − x) dx
(g)
(j)
R
R
cos x
2 dx
π r 2 − x2
x−
dx
R √
x x dx
R
(e)
sin2 x + cos2 x dx
R 1
(h)
sec2 x dx
R 21
(k)
2x dx
(b)
dx
(c)
(f )
(i)
(l)
R
(sin x − 2 cos x) dx
R 2
√3
+
dx
x
x
R
x + 1 + x1 dx
R
2πr dr
5.7. Chapter Exercises
§5.2
97
Area
1. Evaluate the integrals.
(a)
(d)
R2
5
−1 x dx
R3 x
1 e dx
(b)
(e)
R3
− 4x3 dx
1 1 + 2x
R3 1
1 x dx
(c)
R1
(c)
R3
0
x2/3 dx
2. Find the area under the curve.
(a) y = x3 between x = −1 and x = 3
(b) y = (x + 1) (x − 1) between x = −2 and x = 3
3. Find the integral and evaluate according to the limits.
(a)
(d)
R2
(x − 1) (x
R01 x
0 3e dx
§5.3
(b)
+ 5) dx
(e)
x3 −6
1 x2 dx
R2 4
−2 x dx
R2
(f )
R21
dx
3x
−1 x
3
dx
Volume
1. Determine the volumes of the solids of revolution which are generated by rotating the following
areas for one complete revolution about the x-axis.
(a)
(c)
y = 2x + 1 between x = 0 and x = 3
y = −3x between x = −2 and x = 0
(b) y = 6x2 + 1 between x = 1 and x = 4
√
(d) 5 x for 1 ≤ x ≤ 2
2. Determine the volume found when the function is rotated one revolution around the y-axis.
(a)
(c)
y = x − 1 between y = −1 and y = 3
√
y = x for 0 ≤ y ≤ 2
(b) y = x2 − 3 between y = −3 and y = 0
√
(d) y = x + 4 from y = 0 to y = 4
3. The volume for a cylinder with radius r and height h is given by V = πr2 h Use integration to
show that this formula is correct. Hint: Which curve y = f (x) is rotated around the x-axis to
form a cylindrical solid of rotation?
§5.4
Integration by Substitution
1. Use the given substitution to find the integral.
(a)
R
e4x dx,
(c)
R
esin θ cos θ dθ,
(e)
R
2
xex
(g)
R
2x x2 + 2
(i)
R
√x
1+x
u = 4x
dx,
dx,
u = sin θ
u=
3
dx
x2
u = x2 + 2
u=1+x
(b)
R
tan x dx,
(d)
R
(4x − 3)15 dx,
(f )
R
(h)
R
x
x2 +1
2
sin x cos x dx,
(j)
R
√dx ,
x+x
dx,
u = cos x
u=
u = 4x − 3
x2
√
u= x
+1
u = sin x
5.7. Chapter Exercises
98
2. Integrate the following. The substitution has not been given.
(a)
R√
(d)
R
(g)
R
2x + 1 dx
ln x
x
dx
cos 2x dx
(b)
R
(e)
R
(h)
R
√dx
x+1
cos x
sin x+1
ex +e−x
2
(c)
dx
(f )
dx
(i)
ex
ex +1 dx
R
cos x2 dx
R2
11
0 (x − 1)
R
dx
3. Use an appropriate substitution and then evaluate the integral.
(a)
(c)
(e)
§5.5
R2 √
x 2x2 + 1 dx
R01
4
0 (3 − 2x) dx
R 9 dx
(b)
(d)
(f )
4 2x+1
π
π
4
0 sin x + 2
R −1 dx
−2 (2x+1)2
R
R
π
4
0
dx
tan x dx
Integration by Parts
1. Find the integral using integration by parts.
(a)
(c)
(e)
R
x ln x dx
R
x cos 3x dx
R
ln (2x) dx
R
(b)
x sin x dx
R x
(d)
e sin x dx
2. By writing sin2 x as sin x · sin x and using integration by parts show that
Z
1
sin2 x dx = (x − sin x · cos x) + C
2
3. Integrate.
(a)
(c)
(e)
R −x
(b)
xe dx
R
(d)
(x − 1) ex dx
R 2x
xe dx
R 2
x sin x dx
R ln x
x dx
4. Integrate and evaluate.
(a)
(d)
(g)
(j)
§5.6
R5
−x dx
0 xe
R3
x ln x dx
R11 2x
xe dx
R04 ln x
2 x dx
(b)
(e)
(h)
(k)
R1
2 x
0 x e dx
Rπ 2
sin x dx
R07
x cos x dx
R53 2
0 x sin x dx
(c)
Rπ
(f )
R
(i)
0
ex sin x dx
π
2
0 x sin x dx
R2
x
1 (x − 1) e dx
Applications
1. A particle is moving in a straight line with an acceleration given by a(t) = 6t + 4. Its initial
velocity is v(0) = −6cm/s and its initial displacement is s(0) = 9cm. Find the position function
s(t).
5.7. Chapter Exercises
99
2. A ball is thrown upwards with an initial velocity of 15m/s from the edge of a cliff 140m above
the ground. Find its height above the ground t seconds later. When does it reach its maximum
height? When does it hit the ground?
Hits the ground after t ≈ 7.1s
3. A particle is moving along a straight line with an acceleration of a(t) = 3 + 4t − 12t2 . Its initial
velocity, v(0) is 4m/s and its initial displacement, s(0) is 5m. Find the position function, s(t).
4. A stone is dropped off a cliff and hits the ground with a speed of 40m/s. What is the height of
the cliff?
5. A car is travelling at 50km/h when the brakes are firmly applied producing a constant deceleration of 5 m/s2 . How far will the car travel before coming to rest?
6. A car is travelling at 100km/h when the driver sees a railway crossing 80m ahead and slams on
the brakes. What constant deceleration is required so that the car will stop in 80m?
7. Two balls are thrown upwards from the edge of the cliff in exercise 2 above. The first is thrown
with a speed of 15m/s and the second is thrown one second later with a speed of 8m/s. Do the
balls ever pass each other?
8. A particle moves along a straight line with a velocity function v(t) = sin t − cos t and its initial
displacement is s(0) = 0m. Find its position function.
9. A car braked with a constant deceleration 5m/s2 , producing skid marks measuring 60m before
coming to rest. How fast was the car travelling when the brakes were first applied?
10. A particle is moved along the x-axis by a force given by the equation f (x) = 6/ (1 + x)2 N at a
point x metres from the origin. Find the work done in moving the particle from the origin to
a distance of 5 metres.
11. When a particle is located a distance of x metres from the origin a force of cos (πx/3) Newtons
acts on it. How much work is done in moving the particle from x = 1 to x = 2? Interpret your
answer by considering the work done from x = 1 to x = 1.5 and from x = 1.5 to x = 2. By
drawing the graph of y = cos (πx/3) the reason should become apparent.
12. The graph shows a force function . (Force in
Newtons against distance in metres.) The
force increases at a constant rate until it
reaches its maximum value then remains
constant. How much work is done by the
force in moving an object a distance of 7m?
13. A spring has a natural length of 20cm. If a 25N force is required to keep it stretched to a length
of 30 cm how much work is done stretching it from a length of 20cm to 25cm?
14. Suppose that 2J of work is needed to stretch a spring from its natural length of 20cm to a
length of 30cm. How much work is needed to stretch it from a length of 22cm to 28cm?
5.7. Chapter Exercises
100
15. If 6J of work is needed to stretch a spring from 10cm to 12cm and 10J is needed to stretch it
from 12cm to 14cm, what is the natural length of the spring?
6 | Differential Equations
In the last chapter we said that any equation involving derivatives of a function is called a differential
equation. In this chapter we explore this idea further and look at some mathematical models that
take the form of differential equations. Change is a concept that we are familiar with. When we
are imagining what something we are familiar with today will be like in the future we can be pretty
certain that it will be different. Differential equations can arise when we formulate mathematical
models. We can develop our understanding of this process by considering the mathematical models
of some physical phenomena.
6.1
Basic Differential Equations
One model of population growth arises from the assumption that the rate at which the population
grows is proportional to the size of the population. Let N be the size of the population at time t
then the rate of change of population with respect to time is dN
dt . So the model can be expressed
as a differential equation
dN
= kN
dt
where k is the constant of proportionality.
To make any sense of this model we need to explore the range of values of N and k. N cannot be
zero (otherwise there would be nothing to change). We assume that N is a function of t and that
N (t) > 0. Similarly for us to have population “growth” k > 0
∴
dN (t)
>0
dt
We have already shown the properties of the exponential function. The general exponential function
is
N (t) = N0 ekt
Let N (t) be N0 when t = 0. i.e. N (0) = N0 . N0 is called the initial value of N (t).
Now
dN (t)
= N0 ekt × k
dt
= kN (t)
Thus we have shown that N (t) = N0 ekt is a solution of the differential equation
dN (t)
dt
= kN (t)
This solution arises because we are familiar with the behaviour of the exponential function. Our
ability to “guess” the answer is limited so the subject of differential equations involves developing
techniques to handle physical situations that are more and more realistic and more and more
complex. The answer to this problem came so simply to us we have to wonder if there are other
equations for N (t) that give the same answer.
6.1. Basic Differential Equations
102
The Order of a Differential Equation
The differential equation dNdt(t) = kN (t) is referred to as a first order differential equation because
the order of the highest derivative is one.
Here is an example of a second order differential equation
m
d2 x
= −kx
dt2
d2 x
dt2
is a second derivative. This is the differential equation that arises from Hooke’s Law. This
subject will be developed further in later courses.
The Solution of a Differential Equation
When you are asked to “solve” a differential equation you are expected to find all possible solutions
of the differential equation. In chapter 4 we met a range of simple differential equations and showed
how to express all possible solutions.
Example Find all possible solutions of the differential equation
tion could also be expressed as y 0 = 2x.
Solution
dy
dx
= 2x. This differential equa-
By antidifferentiating we obtain
y = x2 + C
Where C is the constant of integration. This is an arbitrary constant and gives us a family of
functions all of which are solutions of the differential equation y 0 = 2x. This family of solutions is
often referred to as the general solution.
In a physical situation we are often provided with additional information and this will allow us to
find a particular solution. This is easy to visualise when considering y 0 = 2x. The general solution is
y = x2 + C and if you are also told that the curve passes through (2, 6) you can use this information
to evaluate C.
6 = 22 + C
C=2
So the particular solution is y = x2 + 2. To appreciate the concept of this particular solution you
only need to use desmos to draw y = x2 + C for various values of C.
Initial-Value Problems
In a physical problem when you are given the conditions for the particular solution in the form
y(t0 ) = y0 where t0 is the initial value of t and y0 is the initial value of y(t), the point (t0 , y0 ) is
called an initial condition and the problem of finding the particular solution given the differential
equation is referred to as an initial-value problem.
Example
For the differential equation y 0 = −y 2
(a) Verify that y =
1
t+C
is the general solution
6.2. Separable Equations
103
(b) Find the solution of the initial-value problem y 0 = −y 2 and y (0) = 0.5
Solution
(a) Given y =
1
t+C
= (t + C)−1
y 0 = − (t + C)−2
−1
=
(t + C)2
2
1
=−
= −y 2
t+C
So y =
1
t+C
is the general solution of y 0 = −y 2
(b) Substitute (0, 0.5) in y =
1
t+C
1
0+C
1
1
=
2
C
C=2
0.5 =
The particular solution is y =
6.2
1
t+2
Separable Equations
In some special cases we can find explicit solutions of differential equations. One type of equation
can be written in the form
dy
f (x)
=
dx
g(y)
Expressed in this form we simply have to recognise that f (x) is a function without any y’s in it
and g(y) is a function without any x’s in it.
To solve
dy
dx
=
f (x)
g(y)
we rewrite the equation as
g(y)dy = f (x)dx
Then we integrate both sides
Z
Z
g(y)dy =
f (x)dx
This procedure can be verified by differentiating both sides with respect to x
Z
Z
d
d
g(y)dy =
f (x)dx
dx
dx
By the chain rule the left hand side becomes
Z
d
dy
g(y)dy ×
dy
dx
So
d
dy
Z
dy
d
g(y)dy ×
=
dx
dx
Z
f (x)dx
6.2. Separable Equations
104
g(y) ×
or
Example
Solve
dy
= f (x)
dx
dy
f (x)
=
dx
g(y)
dy
x2
= 2
dx
y
Solution
Z
y 2 dy = x2 dx
Z
2
y dy = x2 dx
y3
x3
=
+C
3
3
y 3 = x3 + 3C
= x 3 + C1
Where C1 is a new arbitrary constant
y=
p
3
x 3 + C1
Find C1 given y(0) = 2
2=
p
3
03 + C1
C1 = 8
p
3
y = x3 + 8
Example
Solve y 0 = 3x2 y
Solution
First write y 0 =
dy
dx
dy
= 3x2 y
dx
Separate the variables
1
dy = 3x2 dx
y
Integrate
Z
Z
1
dy = 3x2 dx
y
ln |y| = x3 + C
(1)
We usually write this by writing in exponential form
|y| = ex
3 +C
3
= ex eC
3
y = ±eC ex
And write A = ±eC where the value of A is used that satisfies the particular problem
y = Aex
3
In practice we usually jump from line (1) to line (2) and leave out the intermediate steps.
(2)
6.3. Chapter Exercises
6.3
§6.1
105
Chapter Exercises
Differential Equations
1. Solve the following differential equations and use the given conditions to find the particular
solution.
(a)
(c)
dy
t
dt = e + 2t
2
t dy
dt = 1 − t
2. Show that y = x +
1
x
y(0) = 2
(b)
y (0) = 2
(d)
dy
dx =
x2 dP
dx
sin 2x + cos 2x
=2+x
y(0) = 0
P (1) = 2
is a solution of the differential equation xy 0 + y = 2x
3. Verify that y = sin x cos x − cos x is a solution of the initial-value problem
y 0 + (tan x) y = cos2 x
on the interval
−π
2
y(0) = −1
< x < π2 .
4. Which of the following functions are solutions of the differential equation y 00 + y = sin x
(d) y = − 21 x cos x
(a) y = sin x
(b) y = cos x
(c) y = 12 x sin x
5. A mass falling to earth with a constant acceleration of 9.8m/s2 satisfies the differential equation
dv
= 9.8
dt
where v is the velocity at time t. Find an expression for v if
(a) The mass is dropped from a stationary position.
(b) The mass is fired towards earth with an initial velocity of 100m/s.
6. The angular velocity ω of a flywheel under constant braking torque of N is given by the
differential equation
dω
I
+ N = 0.
dt
where I is the moment of inertia (I is a constant)
(a) Find ω in terms of t given that ω = ω0 when t = 0.
(b) Calculate the time taken to bring the flywheel to rest from an initial speed of 60πrad/s
given that the moment of inertia is 100kg m2 under a braking torque of 40N m.
§6.2
Separable Equations
1. Solve the following differential equations
(a)
dy
y dx
=x
given y = 4 when x = 0
(b)
1 dv
t dt = 2
dy
ex dx
+2
given v = 2 when t = 1
(c)
=0
given y = 5 when x = 0
6.3. Chapter Exercises
106
dy
(d) (x + 2) dx
=y+3
(e)
(f )
di
dt + i = 1
dy
dx − 2x =
given y = 0 when x = 0
given i = 10 when t = 0
0
y(0) = 2
2. The rate at which the atoms of a radioactive substance split up is given by
N is the number of atoms present after t seconds and λ is a constant.
dN
dt
= −λN , where
(a) Show that N = N0 e−λt where N0 is the number of atoms present initially (i.e. when
t = 0).
(b) Find the time in years for half of the atoms of a given mass of radium to disintegrate if
λ = 1.37 × 10−11 for radium.
3. A body falling in a medium where the resistance is proportional to the velocity v at time t
obeys the differential equation dv
dt = 10 − 0.2v
(a) If the body falls from rest find v in terms of t and show that as t → ∞ the velocity
approaches 50m/s.
(b) How long would it take for the body to reach a velocity of 25m/s?
4. On a hot summer’s day a bottle of beer is placed in a fridge set at 4◦ C. The rate of change of the
bottle’s temperature θ◦ C after t minutes is given by Newton’s Law of Cooling, dθ
dt = −k (θ − 4)
where k is a constant that is specific to the beer.
(a) Find θ in terms of t. If the initial temperature (t = 0) of the bottle of beer was 22◦ C and
it took 10min to cool to 20◦ C, find the value of k, and the particular solution.
(b) How long would it take the bottle to reach the ideal drinking temperature of 8◦ C?
5. Brine containing 2 grams of salt per litre flows into a tank initially filled with 50 litres of water
containing 10 grams of salt. The brine enters the tank at 5 litres/min, the concentration is kept
constant by stirring and the mixture flows out through a tap at the same rate so that the tank
at all times continues to contain 50 L of mixture. Let there be Q grams of salt in the tank after
t minutes.
(a)
(b)
(c)
(d)
(e)
Q
Show that dQ
dt = 10 − 10
Solve this differential equation to obtain an expression for Q in terms of t.
How much salt is in the tank after 10 minutes?
Sketch the graph of Q against t.
What happens to Q as t → ∞?
6. A rectangular tank is divided into two equal compartments by a vertical porous membrane.
Liquid in one compartment, initially at a depth of 40cm, passes into the other compartment
which is initially empty, at a rate proportional to the difference in levels
(a) If the depth of the liquid in the second compartment is xcm after t minutes show that
dx
dt = k (40 − 2x) .
(b) Show that the solution of this differential equation is x = 20 1 − e−2kt .
6.3. Chapter Exercises
107
(c) If the level in the second compartment rises 2cm in the first 5 minutes, after how much
time will the difference in levels be 2cm?
(d) Interpret what is happening.
7. Under certain conditions the relative density ρ of a gas and its temperature T ◦ C satisfies the
dρ
differential equation dT
= −n2 ρ2 (where n is a constant). Express ρ in terms of T given that
ρ = 0.002 when T = 30◦ C and ρ = 0.0016 when T = 45◦ C.
7 | Answers
Answers to selected exercises.
CHAPTER 1
§1.1, p.11, Introductory Algebra
§1.2, p.12, Functions
1. (a)
t=
(b) t =
1. (a)
(b)
(c)
(d)
−x − y
−15x + 6y
x2 + 3x + 2
4x2 − 1
2. (a) 1
(b) 0.01
(c) 36.462
(d) 27
3. (a)
(b)
(c)
(d)
2. (a)
v−u
a
1
α
(c)
(d) 64
3. 79.4m3
4. (a)
y = 4x − 4
x
3 +1
2
x −3
(c)
6. (a)
(b)
(c)
(d)
(x + 7) (x + 4)
(2x + 3) (x − 4)
(b − 5)(b − 4)
(3x − 1)(x − 2)
7. (a)
(b)
(c)
(d)
(e)
(f )
3.16
−1.87 or 5.87
4 or −2
0.637 or −3.137
x = 2.547
x = {−2.055, 0, 1.055}
27
x6
27
a3
2b
x3
12
7(y 2 − 2z 2 )
x[(y − 2) + x]
(a + c)(−3a − 3c)
2πr(h + r)
−1
(b) 4
(b) 2y =
5. (a)
(b)
(c)
(d)
l
l0
4
9a4 b2
4. (a) 1.28
(b) 1.516
y=
(d) y = −(x + 1)2
5. x−intercepts 1.73 and −1.73 and y−intercept −3
6. (a) Show the points A (3, 5) and B (−2, −5) on
the graph.
(b) 2
(c)
y = 2x − 1
(d) y = 2x + 7
(e)
draw a line with a slope of
7. Table of Values
x
g(x) = x3 − x
−1.5
−1.875
−1
0
−0.5
0.375
0
0
0.5
−0.375
1
0
1.5
1.875
4
5
109
§1.5, p.14, Word Problems
8.
1. 12cm by 15cm
2. The number of children admitted was 1500
and the number of adults was 700.
(a)
3. Run 5mi/h, cycle 20mi/h
4. 2.5 pounds of Kenyan coffee and 0.5 pounds
of Sri Lankan coffee should be mixed.
CHAPTER 2
(b)
§2.1, p.31, Unit Circle
1. (a)
(b)
(c)
(d)
9.
§1.3, p.13, Polynomials
1. (a)
(b)
(c)
(d)
2
3
99
4
2. (a) No, there is no variable term
(b) No, a variable in the exponent is an
exponential equation
(c) Yes
(d) No, the powers must be integers
π
5 ≈ 0.628rad
−8π
3 ≈ −8.378rad
π
3 ≈ 1.047rad
−3π
4 ≈ −2.356rad
2. (a) 135°
(b) 150°
(c) −270
π ≈ −85.9°
(d) −15°
3. (a) 55π
9 ≈ 19.2
(b) 4
4. 4mi
5.
36
π
≈ 11.459m
6. 330π ≈ 1037mi
7. 1.6 million mi
§2.2, p.32, Right Angled Triangles
§1.4, p.13, Systems of Equations
(− 13 , 5 31 )
1. (a)
(b) (0.571, −0.571)
(c) infinite solutions
(d) (1.23, 3.87) , (−0.35, −4.21)
1. sin θ = 45 , cos θ = 35 , tan θ =
2. (a)
sin θ =
(b) sin θ =
√
3. (a) 12 2
40
9
40
41 , cos θ = 41 , tan θ = 9
√
√
2 13
3 13
2
13 , cos θ = 13 , tan θ = 3
√
2. (a)
(b)
(c)
(d)
(3, 1)
no solution; parallel lines
(18.29, −10.29)
(−25, 5), and (−25, −5)
3. (a)
(b)
(c)
(d)
(1, 2)
(−3, 4) and (3, 4)
(−2, −1) , (−2, 1) , (2, −1) , (2, 1)
(−1.5, 0)
(b) 132 3
(c) 16.51658
√
4. (a) 45◦ , 16, 16 2 ≈ 22.63
(b) 38◦ , 44.79, 56.85
5. 1026ft
6. (a) 2100mi
(b) No
4
3
110
(b) 3, 2π
3
7. 415ft, 152ft
8. 30.0
9. 66.1°
§2.3, p.33, Trigonometric Functions
(c)
1.
10, 4π
(a)
(d) 1, 6π
(b)
(e)
3, 23
(a)
1, 2π, π2
(c)
3.
(d)
(b) 2, 2π, π6
(e)
2.
(a)
1, π2
(c)
2, 3π, π4
111
(d) 3, 2, − 21
8. (a) 2
(b) 84.6°
9. 23.1mi
10. 2179mi
11. 3835ft
12. 3.85cm2
13. 14.3m
CHAPTER 3
(e)
1
π
2 , π, 6
§3.1, p.49, ex functions
1. (a)
(b)
(c)
(d)
54.59815
0.99317
22.19795
15.15426
2. Use desmos to verify
(f )
π
1, 2π
3 ,−3
3. (a)
(b)
(c)
(d)
f (x) = 3x
f (x) = 5x
x
f (x) = 41
x
f (x) = 21
4. Use desmos to verify
§3.2, p.51, Logarithmic Functions
§2.4, p.34, Applications
1. (a) 318.8
(b) 44°
2. ∠C = 114◦ , a ≈ 51, b ≈ 24
3. (a) ∠C = 62◦ , a ≈ 200b ≈ 242
(b) ∠A = 100◦ , a ≈ 89, c ≈ 71
4. 219ft
5. (a) 1018mi,
(b) 1017mi
6. 155m
7. (a) 28.9
(b) 28.89°
1. (a)
(b)
(c)
(d)
(e)
(f )
52 = 25
50 = 1
81/3 = 2
2−3 = 18
ex = 5
e5 = y
2. (a)
(b)
(c)
(d)
(e)
(f )
log5 125 = 3
log10 0.0001 = −4
log8 18 = −1
log2 18 = −3
ln 2 = x
ln y = 3
3. (a)
(b)
(c)
(d)
(e)
(f )
(g)
1
0
2
2
2
10
−3
112
(h) 12
(i) −1
(j) 37
(k) 8
√
(l)
5
(m) − 23
(n) 4
(o) −1
4. (a)
(b)
(c)
(d)
(e)
(f )
32
4
5
100
2
4
5. (a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
(i)
1 + log2 x
log2 x + log2 (x − 1)
10 log 6
log2 A + 2 log2 B
log3 x + 12 log3 y
1
2
3 log5 x + 1
1
2 (ln a + ln b)
ln x + 12 (ln y − ln z)
1
2
2
4 log x + y
6. (a)
(b)
(c)
(d)
(e)
(f )
3
2
1
3
ln 8
16
4 + log 3
7. (a) log3 160
(b) log2 AB/C 2
h
3 i
(c) ln 5x2 x2 + 5
§3.3, p.52, Logarithmic Equations
(a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
±1
0, 43
1
2 ln 3 ≈ 0.5493
e10 ≈ 22026
95
3
3 − e2 ≈ −4.3891
5
13
12
§3.4, p.52, Modelling
1. (a) 13kg
(b) 6.6kg
2. (a)
(b)
(c)
(d)
0 fts
50.6 fts , 69.2 fts
Use desmos to plot the graph
80 fts
3. (a) 200
(b) Use desmos to plot the graph
(c) 11, 200
4. 13 days
5. (a) 7337
(b) 1.73 years
CHAPTER 4
§4.1, p.69, Differentiation from First Principles
1. (a) 23
(b) -12
(c) 4
2. Use Desmos to plot the graph
and compare to yours
3. Use Desmos to check
4. Use Desmos to check
5. Use Desmos to check
1. (a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
(i)
2.
2.7726
0.3495
1.2040
0.0767
1.9349
−43.0677
6.2126
−2.9469
14.0055
6. (a)
f 0 (x) = −2x−3 =
1
(b) f 0 (x) = 32 x− 3 =
(c) y 0 = 5x4
(d) y 0 = −3x−4
(e) y 0 = −4x−5
1
(f )
y 0 = 34 x− 4
(g)
y 0 = − 12 x− 2
3
−2
x3
2
√
33x
113
§4.2, p.71, Standard Derivatives
1. (a)
(b)
(c)
(d)
(e)
2. (a)
(b)
10. (a)
(b)
(c)
(d)
(e)
(f )
(g)
2x4
0
−12x3
−8x−5 = −8
x5
−x−4 = −1
x4
1
2t
− 12
3 12
2t
(h) 6 (x − 1) sin2 (x − 1)2 cos (x − 1)2
(i) 4 tan 2x sec2 2x
1
√
2 t
√
3 t
2
√
3
5 z2
3
=
=
5 23
3z
=
(c)
2.2
(d) 6.4x
§4.3, p.72, Maximums, Minimums, and Tangents
1. y = 7x − 3 is a linear function and so has
no maximum or minimum value. Its derivative is constant.
3. (a) f 0 (x) = 6x2 − 6x + 4
(b) f 0 (x) = 2x + 1 − x12
(c) 8x7 + 60x4 − 16x3 + 30x2 − 6
4. (a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
5. (a)
f 0 (x) = −3x−4 + 43 x
3
(c)
(d)
f 0 (x)
= 2ex +
=
1 − 32
3x
2ex
=
+
x+x sin x
(d) cos cos
2x
(e) −ω sin (ωt + δ)
8. y − x − 1 = 0
dx
dθ
= 5m/rad
−1 34
e2
1
√
3
3 x2
7. (a) x2 cos x + 2x sin x
√
(b)
x cos x + 2√1 x sin x
5 sec2 5x
4. There is a local minimum at (−1, −0.368)
If you know how to use the product rule
for differentiation you can find out
without using desmos.
1 − 11
12
12 x
6. (a) Check with Desmos
(b) Check with Desmos
(c) Check with Desmos
9.
3. (a) (3, 4) is a local maximum
(b) (−1, 0.667) is a local maximum and
(1, −0.667) is a local minimum
(c) (−3.67, −14.82) min and (1, 36) max
Df (x) = 2
f 0 (x) = ex − 1
f 0 (x) = 81x2
g 0 (x) = 15x14
f 0 (x) = ex − exe−1
(b) f 0 (x) = 14 x− 4 +
(c)
2. (a) (2, 0) and (−2, 0)
(b) (2, 12)
(c) (0, 1) and (1, 0) and (2, 1)
ds
dt = 8t − 7
d(3x)
dx = 3
d(3u4 )
= 12u3
du
g 0 (x)
4 cos 4x
2 cos πx
−15 sin 3x
3 sec2 3x
3 sec2 (x + 2)
3 sin2 x cos x
6 sin 3x cos 3x
= − x34 +
1
√
+
4
4 x3
2
e2 xe −1
=
+
3
√
4 7
x
1
√
12
12
x11
4
5. Tangent line is 2y − 3x + 1 = 0
Normal line is 3y + 2x − 5 = 0
√
√
6. (0, 4) ,
3, −5 ,
− 3, −5
7. (ln 2, 2) ≈ (0.69, 2)
8. a = − 13 or −1
q
9. x = ± 23
10. See Desmos
11. See Desmos
§4.4, p.73, Product, Quotient, & Chain Rules
1. (a) f 0 (x) = xex + ex = ex (x + 1)
(b) g 0 (x) = x2 ex + 2xex = xex (x + 2)
2. (a) x3 ex 0 = x3 ex + 3x2 ex = x2 ex (x + 3)
0
(b) x−3 ex = x−3 ex − 3x−4 ex
x
= xe 4 (x − 3)
(c) ((x + 1) ex )0 = ex (x + 2)
(d) ((x + 2) (x − 2) ex )0 = ex x2 + 2x − 4
114
3. (a)
(b)
(c)
(d)
4. (a)
(b)
(c)
(d)
2
−5
(2x−1)2
xex
y 0 = (x+1)
2
df
2−2t2
dt = (1+t2 )2
−ACex
f 0 (x) = (B+Ce
x )2
y0 =
(g)
(h)
§4.5, p.74, Parametric Differentiation
1. Check with desmos
−3x(x−2)
(x−1)2
2
(x+1)2
√ 2−x
2 x(x+2)2
1 + x − 2x2 e−x
5. Tangent y = x
Normal y = −x
6. (a) 2y + x − 2 = 0
(b)
7. (a) Horizontal line y =
(b)
√
8. (f ◦ g) (x) = x3
√ 3
(g ◦ f ) (x) = ( x)
x2
9. (h ◦ j) (x) = e 2 = ex
2x
(j ◦ h) (x) = e2
10. (a)
F 0 (x) =
1
2
√ x
1+x2
11. (a) 9 (3x + 2)2
3
(b)
2/5
(d)
(e)
(f )
(g)
(h)
dy
2. (a) dx
= 1t
(b) Lett = 1
(c) y − x = 0
√
3. y + x = 2, horizontal when t = π2 or 3π
2 ,
Vertical when t = 0 or π
√
√
4. 2y + 4 3x = 6 + 3, horizontal when
5π 7π
t = π4 , 3π
4 , 4 , 4 vertical when t = 0, π
5. (a)
(b)
(c)
(5x+3)
−2
(2x+1)2
9
(4−x)4
√ 1
2x−5
−2
√
3
3 (5−x2 )2
√ −1
2 (x+2)3
3
6x2 e2x
t(3t2 +1)
2t+1
cos t+sin t
cos t−sin t
1
t
− ab tan θ
(d)
(e) −2 sin θ because sin 2θ = 2 sin θ cos θ
2 /2
4
dy
(b) dx
= −10x 1 + x2
2 0
2
(c)
ex
= 2xex
x 0
x
x
(d) ee = ee · ex = ee +x
(c)
ex (2x+1)2
√
2 x+1
x−3
(x+1)3
sin θ
1−cos θ
(f )
§4.6, p.75, Related Rates
1. 3x2 dx
dt
2.
dA
dt
= 48cm2 /s
3. (a)
(b)
dA
dt
dA
dt
=
dr
dt
·
dA
dr
= 60πm/s
1
4. − 20π
decreasing at a rate of
5.
6.
1
20π cm/min
√215 ≈ 21.4m/h
101
√
800 5
≈ 596.3km/h
3
§4.7, p.75, Optimisation
1. 25 and 25
2.
1
2
3. 1200m by 600m
12. (a) 2xe−2x (1 − x)
(b) −e−x (1 − 2x) (5 − 2x)
1−p
√
(c)
(p2 +1) p2 +1
(d) x2 + 3 5x2 − 16x + 3
(e)
(x − 3)2 (2x − 1)
(f )
(x−1)(5x+3)
√
2 x+1
4. 150m by 150m
5.
20
3
cm
6. $20,032.00 for 399 people
7. Area = r2
8. 20 by 20 by 10, Volume = 4000cm3
115
CHAPTER 5
§5.1, p.96, Standard Integrals
(g)
(h)
(i)
1. (a) 57π
(b) 36828
5 π
(c) 24π
(d) 75π
2
2x3/2
3 +C
2.2
x
2.2 + C
xπ+1
π+1 + C
−12
√
3x + C
√
(j) 3 3 x + C
(k) 34 x3 + 2x2 + x + C
(l)
2. (a)
ex +
xe+1
e+1
x2
2 +
√
2 x5
5
x−1 + C
+C
(b)
+C
(c) − cos x − 2 sin x + C
3
(d) 2x − 32 x2 + x3 + C
(e) x + C
√
(f ) 2 ln |x| + 6 x + C
(g) 12 sin x + C
(h) 12 tan x + C
(i)
x2
2
1
2
§5.3, p.97, Volume
1. (a) x8 + C
2
(b) 10x + C
(c) 23 x4 + C
(d) 3.2ex
(e) 21 ln |x| + C
(f )
(d) 3 (e − 1) ≈ 5.155
(e) 12 45
(f ) 14 − 14 = 0 Area =
+ x + ln |x| + C
3
(j) πr2 x + πx3 + C
(k) 12 ln |x| + C
(l) πr2 + C
§5.2, p.97, Area
1. (a) 21
2
(b) −70
(c) 35
(d) e3 − e
(e) ln 3
2. (a) 20 12
(b) 9 31
3. (a) 23
(b) −1 12
(c) 13 ln 32 ≈ 0.135
2. (a) 64
3 π
(b) 4.5π
(c) 32π
5
(d) ≈ 308
§5.4, p.97, Integration by Substitution
1. (a) 41 e4x + C
(b) ln |cos x| + C
(c)
esin θ + C
(d)
16
1
64 (4x − 3) + C
1 x2
2e + C
1
2
2 ln x + 1 + C
(e)
(f )
as x2 + 1
is always positive
4
(g) 41 x2 + 2 + C
(h) 13 sin3 x + C
√
√
2 (1+x)3
(i)
−
2
1+x+C
3
√
√
(j) 2 ln (1 + x) + C as 1 + x
is always positive
2. (a) 1 (2x + 1)3/2 + C
3√
(b) 2 x + 1 + C
(c) ln (ex + 1) + C as ex + 1
is always positive
(d) 12 (ln x)2 + C, x > 0 because
does not exist when x ≤ 0
(e)
(f )
(g)
ln |sin x + 1| + C
2 sin x2 + C
1
2 sin 2x + C
x
−x
(h) e −e
+C
2
1
(i) 0, Area = 12
×2=
3. (a)
(b)
(c)
4 13
√
2
2
121
5
= 24.2
1
6
ln x
x
116
5
(d) − 12
1
(e) 2 ln 19
9 ≈ 0.374
9.
(f )
ln
600 ≈ 24.5m/s
10. 5J
√
2
2
√
= −0.347, Area = 0.347
units2
11.
3
π
2−
√
3 ≈ 0.256
12. 275J
§5.5, p.98, Integration by Parts
13. 0.3125J
x2
2
x2
4
1. (a)
ln x − + C
(b) sin x − x cos x + C
(c) x3 sin 3x + 19 cos 3x + C
(d) 12 ex (sin x − cos x) + C
(e) x ln |2x| − x + C
14. 1.2J
15. 8cm
CHAPTER 6
§6.1, p.105, Differential Equations
2.
1
1 2x
2x
2 xe − 4 e + C
−e−x (x + 1) + C
3. (a)
(b)
(c) −x2 cos x + 2x sin x + 2 cos x + C
(d) ex (x − 2) + C
(e) 12 (ln x)2 + C, x > 0 because
does not exist when x ≤ 0
(j)
e ≈ 2.718
h
i
2
2
1
(ln
4)
−
(ln
2)
≈ 0.721
2
(k) 5.777
§5.6, p.98, Applications
1. s(t) = t3 + 2t2 − 6t + 9
2. s(t) = −4.9t2 + 15t + 140, hits ground after
t ≈ 7.1s
3. s(t) = 5 + 4t + 23 t2 + 23 t3 − t4
4. ≈ 81.6m
5. 19.3m
6. ≈ −4.8m/s2 or −62500km/h2
7. Pass after approximately 4.6s
8. s(t) = − cos t − sin t + 1
(c)
(d)
ln x
x
4. (a) −6e−5 + 1 ≈ 0.9596
(b) e − 2 ≈ 0.718
(c) 21 (eπ + 1) ≈ 12.07
(d) 92 ln 3 − 12 ln 1 − 2 ≈ 2.9
(e) π2 ≈ 1.57
(f ) 1
(g) 41 e2 − 1 ≈ 1.597
(h) 7 sin 7 + cos 7 − (5 sin 5 + cos 5)
≈ 9.864
(i)
1. (a)
(b)
y = et + t2 + 1
y = 12 (sin 2x − cos 2x + 1)
2
y = ln |t| − t2 + 1
P = − x2 + ln |x| + 4
2.
3.
4. y = − 12 x cos x
5. (a) v = 9.8t (b) v = 9.8t + 100
6. (a) ω = ω0 −
(b) 47.1s
N
I t
§6.2, p.105, Separable Equations
1. (a)
(b)
(c)
(d)
(e)
(f )
y 2 = x2 + 16
v = t2 + 1
y = 2e−x + 3
y = 32 x
i = −(9e)−t + 1
y = x2 + 2
2. 5 × 1010 years
3. (a) v = 50 − 50e−0.2t
(b) 3.47s
4. (a) θ = 4 + 18e−kt
k = 0.011778303
(b) 127.6989838 ≈ 128 min (3sf)
5. (b) Q = 100 − 90e−0.1t
(c) 66.9g
(e) Q → 100g
6. (c) 142.165794 ≈ 142min
−1
7. ρ = 8 13 t + 250
Source Exif Data:
File Type : PDF File Type Extension : pdf MIME Type : application/pdf PDF Version : 1.5 Linearized : No Page Count : 120 Page Mode : UseOutlines Author : Title : Subject : Creator : LaTeX with hyperref Producer : pdfTeX-1.40.19 Create Date : 2019:02:18 15:46:21+13:00 Modify Date : 2019:02:18 15:46:21+13:00 Trapped : False PTEX Fullbanner : This is pdfTeX, Version 3.14159265-2.6-1.40.19 (TeX Live 2018/W32TeX) kpathsea version 6.3.0EXIF Metadata provided by EXIF.tools