Calculus Online Textbook Instructor's Manual Chapter 6 Instructors 06
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6.1 An Overview CHAPTER 6 6.1 (page 234) EXPONENTIALS AND LOGARITHMS An Overview (page 234) In lo4 = 10,000, the exponent 4 is the logarithm of 10,000. The base is b = 10. The logarithm of lom times 10" is m n. The logarithm of 10m/lOn is m - n. The logarithm of 10, OWx is 4x.If y = bz then x = logby. Here x is any number, and y is always positive. + A base change gives b = alOgab and bx = ax logab. Then 85 is 2''. log8 y. When y = 2 it follows that log2 8 times log8 2 equals 1. In other words log2 y is log2 8 times + On ordinary paper the graph of y = m x b is a straight line. Its slope is m. On semilog paper the graph of y = AW is a straight line. Its slope is log b. On log-log paper the graph of y = Ax k is a straight line. Its slope is k. The slope of y = bx is dyldx = cbx, where c depends on b. The number c is the limit as h -4 0 of bh- 1 ha Since x = logby is the inverse, (dx/dy)(dy/dx) = 1.Knowing dy/dx = cbx yields dx/dy = l/cbX. Substituting bx for y, the slope of logb y is l/cy. With a change of letters, the slope of logbx is l/cx. 15; -5; -1.1. 3.2 1 5' 2 ' 51-10;80;1;4;-1 7nlogbx 9 ' 30 .10 1 13 lo5 1 7 A = 7,b = 2.5 1 5 0; ISF= lo710; 8.3 log104 19 A = 4, k = 1.5 23 y - 1= cx;y - 10 = c(x - 1) 25 ( . I - ~ - I)/(-h) = (loh - I)/(-h) 2;10g2 21 = c2b~; = - l / c 3 29 Logarithm 27 + &; (f) 3 2 (a) 5 (b) 25 (c) 1 (d) 2 (e) 10' 4 The graph of 2-' goes through (0, I),(1, (2, logll2 2 = -1 and logl,2 4 = -2. i), i).The mirror image is x = logt y (y is now horizontal): ! (b) 3 (c) (d) (e) & (f) 5 8 logba = (logbd) (logda) and (logbd) (logdc) = logbc. Multiply left sides, multiply right sides, cancel logbd. 1 0 Number of decimal digits M logarithm to base 10. For 2'"' this logarithm is 1000loglo 2 pi 1000(.3) = 300. 1 2 y = loglo x is a straight line on "inverse" semilog paper: y axis normal, x axis scaled logarithmically (so x = 1,10,100 are equally spaced). Any equation y = loge x C will have a straight line graph. increases with slope 1 4 y = 10'-~drops from 10 to 1 to .I with slope -1 on semilog paper; y = $fix fromy=~atx=0toy=5atx=2. 1 6 If 44O/second is the frequency of middle A, then the next A is 880/second. The 12 steps from A to A are approximately multiples of 2l1l2. So 7 steps multiplies by 27/12 pi 1.5 to give (1.5) (440) = 660. The seventh note from A is E. 1 8 log y = 2 log x is a straight line with slope 2; logy = $ log x has slope 20 g(f(y)) = y gives g ~ ( f ( ~ )=) g1 or ~ g ( f ( ~ )=) 1g or ey$ = 1 or 1 =1 dy cy ' 22 The slope of y = 10' is $ = clOx (later we find that c = In 10). At z = 0 and x = 1 the slope is c and 10c. So the tangent lines are y - 1 = c(x - 0) and y - 10 = 10c(x - 1). 6 (a) 7 + 2 i. 6.2 The Exponential ex (page 241) 2 4 h = 1 gives c = 9; h = .1 gives c = 2.6; h = .O1 gives c = 2.339; h = .001 gives c = 2.305; c = In 10 = 2.3026. bl/' = 1or b114 = (1 2 6 (The right base is b = e.) With h = we pick the base so that ,174 or b = (1 = Generally b = (1 h)'lh which approaches e as h -+ 0. + i)' g. 10'"- 28 c = l i ~ n ~7 , ~ = limh,o 6.2 + i) + 10"-1 = the limit is 100"- 1 ? limhdo 7 =2 The Exponential ex (page 241) The number e is approximately 2.78. It is the limit of (1 1)n . h = .01. An equivalent form is e = lim ( 1 5 + + h) to the power l/h. This gives l.O1loO when When the base is b = e, the constant c in Section 6.1 is 1. Therefore the derivative of y = ex is dyldz = ex. The derivative of x = log, y is dxldy = l / y . The slopes at x = 0 and y = 1 are both 1. The notation for log, y is In y, which is the natural logarithm of y. The constant e in the slope of bx is e = In b. The function bz can be rewritten as ex ln b. Its derivative is (In b)ex ln = (In b)bx. The derivative of e"(x) is e U ( X ) g . The derivative of esinz is esinx c o s x. The derivative of eCxbrings down a factor c. + + C. The integral of e u ( ~duldz ) is eU(X)+ C. In The integral of ex is ex C. The integral of eCx is )ecx general the integral of eU(")by itself is impossible to find. 5 3x1n3 7 (f)xlnf g -(l+eS)% e f 112 lSzex 1 5 i&q5 19 .1246, .0135, .0014 are close t o & 21 e, e 2 5 ( l + ~ ) x < e < e Z < e 3 z / 2 < e 2< x 10%< zx 23Y(h)=1+&;Y(1)=(1+&)10=2.59 3s 7s z2 e-s' 31 &r 27!$-+% 29z+&+& 2ex 33 2 - 2 149e7x 3 8egX 1 7 esin x cos x ex sin ex ' + + 3 5 2exl2 + J ! s < e-9/e-3 < &j for z > 3 3 7 e-z drops faster at z = 0 (slope -1); meet at z = 1; e-"/e-~ 3 9 y - ea = ea(x - a); need -ea = -aea or a = 1 4 1 y' = x x ( l n x + 1) = 0 at xmin = $;y" = x z [ ( l n x + 1)2 $1 > 0 4 3 $(e-" y) = e-"*d z - eqXy = 0 so e-%y = Constant or y = Cex + 4 5 S 2L2 ] ; = & 55 59 $q 2 d z = -e-U 4 7 z 1 ln2]-1 + C; = (n - x)zn-'/ex =1L:=_k Ln2 2in2 49-e-x]r=1 + J(eu)2$dz = ?eZU C 5 1 e 1 + "lo =e2-e 5 7 yy' = 1gives < 0 for z > n; F(2x) < --+ 0 iy2 =z 61 A12rr !-F* 117; - 5 3 2~inx]; = 0 + C or y = Jm (3)6 116; 7 digits 2 49e-7x 4 8e8' 6 (ln 3)exl n 3 = (In 3)3% 8 4(1n 4)4" 10 A ( 1e l+l (~l )+~x ) 1 2 (-$+l)el/x 14z2eZ 1 8 x-lIx = e-(lnx)/x has derivative + kx2) e - ( l n z ) / x= ( v ) x - l / x 16 x2 + x2 has derivative 43,+ e 2 M 7.7 and (1 $)fi-+ 1.Note that (1 $)fiis squeezed between 1 and e l / f i which 20 (1 + approaches 1. 22 ( 1 . 0 0 1 ) ~ =~2.717 ~ ~ and ( 1 . 0 0 0 1 ) ' ~= ~ 2.7181 ~~ have 3 and 4 correct decimals. ( 1 . 0 0 0 0 1 ) ~ =~2.71827 ~~~~ has one more correct decimal. The difference between (1 !)" and e is proportional to (-3 + + + i. 6.3 Growth and Decay in Science and Economics (page 250) + 24 y = e-X solves = -y. The difference equation Y (x f ) = Y (x) - f Y (x) with Y (0) = 1gives Y ( f ) = 3 4 and Y(l) = (z) (Compare e-' = .37 with ( Q ) 4 = .32. See the end of Section 6.6.) . 26 1 \/E; is the same as s l 2 . Its graph at x = -2,0,2 has the same heights f , 1,e as the graph of ex at x = -1,o, 1. 28 (e3')(e7") = elox which is the derivative of & e l k 30 2-x = ,-x 1n2 which has antiderivative d e - ln~ - - in2 34 -ecos x + eein x x - ~has antiderivative -e-= 36 xex - ex 32 e-" 38 ex meets xx at x = e. Their slopes are ex and xx(l In x) by Example 6. At x = e those slopes are ee and 2ee. The ratio $ = (f )% approaches infinity. 40 At x = 0 equality holds: e0 = 1 0 and e-O = 1- 0. (a) Beyond x = 0 the slope of ex exceeds the slope of 1 x (this means ex > 1).So ex increases faster than 1 x. (b) Beyond x = 0 the slope of e-l is larger than the slope of 1- x (this means -e-' > -1). Since they start together, e-' is larger than 1- x. =xlp(l-lnx This slope is zero at x = e, when l n x = 1. 42 xl/' = e(lnx)/xhas slope e(lnX)/'($ - + + + + + + 9) . 7) The second derivative is negative so the maximum of xl/" is elle. Check: &,(In x ) / x ( y )= e ( l n ~ ) / ~ [ ( l - l t+ ~ )L 2 . L ] = at x = e. x 44 xe = ex at x = e. This is the only point where xee-" = 1 because the derivative is xe(-e-") + exe-le-' ( i- l)xee-%. This derivative is positive for x < e and negative for x > e. So the function xee-% increases to 1 at x = e and then decreases: it never equals 1again. = -e-l e. 46 1: sin x eC08xdx= [-eCoSx]~ 48 2-'dx = (by Problem 30) [&2-']L1 = &(+ - 2) = 2 .- 3 50 zl 1; = + - 00 ,-u& ze-"'d~ = J0 -u - [-%IF = z11 - e')lOe~dx = [- 1:=11 52 1: el+xlx dx = [+el+x2];= z(" - e) 54 $', 56 y' (x) = 5 y(x) is solved by y = ~ e "(A is any constant). Choose A = 2 so that ~ ( x = )2e5~ h as ~ ( 0 = ) 2. 58 The asymptotes of (1 !)x = (?)% = are x = -1 (from the last formula) and y = e (from the first formula). 6 0 The maximum of x6e-" occurs when its derivative (6%' - x6)e-" is zero. Then x = 6 (note that x = 0 is a minimum). 6xS 30s' 720s = limT = lim?120x3 = lim7360x2 = limT = l i m =~0. 6 2 lim$ = limF + 6.3 (A)-' Growth and Decay in Science and Economics (page 2 5 0 ) ln2 If y' = cy then y(t) = yoect. If dyldt = 7y and yo = 4 then y(t) = 4e7t. This solution reaches 8 at t = 7. l n 2 If y' = 3y and y(1) = 9 then yo was ge-'. If the doubling time is T then c = T. solution approaches zero as t When c is negative, the -+ oo. The constant solution to dyldt = y + 6 is y = -6. The general solution is y = Aet - 6. If yo = 4 then A = 10. The output from the source The solution of dyldt = cy s starting from yo is y = Aect B = (yo f)ect is f (ect - 1).An input at time T grows by the factor ec(t-T) at time t. + + + e. At c = lo%, the interest in time dt is dy = .O1 y dt. This equation yields y(t) = yOe-O1t. With a source term instead of yo, a continuous deposit of s = 4000/year yields y = 40,00O(e - 1)after ten years. The deposit 6.3 Growth and Decay in Science and Economics (page 250) required to produce 10,000 in 10 years is s = yc/(ect - 1)= 1000/(e - 1). An income of 4000/year forever (!) comes from yo = 40,000. The deposit to give 4000/year for 20 years is yo = 40,000(1- e-2). The payment rate s to clear a loan of 10,000 in 10 years is 1000e/(e - 1) per year. The solution to y' = -3y 1 ( + s approaches y, 16 C = g ; t = ah(!) = s/S. 17t= p(10) = 1013 eloc = 1 0 1 3 4 z = ,/- 2 1 p = 1013 ech; 50 = 1013 eaoc; c = $ In(&); 2 3 = Ina.,(f)3 = ) 2 5 y = yo 29A=& 38 B = - ; slet-1 $9 ye-'; y(t) = tet 5 7 4;4- $;4 19 e c = 3 s o y 0 = e - 3 ~ l o o o = - at reaches yl v; at t = then y = 27 F; F; T; T SSl-e-' S56;6+Ae-2t;6-6e-2t,6+4e-2t;6 4 1 A = 1,B = -1, C = -1 4 s e.07a6 > -075 45 s(e- 1); 47 (1.02)(1.03) + 5.06%; 5% by Problem 27 4 9 20,000 e(20-T)(-5)= 34,400 (it grows for 20 - T ears) 5 1 s = -cyoe"/(ed - 1) = - ( . 0 1 ) ( 1 0 0 0 ) e ~ ~ ~ / (-e -1) ~ ~ 65 yo = -006 - e--005(48)1 Sbe4c=1.20soc=~ S724eS6s5=? WTo-oo;constant;to+oo = 60(-Y 5); still Y , = 5 61 = 6OcY; + 22=-tgivesdy=-tdtandy=-ft2+~.Theny=-2t 1 2+landy=-it2-lstartfromland-1. 4 = -y gives Y = -dt and In y = -t C and y = Ae-' (where A = eC). (Question: How does a negative y appear, since eC is positive? Answer: = In lyl leads to lyl = Ae-' and allows y < 0.)To start from 2 + $9 1 and -1 choose y = e-t and y = -e-t. 6 = 4t gives dy = 4t dt and y = 2t2 C = 2t2 2 - + + 10. This equals 100 when 2t2 = 90 or t = m. 8 $ = e - 4 ' h a s y ( t ) = k + ~ = 7+ loa1 This only increases from 10 to 10; as t -4 t In 5 7 find y(t) = I when 4 =9! ore-" = 37or t = .T. 4t + oo. + Before t = 0 we 10 The solutions of y' = y - 1 (which is also (y - 1)'= y - 1) are y - 1= Aex or y = Aex 1. Figure 6.7b is raised by 1 unit. (The solution that was y = ex is lifted to y = ex 1. The solution that was y = 0 is lifted to y = 1.) 1 2 To multiply again by 10 takes ten more hours, a total of 20 hours. If eloc = 10 (and e2OC= 100) then 10c = In 10 and c = In 10 ar .2S. 1 4 Following Example 2, the ratio e" would be 90% or .9. Then t = = ( 3 ) 5 5 6 8 = (In 1.8)5568 = 3273 1 years. So the material is dated earlier than the year 0. 16 ge-O1t = 6e.014t gives8 = e-O04t and t = In = 2501n = 72 years. 6 1 8 At t = 3 days, eSc = 40% = .4 and c = = -.S. At T days, 20% remember: e-mST = 20% = .2 at T = = 5.56 days. (Check after 6 days: (.4)' = 16% will remember.) 20 If y is divided by 10 in 4 time units, it will be divided by 10 again in 4 more units. Thus y = 1 at t = 12. Returning to t = 0 multiplies by 10 so y o = 1000. 22 Exponential decay is y = Aect. Then y(0) = A and y(2t) = AeDd. The square root of y(O)y(2t) = A2eact is y(t) = Aect. One way to find y(3t) = AeSCtis y(0)(9H)3/2. (A better question is to find y(4t) = + 8 AedCt= y ( ~ )4 0 ) = w,.) 6.3 Growth and Decay in Science and Economics (page 250) In 1 24 Go from 4 mg back down to 1mg in T hours. Then e--OIT = f and -.01T = In f and T = =-& = 139 hours (not so realistic). - (c+C) $+ccy = 0. Check the solution y = ~ e " +~ e 26 The second-order equation is (& -c)(-$ -C)y = by substituting into the equation: c2Aect C2BeCt - (c C)(cAect CBeCt) cC(Aect ~ e ~does ' ) equal sero. 28 Given mu = mu - vAm mAv - (Am)Av Am(v - 7); cancel terms to leave mAv - (Am)Av = 7Am; = 7. Then v = 7 in m C.At t = 0 this is 20 = 7 in 4 + C divide by Am and approach the limit m so that v = 7 l n m + 2 0 - 7 l n 4 = 7 l n ~ + 2 0 . 30 Substitute y = Ae-t B into y' = 8 - y to find -Ae-t = 8 - Ae-t - B. Then B = 8. At the start yo = A B = A 8 so A = yo - 8. Then y = (yo - 8)e-t 8 or y = yoe-t 8(1- e-'). 52 Apply formula (8) to $ = y - 1with yo = 0. Then y(t) = (et - 1) = 1- et. 34 Formula (8) applied to = -y - 1with yo = 0 gives y = *(e-' - 1)= e-t - 1. 56 (a) =3y+6givesy-+oo (b) $=-3y+6gives y - + 2 (c) =-3y-6gives y-+-2 -00. (d) $ = 3y - 6 gives y 38 Solve y' = y et by adding inputs at all times T times growth factors et-T : y(t) = e t - = e T d ~= etdT = tet. Substitute in the equation to check: (tet)' = tet et. 40 Solve y' y = 1by multiplying to give ety' ety = et. The left side is the derivative of yet (by the product rule). Integrate both sides: yet - yoeO= et - e0 or yet = yo et - 1 or y = yoe1- e-t. 42 $1000 changes by ($1000) (-.04dt), a decrease of 40dt dollars in time dt. The printing rate should be s = 40. 44 First answer: With continuous interest at c = .09 the multiplier after a year is emo9 = 1.094 and the effective rate is 9.4%. Second answer: The continuous rate c that gives an effective annual rate of 9% is eC= 1.09 or c = In 1.09 = .086 or 8.6%. 4 6 yo grows to yoe(.l)(ZO)= 50,000 so the grandparent gives yo = 50,000e-~rr $6767. A continuous deposit s grows to +(e(-l)(zO- 1) = 50,000 so the parent deposits s = & j 1 50,000 = $785 p e r year. Saving s = $1000/yr grows to ?(eelt - 1)= 50,000 when eelt = 1 - or .It = In6 or t = 17.9 years. 4 8 The deposit of 4dT grows with factor c from time T to time t, and reaches e ~ ( ~ - ~ ) With 4 d ~ t. = 2 add 4e~('-T) 1 4 e c 4 e 2 ~ = = deposits from T = 0 to T = 1 : 9 + + + + + + + & + + + + + 2 % + % -+ 5,' + + 5: + + + + + 5; ,e -C + - 5 0 y(t) = (5000 - S ) e s o 8 ' is sero when emo8'= = 5. Then .08t = In 5 and t = rr 20 years. (Remember the deposit grows until it is withdrawn.) 52 After 365 days the value is y = e(.01)965= e3-65= $38. 54 (a) Income = expense when IOeLICt = EOeCtor eCt= or t = (b) Integrate Eoect - Ioezct v. 9 until eC' = 2.At the upper limit the integral is $eCt % - hezct zc =L c (I 0g - In!$) 2 I, + = x.Lower limit is 2cIo t = 0 so subtract - $ : Borrow 2cIo - ib c 2c' 56 After 10 years (halfway through the mortgage) the variable rate -09 .001(10) equals the fixed rate 10% = .l. Since the variable was lower early, and therefore longer, the variable r a t e is preferred. 7 then $ is zero at y, = 7 (this is = The derivative of y - y, is $, 58 1f $ = -y so the derivative of y - 7 is -(y - 7). The decay rate is c = -1, and y - 7 = e-t(yo - 7). 6 0 All solutions to = c(y - 12) converge to y = 1 2 provided c is negative. = 62 (a) False because (yl y2)' = cyl s cy2 s. We have 2s not s. (b) T r u e because 1 5cyl i s icy2 i s . (c) False because the derivative of y' = cy s is (y')' = c ( ~ ' ) and s is gone. 6 4 The solution is y = AeCt B. Substitute t = 0,1,2 and move B to the left side: 100 - B = A, 90 - B = AeC,84 - B = Ae2'. Then (100 - B)(84 - B) = (90 - B)(90 - B); both sides are Aze2c. Solve for B : 8400 - 184B BZ= 8100 - 180B Bz or 300 = 4B. The steady state is B = 75. (This problem is a good challenge and was meant to have a star.) -: I). + + + + + + + + + + + + + (iyl + iyZ)' ~ ' 6.4 Logarithms (page 258) + 66 (a) The white coffee cools to y, (yo - y,)e" The milk warms to 2 0 - 10ect. The mixture So it doesn't matter when you add the milk! 6.4 Logarithms = 20 + 40ect. (b) The black coffee cools t o 2 0 + 50ect. has 20 y e " = 20 40ect. + cOffee)+l(milk' 6 + (page 258) $). This definition leads to ln xy = In x + In y and In xn = The natural logarithm of x is $ ' (or n In x. Then e is the number whose logarithm (area under l / x curve) is 1. Similarly ez is now defined as the number whose natural logarithm is x. As x + oo, ln x approaches inRnity. But the ratio (In x)/& approaches Eero. The domain and range of In x are 0 < x < oo, -00 < In x < oo. 1 The tangent approximation to ln(1 + x) at + x) is -=. 1 2. z = 0 is x. The quadratic approximation is x - %x2.T he quadratic approximation to ez is 1+ x + zx The derivative of In x is f1. The derivative of ln(1 The derivative of ln u(x) by the chain rule is -&g.Thus (In cos x)' = - = -tan x. An antiderivative + of tan x is -In cos x. The product p = xeSx has lnp = 6 x In x. The derivative of this equation is p'/p = 6 Multiplying by p gives p' = xeSx(5 = 6xeSx eEx,which is L D or logarithmic differentiation. + 4) + 1 + z. + 4) is 1n(x2 + 4). The integral of l/ex is 9. The integral of 1/cos x, after a trick, is In ( ~ e x c + t a n x ) . We should The integral of u1(x)/u(z) is In u(x). The integral of 2x/(x2 w. 112, + The integral of l/(ct s) is write In 1x1 for the antiderivative of since this allows x < 0. Similarly $ du/u should be written lnlul. 1: s* 61nx 7 - = cot x 9 11$ l n t + c lsln$ 15fln5 17-ln(ln2) 191n(sinx)+C 21-$ln(cos3x)+C 2S$(ln~)~+C 2 9 $ = esin 2 cos x 2 7 In y = f ln(x2 1); = ,/& i 31 $ = ezee' 3 ~ l n ~ = e ~ l n x ; ~ = ~ e ~ ( l n3 x6 1 + n~ y) = - l s 0 y e=9 dz ~ ~ = O 57 0 + 2 39 -f 4 1 sec x 4 7 .l; .095; .095310179 4 9 -.01; -.01005; -.010050335 51 l'H6pital: 1 55 663-21n2 67Rectangulararea?+.--+i<$;y=lnn 69Maximumate 610 6 ~ l o g ~ ~ e o r & 651-x;l+zln2 never equals 1 67bactionisy=1whenln(T+2)-ln2=1orT=2e-2 6gyt=+ --ry=1-L (t+Z) = e'1n2,p' = ln2 ezln2 71 lnp = x l n 2 ; L D P l n 2 ; ~ D decreases after x = e, and the only integers before e are 1 and 2. 75 2' = l a ; y l n x = xln y + = ,+, y; = 2 4 2a*+f 8y=huso%=*=*= u 12In(l+x)from$+. a x2 ln, . I - - 1 (loglo x) 1 - loge10 z - x I n 10' 1Oy=7ln4x=7ln4+7lnxso~7 =~. 6 Use (log, 10)(loglo x) = log, x. Then . d 1 4 f l n ( 3 + 2 t ) ] ~ = f ( l n 5 - l n 3 ) =13 ~ 5 3* Its integral is [;x2 - f ln(x2 + I)]: 16 y = s +1 equals x - ,*. 18$$=-t=[-&l;=-l+=. 1 =2- i ln 6. 6.4 Logarithms + z)];14 = - ln 1+ 0 = 21n 1 2. $ -dz = J = - Inu = 3%dz 22 J a - )In(sin JX) + C. 20 (page 258) fi COS & !. 24 Set u = In in z. By the chain rule $ = Our integral is / 26 The graph starts at -oo when z = 0. It reaches gero when z = -28 ln y = ln(z2 1) ln(z2 - 1).Then y & = ' x'+1 x'-1 = 30 in = - I lnz and d B = W so =( y ) x - l / x . + +i ? + 2 x2 ydx 2 &. Then 2 2 . + + .K it stops. - - zxS $= = 2%. = exe-l. 32 l n y = e lnz and y dx x 34 l n y = $ l n z + $ i n % + i l n z = l n z and eventually = 1. - In r 3 6 1 n y = - l n zy dz s o ~ x& = ~ dz a n d * = - ~Alternatively we have y = $ and = so % = In u = In x)) + C. 5 and goes down again. At z = - @=i- 2 = --x12 ' 38 [Inz]ir [ln 1~111: = (a - 0) (0 - ln I - 21) = 7r - In 2. 1 Alternatively use $&(z2) - $-&(z) = $. 40 l n z = 51. with u = sec z tan z so the integral is in (sec x tan x). See Problem 41! 42 This is $ 44 &(ln(z-a) - l n ( z + a ) ) = 1=- i x a (x+a) - A x2-al0 -& + + F +J x~-- \ / ~ 1 =+$+ US - d-~ ~ + d a P~Gt + ';a'iT;;i' 46 Misprint! 1 + + + + + + 48 Linear: eS1 m 1 .1 = 1.1.Quadratic: e.' rr 1 . l + $ ( . I ) = ~ 1.105. Calculator: em1 = 1.105170918. 50 Linear: e2 rr 1 2 = 3. Quadratic: e2 rr 1 2 i(2') = 5. Calculator: e2 = 7.389. e* 52 Use l'H6pital's Rule: lim - = 1. 2-0 1 bZlnb 54 Use l'H6pital's Rule: lim -- In b. We have redone the derivative of bx at z = 0. -- - 2-0 - - 1 +% + +5 + +5+ 56 Upper rectangles f rr .7595. Lower rectangles: ) f m .6345. Exact area in 2 rr .693. 58 f is smaller than when 1 < t < z. Therefore $ : <$ : or in z < 2& - 2. (In Problem 59 this fi leads to -+ 0. Another approach is from 5 0 in Problem 6.2.59. If z is much smaller than ex then In z is much smaller than 2.) 60 mom 4 0 we know 4 0. This is 4 0. Since n is fixed we have -+ 0. 62 1 l n $ = -- -+0 as z -+ co. This means y In y -+ 0 as y = $ -+ 0. (Emphasize: The factor y -t 0 x is "stronger" than the factor In y -oo.) 64 Fkom / : th-'dt = we find $ ; t-'dt = limhdo The left side is recognized as In x. (The right side is the "mysterious constant c" when the base is b = z. We discovered earlier that c = In b.) 66 .01- +(.01)~$(.01)~ = .00995033.. . Also ln 1.02 N .02 - +(.02)~ $(.02)~= .01980266. . 68 To emphasize: If the ant didn't crawl, the fraction y would be constant (the ant would move as the band stretches). By crawling v dt the fraction y increases by bani SO = = &. Then 3 % 5 -+ '22 *. -+ + + + + cngngth. 2 + + y = f ln(8t 2) C = f (ln(8t 2) - ln2). This equals 1when 8 = In or 4t 1= e8 or t = i ( e 8 - 1) 70 LD: lnp = z l n z so f $ = 1+ l n z and $ = p ( l + l n z ) = z x ( l + l n z ) . Now find the same answer by ED: d ( e x l n x ) = e x l n xdAx ( ~ l n z=) z x ( l lnz). d~ 72 To compute + $: $ = In2 with error rr the trapezoidal rule needs A z rr Siz Simpson steps: 36r1+++++h+++h+&++++ s,=Lr 3) .693149 compared to in 2 = .693147. Predicted error &(i)'(6 - = 1.6 x actual error 1.5 x 74 I, f - -0877 says that about 877 of the next 10,000 numbers are prime: close to the actual count 879. ,,,,, 76 = -t*.ln(-) ( t ) This equals 7= (t+l)ln(+u) l c'fl)": =+*.Th x = 1, for t near zero. It approaches z = e, y = e as t -+ n e curve = yz is asymptotic to oo. It is symmetric across the 45" 6.5 Separable Equations Including the Logistic &uation (page 266) line (no change by reversing z and y), roughly like the hyperbola (x - - 1) = (e - I ) ~ . Separable Equations Including the Logistic Equation 6.5 (page 266) + The equations dyldt = cy and dyldt = cy s and dyldt = u(y)u(t) are called separable because we can separate y from t. Integration of dy/y = c dt gives In y = c t constant. Integration of / d y / ( ~+ s/c) = $ c dt gives ln(y f) = c t C. The equation dy/dz = -z/y leads to / y dy = - dx. Then y2 z2 = constant and the solution stays on a circle. + + I + + Ix The logistic equation is dy/dt = c y - by2. The new term -b# represents competition when cy represents growth. Separation gives J dy/(cy - b#) = dt, and the y-integral is l/c times In . Substituting yo at I & t = O and taking exponentials produces y/(c - by) = edyo/(c - b y o ) As t + oo, y approaches $. That is the steady state where cy - by2 = 0. The graph of y looks like an 5,because it has an 16. c inflection point at 2 In biology and chemistry, concentrations y and z react at a rate proportional to y times z. This is the Law of M a s s Action. In a model equation dyfdt = c(y)y, the rate c depends on y. The MM equation is K). Separating variables yields y d y = / -c dt = -ct + C. dyldt = -cy/(y + I + 17et - 5 3 (qx2 l)lI3 6z 7 el-coe t ~ ( $ f + f1 1iy ) , =~ O ; t = Lbuo 1 5 ~ = l + e - ~ , y i s i n l S 17ct=ln3,ct=ln9 19 b = 10-',c = 1 3 - I O - ~ ; ~ ,= 13-10'; at y = (10)gives l n b = ct+ln* C- l06b so t = 1900+ 2 1 d dips down and up (a valley) 2 3 sc = 1= sbr so s = ;,1 r = b N +0 27 Dividing cy by y K > 1 slows down y1 26 y = l + , - ~ t ( N - l ) i'2 = 6 = 2091 + *; 2 ~ $ = ~ & > 0 , x -u- +r c~ 31 = multiply eJKB = e-"lKeuOIK(g) by K and take the Kth power to reach (19) S~y'=(3-~)~;&=t+$;~=2att=% 36Aet+D=Aet+B+Dt+t-,D=-l,B=-l;yo=A+BgivesA=l 3 7 y-, 1from yo > 0, y + --from yo < O;y+ 1 from yo > O,y+ -1 from yo < 0 $9 $'f:udy =$dt+ln(siny) = t + ~ = t + l n i .Thensiny= i e t stops at 1 when t = l n 2 2ydy=dtgives t y 2 = t + C . T h e n C = f a t t = 0 . ~ 0 ~ ~ = 2 t + l a n d ~ = ~ ~ . 4 ~=dzgivestan-ly=z+~.~hen~=~atx=~.Soy=tanx. 6 = cos zdz gives ln(sin y) = sin z C.Then C = ln(sin 1) at x = 0. After taking exponentials & + sin y = (sin l)esiP X. No solution after sin y reaches 1 (at the point where (sin I)elin = 1). 8 eudy = etdt so eu = et C. Then C = ee - 1 at t = 0. After taking logarithms y = ln(et + ee - 1). 10 dM = = n. Therefore In y = nln x C. Therefore y = (xn)(eC) = constant times xn. + + n+(&)2 6.6 Powers Instead of Exponentids 12 y' = b$ gives y - q Y = b dt and -: b t + = - C. Then C = -? (page 276) at t = 0. Therefom y = t-f which becomes 4s. infinite when bt = f or t = 14 (a) Compare &< with b+&n. In the exponent c = 1. Then b = d = f . Thus y' = y - s1 y 2 with yo = 1. the exponent gives c = 3. Then also b = d = 3. Thus y' = Sy - sY2 with yo = f. (b) For I6 ~ ~ u a t i(14) o i is s = $0 T c - " ) . Turned upside down this is y = with d = + A.Then = $!&) 3 = cu. So u = u ~ e. 18 Correction: u = to obtain $ = + = * -Y - b h = *.- *g. Substitute 2 = y(c - by) 1 has c = 1and b = -1 with yo = 1. Then y(t) = 7 by formula (12). The denominator 20 y' = y -1+2eis sero and y blows up when 2e-t = 1or t = In 2. = = -2cu 2b. The solution is u = ( u ~ %)e-l" + 22 If u = $ then = Then y = - !)e-l" !]-lI2 solves the equation y' = cy - b# with 'cubic competition". Another S-curve! 24 =rYo and = so ( g ) o = tyh. 9 [(a + 9 % + 9. 2 26 At the middle of the S-curve y = and = c( $) - a(&)' = If b and c are multiplied by 10 then so is this slope which becomes deeper. 28 If = d then cy = dy dK and y = At this steady state the maintenance dose replaces the aspirin being eliminated. SO The rate R = f& is a decreasing function of K becauae $$ = a 9, 5 3. + a. = -r[A][B] = -r[A](bo - E(w - [A])). The changes w - [A] and bo - [B]an in the proportion rn to n; we solved for [B]. 56 To change cy - by2 (with linear term) to a2 - x2 (no linear term), set x = dy - and a = 4 2 6 . (We completed the square in cy - bd.) Now match integrals: The factor is times 4 (from dx = 4 dy). The ratio =G *, is S4 A ,b & ! A. - 18 The y line shows where y increases (by y' = f (y)) and where y decreases. Then the points where f (y) = 0 are either approached or left behind. 40 y' = cy(1- %) agrees with y' = cy - bd if K = $. Then y = K is the steady state where y1 = 0 (this agrees with y, 6.6 K where y' = c Q ( 1 = t). The inflection point is halfway: y = 5 Powers Instead of Exponentials f ) = ZK and y" = 0. (page 276) 1 2 + gx 1 s + . . Its derivative is p. The denominator n! is called h + + ax 1+ 8 1+ .. (1).A t x = 1the series for e is 1+ 1+ 5 The infinite series for e' is 1 x factorialn and is equal to n(n - 1) To match the original definition of e, multiply out (1+ l/n)" = 1+ n(B) + (first three terms). As n -r oo those terms approach 1+ 1+ in agreement with e. The first three terms of (1 + ~ / n are ) ~ 4 + 1+ n(H) + P+(H)'. As n + oo they approach 1+ x 4x2 in agreement with ez. Thus (1 + ,/,In approaches ex. A quicker method computes ln(1 ~ / n ~l) x~(first term only) and takes the exponential. + + g)n.As n +,, Compound interest (n times in one year at annual rate z)multiplies by (1 continuous 6.6 Powers Instead of Exponentids (page 276) compounding multiplies by ex. At z = 10% with continuous compounding, $1 grows to e - l rr $1.105 in a year. The difference equation y(t + 1) = ay(t) yields y(t) = at times yo. The equation y(t + 1) = ay(t) + s is solved + s [ l + a + - - + at-']. --. I-at The sum in brackets is . = or at-1 When a = 1.08 and yo = 0, annud 1 . 0 8 -~1 after t years. If a = and yo = 0, annual deposits of s = 6 leave deposits of s = 1 produce y = 7 1 after t years, approaching y, = 12. The steady equation y, = ay, 1 2 ( 1 - $) s gives y, = s / ( l - a). by y = at yo ? + + When i = interest rate per period, the value of yo = $1 after N periods is y(N) = (1 ilN. The deposit to produce y(N) = 1 is yo = (1 i)-N. The value of s = $1 deposited after each period grows to y(N) = 1 - (1 i)-N). $((I i)N - 1).The deposit t o reach y(N) = 1 is s = ;(I 1 + + + + Euler's method replaces y' = cy by Ay = cyAt. Each step multiplies y by 1 c A t . Therefore y at t = 1 is (1+ c ~ t ) ' / * ' ~ which ~, converges to yoec as At -+ 0. The error is proportional to A t , which is too large for scientific computing. ~ l & z + $ f$ + . . . 1 1 - 2 + $ - d + 6. . . 5 1050.62; 1050.95; 1051.25 i)"; 1 9square of (1+ set N = 2n 7 1 + n ( 2 ) + w ( + ) 2 1- 1 + 2 11Increases; ln(1 >O 13y(3)=8 15y(t)=4(3') 17y(t)=t lgy(t)=+(3'-1) 21~(9)ifa#l;stifa=l 23yo=6 25yo=3 17 41 31 10.38% 29 P = 2 7 -2,-10,-26 --+ - 0 0 ; - 5 , - 7 , - T + -12 33 100(1.1)~~ = $673 -+ + i) 35 5 ~m 5 7 y(l.lao - 1) = 57,275 = 965 4 3 1.0142'~= 1.184 + -+ 3 9 y, = 1500 = 2.69; e 4 1 2; Visa charges 18.4% + + + + + +k+ 2 y = 1 22 + i ( 2 ~ ) i~( 2 ~ ) .~. . Integrate each term and multiply by 2 to find the next term. . - . = 3. This is greater than 1 1 - = e. 4 A larger series is 1 1 6 ln(1= n ln(1rr n ( - i ) = -1. Take exponentials: (1- :ln rr e - l . Similarly ln(1 + ;)" = nln(1 ): w n(:) = 2. Take exponentials: (1 rr e2. 8 The exact sum is e-' rr .37 (Problem 6). After five terms 1 - 1 = = .375. i)" + + +5+i+ i) + i)n +i i+ & 0. So (1+ 3)"-+ e0 = 1.Similarly l n ( l + rr n ( 5 ) 1 0 By the quick method l n ( l + SO (1 i ) n ' + 00. 1 2 Under the graph of the area from 1to 1 f is ln(1 so this is below l n ( l + Its a r e a is height -+ + 4. I+, 2, + :)"I rr n 2 ( i ) -+ oo + i).The rectangle inside this area has base f and i). x+f 1 4 y(0) = O,y(l) = 1,y(2) = 3,y(3) = 7 (and y(n) = 2n - 1). 16 y(t) = 1 8 y(t) = t (Notice that a = 1). 2 0 y(t) = 3 t s[-1.5 t - 1 22 y(t) = 5at + (i)t. + s[%]. 24 Ask for ky(0) - 6 = y(0). Then y(0) = -12. 2 6 Ask for - i y ( 0 ) 6 = ~ ( 0 )Then . y(0) = 4. 1 2 8 If -1 < a < 1then approaches E. 3 0 The equation -dP(t 1) b = cP(t) becomes -2P(t 1) 8 = P ( t ) or P ( t 1) = - i P ( t ) 4. Starting + + from P(0) = 0 the solution is P ( t ) = 4[ + s)365 -1 (-qt-1 + + + ] = !(I-(-%) I t ) + + + + f. i(.~)~ 5 2 (1 = 1.105156. - (Compare with em1 rr 1 . l + = 1.105.) The effective rate is 5.156%. 34 Present value = $1,000 (1.1)-~'rr $148.64. 36 C o r r e c t i o n to formulas 5 and 6 on page 273: Change .05n to .05/n. In this problem n = 12 and N = 6(12) = 72 months and .05 becomes .1in the loan formula: s = $10,000 (.1)/12[1- ( I + w $185. - 6.7 Hyperbolic finctions (page 280) 38 Solve $1000 = $8000 [l-li!l,-m] . , for n. Then 1- (1.1)-" = .8 or (1.1)-" = .2. Thus 1.1" = 5 and In 5 n= n 1.1 w 1 7 years. 40 The interest is (.05)1000 = $50 in the first month. You pay $60. So your debt is now $1000 - $10 = $990. Suppose you owe y(t) after month t, so y(0) = $1000. The next month's interest is .05y(t). You pay $60. So y(t 1) = 1.05y(t) - 60. After 12 months (1000 (1.05) l2 M $841. This is also y(12) = (1.05)121000- 60[=]. Its logarithm is nln(1 w n[i 42 Compounding n times in a year at 100% per year gives (1 = 1Therefore (1 w e(e-'I2") M e(1- &). 44 Use the loan formula with .O9/n not .O9n: payments s = 80,000w $643.70. Then 360 payments equal $231,732. + + i)" &. 6.7 + Hyperbolic Functions Cosh x = 1(ex + e-X) and sinh x = g) + i)". + i) &] (page 280) 4(9- e-X) and cosh2x - sinh2 x = 1.Their derivatives are s i n h x and cosh x and zero. The point (x, y) = (cosh t, sinh t) travels on the hyperbola x2 - y2 = 1.A cable hangs in the shape of a catenary y = a cosh:. ' x are equal to ln[x + d z ]and 4 in E.Their derivatives are 1 / d Eand &. So we have two ways to write the antiderivative. The parallel to cosh x + sinh x = ex is Euler's formula cos x + i sin x = ex. The formula cos x = i(eiX+ e-ix) involves imaginary exponents. The ' The inverse functions sinh- x and tanh- parallel formula for sin x is 1e ~e-x , , e'=-e-2z (ejx - e- jx) . + - isinh2x 7 sinh nx 9 3sinh(3x 1) 1 7 6 sinh5 x cosh x 1 3 4cosh x sinh x 1 5 *(sechJ&T)' 1 9 c o s h ( l n x ) = ~ ( x + ~ ) = 1 a t ~ = 1 2 1 d1 3' 3, 5- 1= - 5' 3' - ' 1 2 , 12 i 5 + + 11 -cosh2 s i n h x = - tanh x sech x 2Q0,0,1,00,00 29 ln(1 cosh x) 25 sinh(2x 1) 27 cosh3x 31 ex 33 dx = ssinh t(sinh t dt);A = isinh tcosht dz;A1 = ;;A = 0 at t = 0 so A = i t . 4 1 e ~ = x + d ~ , ~ = l n [ x + d X C i 4] 7 : l n l E I 4 9 sinh-' x (see 41) 5 1 -sech-'x 53iln3;oo 55y(x)=~coshcx;~coshcL-$ 57 y"= y - 3 3 ; = - y3 is satisfied by y = isech2: sy sy i(y1)2 iy2 2 cix (eZ;e-= d sinhx dx(Z&G) ) = e"-e-" -- s i n h x ; $ ( ~ ) = ~ = c o s h x . leoshz)'-(~inhx)~ 1 2 (coah x)2 (cosh x ) t = sech x. 7 6 T h e factor should b e removed from Problem 5 Then the derivative of Problem 5 is 2 cosh x sinh x 2 sinh x cosh x = 2 sinh 2s. Therefore sinh 22 = 2 sinh x cosh x (similar to sin 22). + 8 (-)(-) 2 + (w 2 ()w =) f (2ex+y - 2eVx-Y) = sinh(x + y). 2 + i cosh(x + y) = cosh x cosh y sinh x sinh y. 1 2 sinh(1nx) = (eln - e- ln =) = (x 1 0 2x cosh x2 i The x derivative gives 1 . t)with derivative 1(1+ x2) 1 4 cosh2x - sinh2 x = 1with derivative zero. 16 1 lTtanh t a n h x = e2 s by the equation following (4). Its derivative is 2 e 2 ~ More . directly the quotient rule gives 6.7 Hyperbolic finctions (page 280) jl-tanh x)sechax+(l+tanhx)sechax (1-tanh x)' 2 sechax (1-tanh x)' 2 (coshx-sinh x)' , - 2 - - e-as - 2e2X. f ~ ~ ~ofhthe 3 xminus . sign we do not get sech x. The integral of = s e c h , ~ ~ ~ ~ Because 18 $ In u = sech x is sin-'(tanh x) C. 20 sech x = + 4- r = 4s , c o s h x = f , s i n h x = ( f ) 2 - 1 = ~ , c o t hC Ox S= x~ ~ = ~ , c s c h3 x = ~ . 22 c o s h x = ~ ~ = & , t a n h x -=&,cschx= eln S-=- i , s e c h x = $,cothx= In 4 -e-2 In 5 - 2 = y ; t a n h ( 2 l n 4 ) = e,1n4+ a 26 J' x ~ o s h ( x ~ ) d=x sinh(x2) C. 28 f (tanh x ) ~C. 24 sinh(ln5) = 4 + In 4 - 16-k g. - 255 rn' + 3 2 sinh x + cosh x = ex and J enxdx = Be- + C. + l6+& - 30 J' coth x dx = J' -dx = ln(sinhx) C. 34 y = tanh x is an odd function, with asymptote y = -1 as x -+ -oo and y = +1 as x + +oo. The inflection point is (0,O). = 1 at x = 0. The x axis is the asymptote. But note 36 y = sechx looks like a bell-shaped curve with ,y, that y decays like 2e-' and not like e-". 38 To define y = cosh-' x we require x 2 1. Select the positive y (there are two y's so strictly there is no inverse). For large values, cosh y is close to i e so ~ cosh-' x is close to ln 2%. 4 0 l n ( e ) approaches +oo as x + 1 and -a, as x -+ -1. The function is odd (so is the tanh function). The graph is an S curve rotated by 90'. 42 The quadratic equation for eY has solution eY = x f dm. Choose the plus sign so y -t oo as x oo. Then y = ln(x d-) is another form of y = cosh-' x. = - of sinh-' x. 4 4 The x derivative of x = sinh y is 1= cosh y $ . Then $ = 1 4 A 3 - slope c~shg-d- i - + 4 6 The x derivative of x = sech y is 1= -sech y tanh 4 8 Set x = ou and dx = a du to reach + j' 1 u ) = yg.Then 2 = set,,,-1 1 tanh-' u = ptanh-lg + - + C. -1 xdis' + 50 Not hyperbolic! Just I ( x 2 1)-'I2% dx = (x2 1) C. 52 Not hyperbolic! @ = sin-lx C. 5 4 (a) = (&I2sech fi t = g(1- tanh2 J@) = g - v2. (b) = dt gives (by Problem 48) 1tanh-' - t or tanh-' = f i t or = tanh fit. (c) f (t) = f i t a n h f i t dt = 2 4 J' $$&fidt I fi- + fi = ln(cosh&t) 56 Change t o d x = + C. I5 fi I f:rw=-x- andintegrate: x = l n ( 2 - W ) - l n W = l n ( w ) . T h e n 26 = ex and W = r . (Note: The text suggests W Writing as e-x12sechg is not simpler.) + 4 + + - 2 but that is negative. 58 cos i x = L(ei(ix) 2 c-i(ix)) = (e-' ex) = cosh x. Then cos i = cosh 1= +e (real!). 60 The derivative of eix = cos x i sin x is ieix = i(cos x i sin x) on the left side and $ cos x + i$ sin x on the right side. Comparing we again find (sin x ) = cos x and &(cos x) = i2sin x. 2 + MIT OpenCourseWare http://ocw.mit.edu Resource: Calculus Online Textbook Gilbert Strang The following may not correspond to a particular course on MIT OpenCourseWare, but has been provided by the author as an individual learning resource. For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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File Type : PDF File Type Extension : pdf MIME Type : application/pdf PDF Version : 1.4 Linearized : Yes Tagged PDF : Yes XMP Toolkit : 3.1-702 Producer : Adobe Acrobat 7.1 Paper Capture Plug-in Create Date : 2010:10:08 11:25:15+05:30 Modify Date : 2010:10:08 12:12:01+05:30 Metadata Date : 2010:10:08 12:12:01+05:30 Creator Tool : Adobe Acrobat 7.1 Document ID : uuid:9863A79EA0D2DF11B161BB150FD73E79 Instance ID : uuid:57affd71-7603-4f2f-8027-2a99d4eeaeb9 Derived From Instance ID : uuid:50645686-f4fd-4968-b2ed-fd4d2a95ca0e Derived From Document ID : uuid:CA103DBB7EBADF11A681A420252226BB Format : application/pdf Title : Calculus Online Textbook Instructor's Manual Chapter 6 Creator : Gilbert Strang Page Count : 13 Language : EN Author : Gilbert StrangEXIF Metadata provided by EXIF.tools