{VERSION 6 0 "IBM INTEL NT" "6.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 263 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 264 "" 0 1 255 0 0 1 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 266 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 267 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 268 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 269 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 270 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 272 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 273 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 274 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 275 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 276 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 277 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 278 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 279 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 281 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{PSTYLE "Head ing 2" 3 4 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 8 2 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 3 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 17 "Using Derivatives" }}{PARA 260 "" 0 "" {TEXT 263 265 "Note: In \+ using this, and other worksheets, remember that each time you open the worksheet you have to re-execute all the commands from the beginning \+ to insure that all functions and expressions are defined properly. Oth erwise you will get mysterious error messages!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "As you already know, Mapl e can calculate derivatives with a simple \"" }{TEXT 266 4 "diff" } {TEXT -1 22 "\" comand. For example:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "diff(x^3,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$)%\"xG\"\"#\"\"\"\"\"$" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 84 "There is another command for finding derivatives t hat is very useful, namely, the \"" }{TEXT 267 2 "D(" }{TEXT 264 16 " name of function" }{TEXT 268 2 " )" }{TEXT -1 31 "\" command. It works as follows." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "h :=x->x^3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %\"hGf*6#%\"xG6\"6$%)operatorG%&arrowGF(*$)9$\"\"$\"\"\"F(F(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "D(h);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#f*6#%\"xG6\"6$%)operatorG%&arrowGF&,$*$)9$\"\"#\"\"\"\" \"$F&F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "When you use the \"" }{TEXT 269 4 "D(h)" }{TEXT -1 41 "\" command Maple recognizes the derivative" }{TEXT 270 5 " D(h)" }{TEXT -1 88 " as a function of x, which is convenient if yo u plan to manipulate the derivative later." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 11 "Example 1. " } {TEXT -1 119 " The amount of a certain bacteria in the blood (measured in number of bacteria per cubic centimeter) can be modeled by " }} {PARA 257 "" 0 "" {XPPEDIT 18 0 "f(t) = 60*t^2*e^(-.27*t);" "6#/-%\"fG 6#%\"tG*(\"#g\"\"\"*$F'\"\"#F*)%\"eG,$*&-%&FloatG6$\"#F!\"#F*F'F*!\"\" F*" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{TEXT 257 2 "t " }{TEXT -1 55 "is measured in days since the person first became ill. " }}{PARA 0 "" 0 "" {TEXT -1 143 " (a) How does the amount of b acteria vary with time? Sketch a graph of f(t) from the time the perso n first becomes ill until five weeks later." }}{PARA 0 "" 0 "" {TEXT -1 97 " (b) When does the number of bacteria reach its maximum? What i s the maximum number of bacteria? " }}{PARA 0 "" 0 "" {TEXT -1 171 " ( c) When is the number of bacteria growing fastest and how fast is it \+ growing then? When is the number of bacteria decreasing fastest and ho w fast is it decreasing then?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 175 "We begin by drawing the function f(t) ov er the initial thirty-five day period. Before that, we have to define the function f(t) properly so Maple recognizes it as a function." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "f := t -> 60*t^2*exp(-.27*t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%\"fGf*6#%\"tG6\"6$%)operatorG%&arrowGF(,$*&)9$\"\"#\"\"\"-%$expG6#, $F/$!#F!\"#F1\"#gF(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 " plot(f(t),t=0..35);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 197 " As we see, the number of bacteria increases fast at first, reaches its maximum around day 6 or 7, then begins to decline. To find the exact \+ time of the peak we have to find the derivative of f(t). " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "D(f); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#f*6#%\"tG6\"6$%)operatorG%&arrowGF &,&*&9$\"\"\"-%$expG6#,$F,$!#F!\"#F-\"$?\"*&)F,\"\"#F-F.F-$!%?;F4F&F&F &" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "As you see Maple considers \+ " }{TEXT 272 4 "D(f)" }{TEXT -1 18 " as a function of " }{TEXT 271 1 " t" }{TEXT -1 27 ". Let's plot the derivative" }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(D(f)(t),t= 0..35);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 169 "It seems tha t the zero of the derivative which corresponds to the maximum of f(t) \+ is between 7 and 8. To find it exactly we ask Maple to solve the corre sponding equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 20 "solve(D(f)(t)=0, t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$\"\"!$\"+2uS2u!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 169 "Luckily, Map le gave us both solutions that we can see on the graph. Typically, Map le returns only one solution. If it isn't the one we want, we have to \+ use the command \"" }{TEXT 273 6 "fsolve" }{TEXT -1 234 "\" that allow s us to specify the range in which we want our solution to be. You wil l see such situation in the next example. The number of bacteria reach es its maximum at approximately t = 7.407. What is the maximal number \+ of bacteria?" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "evalf(f(7.407));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+hD[bW!\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 203 "The maxim al number of bacteria is about 446 bacteria per cc of blood. To answer the questions when is the number of bacteria growing and declining fa stest, we have to find points at which the derivative " }{TEXT 274 4 " D(f)" }{TEXT -1 183 " has its maximum and its minimum in the interval . They seem to occur, respectively, between 2 and 3, and between 11 an d 12. To find their exact values, we need the second derivative." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "D(D(f));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#f*6#%\"tG6\"6$%)opera torG%&arrowGF&,(-%$expG6#,$9$$!#F!\"#\"$?\"*&F/\"\"\"F+F5$!%!['F2*&)F/ \"\"#F5F+F5$\"&SP%!\"%F&F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot(D(D(f))(t), t=0..35);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "It is hard to pin-point zeros of the second derivative f rom its graph. Again we ask Maple to solve an equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "solve( D(D(f))(t)=0,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"+*Rz&p@!\"*$\"+ UN_k7!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 344 "The solutions corre spond, of course, to the maximum and minimum of the first derivative. \+ The number of bacteria increases fastest at approximately t= 2.17, dec reases fastest at t =12.645. How fast is it increasing or decreasing t hen? We have to find the values of the rate of change; that is, the va lues of the first derivative at those points:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "evalf(D(f)( 2.17)); evalf(D(f)(12.645));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+$H (zC5!\"(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+60OIN!\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 236 "At t =2.17 the number of bacteria increases at its fastest, at the rate of approximately 102.5 bacteria/cc per day. At t = 12.645, t he number of bacteria decreases at its fastest, at the rate of approxi mately -35.30 bacteria/cc per day." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 110 "We were very lucky in the last example t o have Maple find all zeros of functions that we wanted using simple \+ \"" }{TEXT 275 5 "solve" }{TEXT -1 67 "\" command. The next example sh ows what to do if it does not happen." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 258 13 "Example 2. " } {TEXT -1 499 "We take a spring suspended vertically from a fixed suppo rt. At the lower end of the spring we attach a weight. The spring stre tches slightly and the system remains motionless in the position of a \+ static equlibrium. Then we pull the weight down and release it. The we ight is beginning to oscillate up and down. We assume that the motion \+ is strictly vertical.We take the position of the static equlibrium as \+ 0, downward as the positive direction. The motion of the weight is des cribed by the function:" }}{PARA 258 "" 0 "" {XPPEDIT 18 0 "s(t) = e^( -.3e-1*t)*(10*cos(2*t)+.15*sin(2*t));" "6#/-%\"sG6#%\"tG*&)%\"eG,$*&-% &FloatG6$\"\"$!\"#\"\"\"F'F2!\"\"F2,&*&\"#5F2-%$cosG6#*&\"\"#F2F'F2F2F 2*&-F.6$\"#:F1F2-%$sinG6#*&F;F2F'F2F2F2F2" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 54 "where s is measured in centimeters, time t in sec onds." }}{PARA 0 "" 0 "" {TEXT -1 48 " (a) How does the motion look \+ in the long run?" }}{PARA 0 "" 0 "" {TEXT -1 139 " (b) During the fi rst four seconds, how many times and at what values of t does the weig ht pass through the position of the equilibrium? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "We begin by defining the \+ function s(t) and plotting it on a long time interval, say, from 0 to \+ 100." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "s := t -> exp(-.03*t)*(10*cos(2*t)+.15*sin(2*t));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"sGf*6#%\"tG6\"6$%)operatorG%&arrow GF(*&-%$expG6#,$9$$!\"$!\"#\"\"\",&-%$cosG6#,$F1\"\"#\"#5-%$sinGF9$\"# :F4F5F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "plot(s(t), t =0..100);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 421 "We see a t ypical example of damped oscillations. The weight begins around 10 cm \+ below the equlibrium and then oscillates around it, with the amplitude decreasing to 0. This is, of course, what we would expect given the p hysical setting of the motion. Remember, every time the function s(t) \+ crosses the t axis, the weight is back at the position of equilibrium. Let's examine the situation during the first four seconds. " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(s(t),t=0..4);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "We se e three moments when the weight passes through the position of the equ librium, that is, s(t) = 0. Let's see if we can find them using the \" " }{TEXT 276 5 "solve" }{TEXT -1 10 "\" command." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(s(t)= 0,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+es)*yx!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 249 "This time we didn't get lucky. Maple ret urned a zero, but it isn't the one we want. To find the ones that we w ant, we shall specify the range in which we want Maple to find a solut ion to the equation s(t)=0. Specifying the range is possible with the " }{TEXT 277 7 "\"fsolve" }{TEXT -1 22 "\" command. While the \"" } {TEXT 278 5 "solve" }{TEXT -1 49 "\" command attempts to find exact so lutions, the \"" }{TEXT 279 6 "fsolve" }{TEXT -1 142 "\" command uses \+ numerical methods to find approximate solutions. We see that there is \+ a zero somewhere between 0.5 and 1. Let's try to find it." }}{PARA 0 " " 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "fso lve(s(t)=0,t, 0.5..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+5g(*Gz!# 5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "Maple gave us the first zer o that we wanted, the one somewhere between .5 and 1. Let's do the sam e with the remaining zeros. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "fsolve(s(t)=0,t,2..3); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+G RpjB!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "Note the syntax which allows us to find a zero in each of the corresponding ranges." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "fsolve(s(t),t,3.5..4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+b- \\MR!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "We have found all the times during the f irst four seconds when the weight passes through the equilibrium." }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 4 "" 0 "" {TEXT -1 17 "Ho mework Problems" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 259 11 "Problem 1. " }{TEXT -1 87 "The amount of \+ a certain drug in a patient's blood, measured in milligrams, is given \+ by " }}{PARA 259 "" 0 "" {XPPEDIT 18 0 "g(t) = 150*t*e^(-.3*t);" "6#/- %\"gG6#%\"tG*(\"$]\"\"\"\"F'F*)%\"eG,$*&-%&FloatG6$\"\"$!\"\"F*F'F*F3F *" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 52 "where t is measured in hours following an oral dose." }}{PARA 0 "" 0 "" {TEXT -1 77 " (a) Plot the function g(t) to see how the amount of the drug varies in ti me." }}{PARA 0 "" 0 "" {TEXT -1 84 " (b) When does the amount of the d rug reach its maximum? What is the maximum amount?" }}{PARA 0 "" 0 "" {TEXT -1 176 " (c) When is the amount of the drug increasing fastest a nd how fast is it inceasing then? When is the amount of the drug decre asing fastest and how fast is it decreasing then? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 11 "Problem 2 . " }{TEXT -1 289 "For the motion s(t) in Example 2 find the derivativ e D(s). What is the physical meaning of the derivative? Plot the deriv ative between t=0 and t=5. Find zeros of the derivative in that inter val. What can you say about the motion of the weight at the instants w hen the derivative is zero?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 13 "Problem 3. " }{TEXT -1 108 "For the \+ motion of the weight in Example 2 and the previous problem, find the m aximum speed of the weight as " }{TEXT 262 23 "accurately as you can. \+ " }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }{TEXT 280 5 "Note." }{TEXT -1 16 " If you use th e " }{TEXT 281 7 "restart" }{TEXT -1 74 " command in your homework wor ksheet, you have to redefine the function s. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 265 76 "M TH 141 Maple Worksheets written by B.Kaskosz and L. Pakula, Copyright \+ 1998." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 282 39 "Last mod ified August 1999. Update 2006" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "2 0 \+ 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }