{VERSION 5 0 "IBM INTEL NT" "5.0" } {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 0 1 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 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 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 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 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 1 }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 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }0 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 1 0 0 0 0 0 0 0 1 }0 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 1 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 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 65 "MAT 155 Project VII F all 2002" }}{PARA 0 "" 0 "" {TEXT 269 55 " \+ Derivatives " }{TEXT -1 13 " " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 8 " " }}}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 25 "Do not work on this file." }{TEXT -1 118 " This is just the list of problems. Open your own file and save it regularly \+ as yourname7.mws and as yourname7b.mws." }{TEXT 259 1 " " }{TEXT -1 276 " Sign your name as a comment at the top of your file by backspaci ng in front of the prompt and typing. Also write Project VII and the \+ names of any other students you are working with. Don't forget to num ber your problems and to type restart at the beginning of each problem ." }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 256 89 "If you cannot complete a problem, go on to the next one and re turn to the problem later. " }{TEXT -1 9 " You can " }{TEXT 257 16 "ge t a new prompt" }{TEXT -1 236 " by selecting the prompt button right b elow the word \"spreadsheet\". It has the symbol \"[>\" on it. You m ust hit enter on every line of the problem in order, including the res tart line, to review what you've done for the Maple program. " } {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 " " 0 "" {TEXT 260 31 "If you cannot recall a command " }{TEXT -1 160 "f rom a previous lab you may consult the command index which can be open ed up as a second window. The name of the file with the command index is 155.00.00.html." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 261 10 "Problem 1:" }{TEXT -1 33 " Recall how to define a function:" } {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "f:= x -> x^2;" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "Recall that the derivative of a function is the limit as h approaches 0 of the \"diff erence quotient\":" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "(f(x+h)-f(x)) /h;" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Here we are using an h instead of the usual " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG " }{TEXT -1 2 "x." }}{PARA 0 "" 0 "" {TEXT -1 51 "You can find this li mit by expanding the numerator:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 " expand(f(x+h)-f(x));" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "th en cancelling the h by hand and using direct substitution. Do this an d then check your answer by" }}{PARA 0 "" 0 "" {TEXT -1 24 "using the \+ limit command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "limit( ( \+ f(x+h)-f(x) )/h , h=0 );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "Now i t is easy to take the limit as h goes to zero, and you can see that th e answer is 2x." }}{PARA 0 "" 0 "" {TEXT -1 17 "That is f'(x)=2x." }} {PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 95 "The comp uter can do all this work to find the derivative in one step if you us e the command D: " }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "D(f);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "Noti ce that D(f) is a function. So to graph it we must type D(f)(x):" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plot( D(f)(x), x=-2..2);" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 256 "Now define the function g(x)= x^3-x. Compute its deriva tive using the difference quotient by expanding the numerator, dividin g out the h and using direct substitution. Then check using the limit command and finally use the D command. Then graph D(f)(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 10 "Proble m 2:" }{TEXT -1 717 " Don't forget to restart. Define a function f wh ich maps x to x^5. Graph it. Now we want to find the tangent line at x=2. Instead of using a loop command and graphing a lot of secant li nes, we can just use the fact that the difference quotient (f(x+h)-f( x))/h measures the slope of the secant line through the points (x,f(x) ) and ((x+h),f(x+h)). So the slope of the tangent line can be found b y taking the limit of this difference quotient as h goes to 0 (then x+ h goes to x). In problem 1 we saw that this limit can be done quickly using the D(f)(x) command. So to find the slope of the tangent line to the graph of f at (2,f(2)), we find the derivative of f using the \+ D command and find its value at 2: " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "D(f)(2);" }}{PARA 0 "" 0 "" {TEXT -1 102 "This should be the sl ope of the graph of f at (2, f(2)). So call it m by using the followi ng command:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "m:=D(f)(2);" }} {PARA 0 "" 0 "" {TEXT -1 67 "Now define the line through (2,4) with th e slope m as a function L:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "L:= x -> m*(x-2)+f(2);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "Then graph both f and L together using the command:" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 27 "plot(\{f(x),L(x)\}, x=-5..5);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "You can see \+ that you have found the tangent line from the graph." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 265 "Now examine the graphs near where they meet by plotting f(x) and L(x) together on smaller an d smaller intervals around x=2. Try x in [1,3], x in [1.9, 2.1] and x in [1.99, 2.01]. Note that the two graphs look almost the same! For this reason L(x) is called the \"" }{TEXT 270 37 "Linear Approximatio n of f(x) near x=2" }{TEXT -1 2 "\"." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 265 8 "Problem " }{TEXT -1 0 "" }{TEXT 264 2 "3:" }{TEXT -1 76 " Don't forget to restart. Take the function f which takes x to x^ 2+2x+1" }{TEXT 263 0 "" }}{PARA 0 "" 0 "" {TEXT -1 216 "and find its t angent lines at x=-2, x=2 and x=0. Be sure to use different variables to represent the slope (like firstm, secondm and thirdm) and to give \+ the lines different names (like firstL, secondL and thirdL). " } {TEXT 271 193 "In computers it is common to use entire words for funct ions and variables. In math we cannot do this because when letters are placed next to each other it is assumed that they are multiplied. " }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 314 "Now graph all three tan gent lines together with f(x) by including all of them in the brackets \{\}, and write a comment discussing which lines are increasing and de creasing. Recall that a graph is increasing if it goes up when travel ling from left to right and it is decreasing when it goes down from l eft to right." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 257 "" 0 "" {TEXT -1 12 "Exploration:" }}{PARA 0 "" 0 "" {TEXT 266 6 "A: " }{TEXT -1 215 "A ball is thrown straight up in th e air and its height is described by the formula h(t)= -4.9 t^2 +100t \+ +2 meters where t measures the time in seconds. The velocity is then \+ described by the function v(t)=D(h)(t). " }}{PARA 0 "" 0 "" {TEXT -1 515 "What is the initial velocity v(0)? When a ball is thrown straigh t up in the air, its velocity starts out positive then right before th e ball starts to drop back down the velocity is zero and when the ball goes down the velocity is negative. To find out how high the ball go es before it falls, solve for when the velocity is zero (v(t)=0) and t hen substitute the answer into the height equation h(t). To find out \+ when the ball hits the ground solve for h(t)=0 and only accept a posit ive answer (time starts at 0)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } {TEXT 267 4 "B: " }{TEXT -1 110 "Write a loop program which tells you t, h(t),v(t) for t from 1 to 20 (consult the command index if necessa ry)." }}{PARA 0 "" 0 "" {TEXT -1 52 "Be sure to save your work before \+ running the loop. " }}{PARA 0 "" 0 "" {TEXT 268 4 "C: " }{TEXT -1 45 "The last line of the loop's output should be " }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"#?$\"$?%!\"\"$!$g*F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "This means that at time t=20 seconds, the ball is 42 met ers high and falling downward at 96 meters per second." }}{PARA 0 "" 0 "" {TEXT -1 340 "Using this information you can find the tangent lin e, L(t), to h(t) at t=20 seconds. This tangent line is a linear appro ximation of h and can be used to estimate the values of h near t=20. \+ To see this plot L(t) and f(t) for t in [19,20]. If we fsolve(L(t)=0 , t=19..21), we can see when the linear approximation of the height h its zero." }}{PARA 0 "" 0 "" {TEXT -1 130 "This should be a good estim ate for when the height itself hits zero fsolve(h(t), t=19..20); This idea has been used since Newton." }}}}{MARK "0 1 0" 65 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }