18.01 Calculus Jason Starr Fall 2005

Transcription

1 Lecture 18. October 5, 005 Homework. Problem Set 5 Part I: (c). Practice Problems. Course Reader: 3G 1, 3G, 3G 4, 3G Approximatig Riema itegrals. Ofte, there is o simpler expressio for the atiderivative tha the expressio give by the Fudametal Theorem of Calculus. I such cases, the simplest method to compute a Riema itegral is to use the defiitio. However, this is ot ecessarily the most efficiet method. Ofte trapezoids or segmets uder a parabola give a better approximatio to the Riema itegral tha do vertical strips.

2 . The trapezoid rule. The problem is to fid a approximatio of the Riema itegral, I = I l = y k 1 Δx k. I r = y k Δx k. I trap = b for a fuctio y(x) defied o the iterval [a, b]. Choose a partitio of the iterval [a, b] ito equal subitervals. The poits of this partitio are, The values of these poits are, a ydx b a x k = a + (b a)k, Δx k =. y k = f(x k ). The Riema sum usig always the left edpoit is, k=1 The Riema sum usig always the right edpoit is, The average of the two is, k=1 y k 1 + y k Δx k. k=1 This is usually a better approximatio tha either of the two approximatios idividually. Part of the reaso is that the term (y k 1 + y k )Δx k / is the area of the trapezoid cotaiig the poits (x k 1, 0), (x k 1, y k 1 ), (x k, 0) ad (x k, y k ). I particular, if the graph of y = f(x) is a lie, this trapezoid is precisely the regio betwee the graph ad the x axis over the iterval [x k 1, x k ]. Thus, the approximatio above gives the exact itegral for liear itegrads. Writig out the sum gives, I trap = b a ((y 0 + y 1 ) + (y 1 + y ) + (y + y 3 ) + + (y + y 1 ) + (y 1 + y )). Gatherig like terms, this reduces to, I trap = (b a)(y 0 + y 1 + y + + y 1 + y )/. 3. Simpso s rule. Agai partitio the iterval [a, b] ito equal subitervals. For reasos that will become apparet, must be eve. So let = m where m is a positive iteger. Agai defie, (b a)k (b a)k b a b a x k = a + = a +, Δx k = =. m m

3 Pair off the itervals as ([x 0, x 1 ], [x 1, x ]), ([x, x 3 ], [x 3, x 4 ]), etc. Thus the l th pair of itervals is, ([x l, x l 1 ], [x l 1, x l ]). The idea is to approximate the area of the graph over the pair of itervals by the area uder the uique parabola cotaiig the 3 poits (x l, y l ), (x l 1, y l 1 ), (x l, y l ). For otatio s sake, deote l 1 by k. Thus the 3 poits are (x k 1, y k 1 ), (x k, y k ), ad (x k+1, y k+1 ) (this is slightly more symmetric). The first problem is to fid the equatio of this parabola. Sice the parabola cotais the poit (x k, y k ), it has the equatio, y = A(x x k ) + B(x x k ) + y k, Pluggig i x = x k 1 ad x = x k+1, ad usig that x k+1 x k = x k x k 1 equals Δx, y k+1 = A(Δx) + B(Δx) + y k, Summig the two sides gives, Solvig for A gives, y k 1 = A(Δx) B(Δx) + y k. y k+1 + y k 1 = A(Δx) + y k. 1 A = (Δx) (y k 1 y k + y k+1 ). Similarly, takig the differece of the two sides gives, y k+1 y k 1 = B(Δx). Solvig for B gives, 1 B = (Δx) (y k+1 y k 1 ). Thus, the equatio of the parabola passig through (x k 1, y k 1 ), (x k, y k ) ad (x k+1, y k+1 ) is, y = A(x x k ) + B(x x k ) + y k, A = ( y k 1 y k + y k+1 )/ (Δx), B = ( y k+1 y k 1 )/ (Δx). The ext problem is to compute the area uder the parabola from x = x k 1 to x = x k+1. This is a straightforward applicatio of the Fudametal Theorem of Calculus, ( x k+1 x A B k+1 A(x x k ) + B(x x k ) + y k dx = (x x k ) 3 + (x x k ) + y k (x x k ) 3. x k 1 x k 1

4 Pluggig i ad usig that x k+1 x k = x k x k 1 equals Δx, this is, A (Δx) 3 + y k (Δx). 3 Substitutig i the formula for A ad simplifyig, this is, Δx Δx Δx (y k 1 y k + y k+1 ) + (6y k ) = (y k 1 + 4y k + y k+1 ) Back substitutig l 1 for k ad (b a)/m for Δx, the approximate area for the pair of itervals [x l, x l ] ad [x l 1, x l ] is, ΔI l = b a (y l + 4y l 1 + y l ). 6m Fially, summig this cotributio over each choice of l gives the Simpso s rule approximatio, Writig out the sum gives, m b a I Simpso = (y l + 4y l 1 + y l ). 6m l=1 b a I Simpso = ((y 0 + 4y 1 + y ) + (y + 4y 3 + y 4 ) + (y 4 + 4y 5 + y 6 )+ 6m + (y m 4 + 4y m 3 + y m ) + (y m + 4y m 1 + y m )). Gatherig like terms, I Simpso reduces to, (b a)(y y 1 + y + 4 y 3 + y y 5 + y y m 3 + y m + 4y m 1 + y m )/6m. Example. Approximate l() usig a partitio ito 4 equal subitervals with the Trapezoid Rule ad with Simpso s Rule. The value l() equals the Riema itegral, 1 dx. 1 x The poits of the partitio are x 0 = 4/4, x 1 = 5/4, x = 6/4, x 3 = 7/4 ad x 4 = 8/4. The correspodig values are y 0 = 4/4, y 1 = 4/5, y = 4/6, y 3 = 4/7, y 4 = 4/8. Thus the Trapezoid Rule gives, I trap = b a (y 0 + y 1 + y + y 3 + y 4 ) = ( ) = For Simpso s Rule, because equals 4, m equals. Thus, I Simpso = b a (y 0 + 4y 1 + y + 4y 3 + y 4 ) = ( ) = m

5 Accordig to a calculator, the true value is, l() = ± 10 4 Note that trapezoids overestimate the area, because 1/x is cocave up. The approximatig parabolas cross the graph of y = 1/x, thus the uderestimatio to the left of (x k, y k ) somewhat cacels the overestimatio to the right of (x k, y k ), explaiig the better approximatio. 4. Oe review problem. This is a related rates review problem for Exam 3. A particle moves with costat speed 3 o the parabola y = x. The particle is movig away from the origi. What is the rate of chage of the distace from the origi to the particle whe the distace equals 5? The idepedet variable is time, t. The depedet variables are the x coordiate of the particle, x(t), the y coordiate of the particle, y(t), ad the distace L(t) from the particle to (0, 0). The costat is the speed s = 3 of the particle. The costraits are that the poit moves o the parabola, y = x, ad the Pythagorea theorem, Also, sice the speed is costat, L = x + y. ( ) ( ) dx dy s = +. This plays the role of the kow rate of chage i a typical related rates problem. It is simplest to relate the depedet variables y ad L to x. The first step is to determie x at the momet whe L equals 5. Pluggig y = x ito the equatio for L gives, ) 4 L = x + y = x + (x = x + x. At the istat whe L equals 5, L equals 0. Thus, at that momet, This factors as, 4 x + x = 0. (x 4)(x + 5) = 0. Sice x is oegative, the solutio is x = 4. Assumig the particle is i the first quadrat (this is ot specified i the problem), x is positive. The other choice leads to a symmetric problem ad the same fial aswer. So, at the momet whe L equals 5, x equals. The ext step is to determie the kow rate of chage, dx/ at the momet whe L equals 5. Implicitly differetiatig the equatio y = x gives, dy dx = x.

6 Substitutig this ito the equatio for s gives, ( ) ( ) ( ) dx dx dx s = + x = (1 + 4x ). Sice s is kow to be 3, ad x is kow to be, this equatio ca be solved for dx/, ( ) dx 3 9 = = () 17 Sice the particle is i the first quadrat ad movig away from the origi, dx/ is positive. So, at the momet whe L equals 5, dx/ equals 3/ 17. The fial step is to compute dl/ at the momet whe L equals 5. Implicitly differetiatig the equatio, 4 L = x + x, gives, dl dx L = (x + 4x 3 ). Pluggig i for L, x ad dx/ gives, dl 3 ( 5) = (() + 4() 3 ) 17. Solvig gives, at the momet whe L equals 5. dl = 7/ 85.