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Chapter 4 126 - 388 CHAPTER 4 APPLlCATIONS OF...

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Unformatted text preview: 388 CHAPTER 4 APPLlCATIONS OF DIFFERENTIATlON (b) f(—1) = 13. f(7) = —1939. f(6.810) Rs —1949.07. f(—0.455) z —6.912.f(0.645) m 10.982. Therefore. f(6.810) % —1949.07 is the absolute minimum correct to two decimal places. 36. f(;1:) = :62 + since => f’(x) = 23: + cos x. f’(;c) exists for all as. so to '0 find the minimum of f. we can examine the zeros of f’. From the graph of V f'. we see thatagood choice for 331 is :01 = —0.5. Use g(m) = 2m+c0sm _10 I 10 and g’(;c) = 2 — sinw to obtain 332 z —0.450627. 223 z 70.450184 % $4. Since f”(x) = 2 — sinx > 0 for all it. ‘10 f(~0.450184) % 70.232466 is the absolute minimum. 37. From the figure. we see that y : fix) : etc” is periodic with period 271'. To find the m—coordinates of the IP. we only need to approximate the zeros ofy” on [0.7r]. f’(a:) = —eC°” sinw : f”(w) = eCOS 3” (sin2 ac — cos an). Since 6“)” ¢ 0. we will use Newton‘s method with g(;1:) 2 she2 m — cos 3:. g'(w) : 2sinm cosac + sin x, and m1 = 1. 3:2 % 0.904173. 3:3 % 0.904557 % :34. Thus. (0.904557,1.855277) is the IP. 38. 2 f(m) = —sinx => f’(a:) : , cosm. At :u = a. the slope of the tangent line is ' a. = — cos a. The line through the origin and a. f a ,2” All” _Sma_0 V' is y : was. If this line is to be tangent to f at a: : a, then its —2 _ S. slope must equal f'(a). Thus. ma : , cosa :> tana = a. a To solve this equation using Newton‘s method. let 9(23) 2 tanm * :c, w with $1 = 4.5 (estimated from the figure). 332 % 4.493614. / 2 mzsec :1:~1,andm 1:33 1 g( ) "Jr R sec2 2:“ — 1 223 2 4.493410, 934 z 4.493409 m 105. Thus. the slope of the line that has the largest slope is f’(m5) w 0.217234. 39. 56000 The volume of the silo. in terms of its radius, is V0") : 7rr2(30) + %(§7r'r3) : 3071’7‘2 + §7rr3. From a graph of V. we see that V0") : 15.000 at r w 11 ft. Now we use 7‘ Newton‘s method to solve the equation V(r) — 15.000 : 0. 0 20 dV 30m: + gm: — 15.000 —— : 2 z n 1 . T k‘ 607W + 271'7‘ . so Tn+1 1" 60717171 + 27r7‘2L a mg dr r1 : 11. we get m % 11.2853, 73 % 11.2807 x 7‘4. So in order for the silo to hold 15.000 ft3 of grain. its radius must be about 11.2807 ft. ...
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