{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 4 126

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern