Chapter 3 76 - 226 30 We want to find — when L = 31...

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Unformatted text preview: 226 30. We want to find — when L = 31. WearegiventhatZ—g: =2ft/s.sin0= {80 => $210sin6 => d1: d6 71' 7r d9 — 2 1 6—. = —— = — — 10 dt Ocos dt When 6 4, 2 10 cos 4 dt => d6 2 \/§ x — : ———- = — rad/s dt 10(1/05) 5 . . i . dm , dy 32. P Usmg Q for the orlgin, we are given a = —2 ft / s and need to find E when m z —5. Using the Pythagorean Theorem twice, we have V322 + 122 + y2 + 122 = 39, the total length of the rope. Differentiating Q y _ x dz: y dy With res ecttot, we et ———— + ————— = 0, so p g «9:2 + 122 dt My? + 122 dt dy so My? —— 122 dm ——:f————————<PWWWMHm=#539:\/—52+1P%- 24422213+ 2+122 ¢> dt y GfifiTiii dt ( ) y x/y My? + 122 : 26, and y : V262 - 122 = V532. So when x = —5, dy : (—5)(26) ( 2) — ———10— m —0.87 ft/s. So cart B is moving towards Q at about 0.87 ft/s. dt 4/532(13) x/133 33. (a) By the Pythagorean Theorem. 40002 + y2 2 £2. Differentiating with respect 6 to t, we obtain 2y % = 23:17:. We know that 2—: : 600 ft/s, so when 4000 y : 3000 ft, 2 : \/ 40002 + 30002 : «25,000,000 : 5000 ft and d! y dy 3000 1800 - —-— 600 ————360ft dt 6 t 5000( ) 5 / 2 __y_ i _i_y_ ”ELLE : @26089‘11 (bIIIGBtan6“ 4000 dtaane)" dt(4000) Z? sec dt 4000 t dt 4000 dt dB _ dB dW dL E?_dW’fl;fi dB dt CHAPTER 3 DIFFERENTIATION RULES 18mmg3200mwflflmmw/20HL“? : (0.007. %W_1/3)(0.12 '2'53. L1.53)<_2_0_i> 10,000,000 5 = 01007.§(012.18253)‘“3]0112.253-18153)<——~) % L045><10_8g/yr 107 dy _‘ _ fl _ : When y — 3000 ft‘ E — 600 ft/s. E — 5000 and 0050 ! —_Z # #5000 99 dt “/32 4000 4000 4000 :1 5 (600) 2 0.096 rad/s. ...
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