Question

A winch at the top of a 12-meter building pulls a pipe of the same length to a vertical position, as shown in the figure. The winch pulls in rope at a rate of -0.6 meter per second (ds/dt = -0.6 m/sec). Find the rate of vertical change and the rate of horizontal change at the end of the pipe when y = 1. (Round your answers to three decimal places.) horizontal -0.5 m/sec vertical m/sec

Answer 1

Given {x,y) B 12m -12m- Age so now given that ds = – 0.6 m dx = 1, du = ? When y=1 from, Triangle ABE we can say that (x² + y2 = 122o . so now apply implicit derivative on both sides with respect to time to so ( x + y2) = 5 (122 2x du ty du l=0 av de -zy be Now from Triangle BCE we can say that s² = x²+(12-y)2 So now apply implicit deqi vative on both sides with respect to time t. so I (52) = a (x²7 (12-432)

2s 2x +2412-y) (12-y) 25 = 2x dx +2612–) 10-dan Is de = x de - (12-y) dy from we can say that a y du so sode = y de - (12-y) dy s de =1-4- (12-yl) d4 = -4 -12 ty du s de = 42 dy Now find so when y=1 Now find x: by plugging y=1 in 6.so x+12=122 =).?=144-) = 143 x² = (1 14372 S.O x = V143 1 Now put all the values in 6.so . s² =W143)²+(12-1) 2 = 143+121 s²=264 S=1264 so Now plug all the values in . so

That is 264 (-0.6) = 0.812 40 38 4046 12 y = 0-81240384046 mls. 0.8129 mls = 0.812 mis now put this value in ..so x dx = -y dy 1143 del = (-1)(0.812-40384046) da -0.82403840u6 x = y 143 dx=-0.06793662205 m/s Idne =-0.0689 dr . Aul | dex = -0.0.68 ml