3.8.11-eg-1

# 3.8.11-eg-1 - 100 2 b 2 t = c 2 t(1 so diﬀerentiating...

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Example A hot air balloon is released from its launching area and watched by a crowd in a viewing area 100 feet away from the launch site. What is the vertical speed of the balloon when its distance from the viewing area is 300 feet and that distance is increasing at a rate of 600 ft/hr? Solution: Let b ( t ) denote the changing height of the balloon, and let c ( t ) denote the changing distance from the balloon to the viewing area. We are given that at some instant t = t ? hour, c ( t ? ) = 300 ft , and c 0 ( t ? ) = 600 ft/hr , and we need to determine is the rate of change of b ( t ) at that same instant t ? . That is, we want : b 0 ( t ? ) . To relate the variables, we can use the fact that c ( t ) and b ( t ) are sides of a right triangle, with the base of the triangle of length 100 ft. Using the Pythagorean Theorem,

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Unformatted text preview: 100 2 + b 2 ( t ) = c 2 ( t ) , (1) so diﬀerentiating this equation with respect to t , 2 b ( t ) b ( t ) = 2 c ( t ) c ( t ) . Solving for b ( t ), we have b ( t ) = c ( t ) c ( t ) b ( t ) , so at the time t ? of interest it follows that b ( t ? ) = c ( t ? ) c ( t ? ) b ( t ? ) = 300 · 600 b ( t ? ) . In order to determine the missing value b ( t ? ) we note that equation (1) holds for all time, and so in particular it holds for t = t ? . That is, 100 2 + b 2 ( t ? ) = c 2 ( t ? ) = 300 2 , so b ( t ? ) = √ 300 2-100 2 = √ 80000 . It follows that b ( t ? ) = 300 · 600 √ 80000 = 180000 200 √ 2 = 450 √ 2 ft/hr . 2...
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## This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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3.8.11-eg-1 - 100 2 b 2 t = c 2 t(1 so diﬀerentiating...

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