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Old May 1st 2009, 01:25 AM   #1
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Problem on Application of calculus in physics.

At a distance of 4000 ft from the launch site, a spectator is observing a rocket being launched. If the rocket lifts off vertically and is rising at a speed of 600 ft/s when it is at an altitude of 3000 ft, the distance between rocket and the spectator is changing at that instant at the rate ------- ?
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Old May 1st 2009, 04:01 AM   #2
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hi

This is a related rates problem. First you list some relationship between the sides of a triangle.

Let x be the distance from the spectactor to the rocket, and y the rockets height.

$\displaystyle 4000^{2}+y^{2}=x^{2}$

You know that, when $\displaystyle y = 3000 \; \mbox{ we have }\frac{dy}{dt}=600 $

Now differentiating the first expression with respect to time we get:

$\displaystyle 2\cdot y \cdot \frac{dy}{dt} = 2\cdot x \cdot \frac{dx}{dt} $

Now, solve for $\displaystyle \frac{dx}{dt} $ .

$\displaystyle \frac{dx}{dt} = \frac{y}{x}\cdot \frac{dy}{dt} = \frac{3000}{5000}\cdot 600 = 360 ft/s$
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