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MysticDude
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Homework Statement
A rocket is fired vertically into the air at a rate of 6 mi/min. An observer on the ground is located 4 miles from the launching pad. When the rocket is 3 miles high, how fast is the angle of elevation between the rocket and the observer changing? Be sure to specify units.
Homework Equations
Derivative of tanθ = sec²θ
The Attempt at a Solution
Ok well I know that at that moment that θ is tan-1(3/4) which is 36.870°. Then I went ahead and set up my equation as tanθ = y/x. Taking the derivative of both sides I get:
[tex]\theta'sec^{2}(\theta) = \frac{xy' - yx'}{x^2}[/tex] then I substitute for the values and I get θ'sec²(θ) = 1. Finally, I divide both sides by sec²θ to get my final answer that θ' = .64 degrees per minute.
Thanks for any help guys!