How Fast Is the Angle of Elevation Changing as the Rocket Ascends?

[MysticDude]

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:
\theta'sec^{2}(\theta) = \frac{xy' - yx'}{x^2} 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!

 

[LCKurtz]

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:
\theta'sec^{2}(\theta) = \frac{xy' - yx'}{x^2} 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!

Not sure exactly what your question is. You have the correct answer. You might have come to it easier if you had started with the equation tan(θ) = y/4 instead of y/x. You don't need the quotient rule because x is a constant. This gives you directly that

sec²(θ) θ' = y'/4 = 3/2

 

[MysticDude]

Not sure exactly what your question is. You have the correct answer. You might have come to it easier if you had started with the equation tan(θ) = y/4 instead of y/x. You don't need the quotient rule because x is a constant. This gives you directly that

sec²(θ) θ' = y'/4 = 3/2

Hehe, oopsie. Thanks for the help LCKurtz! :D