# Elementary Calculus: Related Rates with Angle Problem

• February 12th 2011, 06:40 PM
evankiefl
Elementary Calculus: Related Rates with Angle Problem
A rocket is launched vertically upward from a point 2 miles west of an observer on the ground. What is the speed of the rocket when the angle of elevation (from the horizontal) of the observer's line of sight to the rocket is 50 degrees and is increasing at 5 degrees per second?

This is what I have done:

$tan(x) = y/2$

$sec^2(x) * dx/dt = (1/2) * dy/dt$

$dy/dt = 2sec^2(50) * 5$

$dy/dt = 24.2 miles/second$

The actual answer is 0.42 miles/second.

Any help is much appreciated.
• February 12th 2011, 07:10 PM
alexmahone
Quote:

Originally Posted by evankiefl
A rocket is launched vertically upward from a point 2 miles west of an observer on the ground. What is the speed of the rocket when the angle of elevation (from the horizontal) of the observer's line of sight to the rocket is 50 degrees and is increasing at 5 degrees per second?

This is what I have done:

$tan(x) = y/2$

$sec^2(x) * dx/dt = (1/2) * dy/dt$

$dy/dt = 2sec^2(50) * 5$

$dy/dt = 24.2 miles/second$

The actual answer is 0.42 miles/second.

Any help is much appreciated.

You need to convert the angles to radians.

$\displaystyle \frac{dy}{dt}=2sec^2(\frac{50\pi}{180})\times\frac {5\pi}{180}=0.4224$
• February 12th 2011, 07:35 PM
evankiefl
Thank you. Why is that so?