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:

$\displaystyle tan(x) = y/2$

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

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

$\displaystyle dy/dt = 24.2 miles/second $

The actual answer is 0.42 miles/second.

Any help is much appreciated.