# Thread: rates of change using inverse trig

1. ## rates of change using inverse trig

A helicopter leaves a base, rising straight up at a speed of 15m/s. At the same time that the helicopter leaves, an observer starts from a point 100m away from the base and moves on a straight line away from the base at a speed of 80m/s. How fast is the angel of elevation from the observer to the helicopter increasing when the observer is
a) 400m from the base
b) 600m from the base

Thanks i;m not sure which rates i have to find.

2. ?? You are told exactly which rate- "How fast is the angle of elevation from the observer to the helicoper increasing". Draw a picture- a right triangle having the "observer" at one vertex, the helicopter at another, and the right angle directly beneath the helicopter. The "angle of elevation" is the angle of that triangle at the observer's vertex. You want to find $\displaystyle \frac{d\theta}{dt}$. Note that both the height of the helicopter (the "far side" of the right triangle) and the distance to the observer (the "near side" of the right triangle) are functions of t.