Searched through previous threads, cannot find one like this listed.

A Weather Balloon that is rising vertically is being observed from a point on the ground 2 miles from the spot directly beneath the balloon. At what rate is the balloon rising when the angle between the ground and the observer's line of sight is pi/5 and is increasing at 0.2 radians per minute?

x = theta

h = height of balloon

dx/dt = ?

dh/dt = .2 radians per minute

So distance between person and balloon is 2 miles. So...tan(x) = h/2

differentiate: tan(x) = h/2 --> sec^2(x)(dx/dt)=(1/2)(dh/dt)

2sec^2(dh/dt)=dx/dt ---> 2sec^2(.2) = dx/dt...

this is where i get lost if i already didn't mess up bad to begin with...