Posted by Linda on .
A projectile is fired upward 200 ft from an observer. The height in feet of the projectile after t seconds is given by s=16t2 + 180t. What is the rate of change of the angle of elevation of the observer to the projectile after 7 seconds?
I don't understand how to get there!

Math 
Steve,
if θ is the angle of elevation,
y = 16t^2 + 180t
tanθ = y/200
sec^2θ dθ/dt = 1/200 dy/dt = 1/200 (32t + 180)
= 32/200 (t5)
dθ/dt = 4/25 (t5) cos^2θ
= 4/25 (t5) (200^2/(200^2 + y^2)
y(7) = 16*49 + 180*7 = 476
at t=7,
dθ/dt = 4/25 * 2 * 40000/266576 = .048
Seems kinda small, so you better check the details.