A weather balloon is released 60 feet from an observer with a camera. if the balloon is rising at 10 ft. per second how fast is the angle of the camera changing when the balloon is 100 ft. high?
I did the problem and got 14.29 radians per second, but I'm not too sure that i did it right. Thanks in advance
First I made a right triangle with th ebase as 60, the right leg as 100, the hypotnuse as s, and the angle as theta
then i made the equation tanx=(116.62/60)10
116.62 was my hyponuse
I took the derivative of the equation and got sec^2 x (dx/dt)=(116.62/60)10
I then solved for dx/dt and got my answer