Question

A balloon released at point A rises vertically with a constant speed of 5m/s. Point B is level with and 100m distant from point A. How fast is the angle of elevation of the balloon at B changing when the balloon is 200m above A?

Answer 1

let at any time t height of balloon is x m.
and angle of elevation is y...we have to find dy/dt.
tan y=x/100
(sec^2y)dy/dx=1/100
dy/dx=1/(100sec^2y)
(dy/dt)(dt/dx)=1/(100sec^2y)
dy/dt=1/(100sec^2y) .(dx/dt)
put the value from question
dy/dt=1/(100sec^2y).(5)....................1
now sec^2y=1+tan^2y=1+(200/100)^2=5
put this value in equation 1
dy/dt=1/500.(5)
dy/dt=1/100