• anonymous
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?
  • chestercat
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  • anonymous
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

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