Application of differentiation ~~~~~~~

let the length of its shadow is x, the angle between the hypotenuse and theshadow is θ



tanθ=180/x



x=180/tanθ



dθ/dt=0.25



dx/dt



=(dx/dθ)(dθ/dt)



=180(dcotθ/dθ)(0.25)



=45(-csc^2θ)



so, when the elevation of the sun is 30 degree,



the rate of change of the length of its shadow



=45(-csc^230)



=-180 m/min