Question

A man is walking at at the rate of $4.5xkm/hr$ towards the foot of the tower $120xm$ high. At what rate is he approaching the top of the tower when he is $50xm$ away from the tower?

Answer 1

Let distance between the tower and man be $x$

then $\frac{dx}{dt}=4.5km/hr=1.25m/s$

$y^2={120}^2+x^2⟹y\frac{dy}{dt}=x\frac{dx}{dt}$

When $x=50⟹y=130$

$\frac{dy}{dt}=\frac{x}{y}\frac{dx}{dt}=\frac{5}{13}\times 1.25=0.48m/s$

the rate at which the person is approaching the top of tower is $0.48m/s$