Question

A man is walking at the rate of 6.5 km/hr towards the foot of a tower 120m high. At what rate is he approaching the top of the tower when he is 50m away from the tower.

Answer 1

hope it will be helpful to you.

Answer 2

dx/dt = 6.5 km/ hr

dy/dt = ?

x² +120² = y²

2x dx/dt = 2y dy/dt

xdx/dt = ydy/dt

x(6.5) = y dy/dt

dy/dt = 6.5x/y --- (1)

When x = 50 then y² = 50² + 120²

= 25000 + 14400

y² = 16900

y = 130

We will now put the value x = 50 and y = 130 in the equation 1

dy/dt  = 6.5 x 50/130 = 2.5

dy/dt = 2.5km/hr