Question

A 10 cm thick grindstone is initially 200 cm in diameter, and it is wearing away at a rate of LaTeX: 50cm^3/hr. At what rate is it's diameter decreasing

Answer 1

Complete question is;

A 10 cm thick grindstone is initially 200 cm in diameter, and it is wearing away at a rate of 50 cm/hr. At what rate is its diameter decreasing?

Answer:

Diameter is decreasing at the rate of 5/(2πr) cm/hr

Step-by-step explanation:

We are told the stone is wearing away at a rate of 50 cm/hr. This means the volume is decreasing. Thus;

dV/dt = -50 cm/hr

Now, a grindstone is in the shape of a cylinder. Thus, volume of grindstone is;

V = πr²h

dV/dr = 2πrh

Now,to find the rate at which the diameter is decreasing, we'll write;

dr/dt = (dV/dt)/(dV/dr)

dr/dt = -50/(2πrh)

We are given;

Diameter = 200 cm

Radius; r = 200/2 = 100 cm

Thickness; h = 10 cm

Thus;

dr/dt = -50/(2π × r × 10)

dr/dt = -5/(2πr) cm/hr