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

The rate at which grindstone diameter decreases is and this can be determined by using the given data.

Given :

A 10 cm thick grindstone is initially 200 cm in diameter and it is wearing away at a rate of 50 .

The following steps can be used in order to determine the rate at which grindstone diameter decreases:

Step 1 - According to the given data, the rate at which grindstone volume decreases is:

       --- (1)

Step 2 -  The formula of the volume of the cylinder (grindstone) is given below:

Step 3 - Differentiate the above expression with respect to 'r'.

         --- (2)

Step 4 - So, using the expression (1) and (2) the rate at which grindstone diameter decreases is:

So, the rate at which grindstone diameter decreases is .

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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?

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

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