Question

The radius of a right circular cylinder is increasing at the rate of 7 in./sec, while the height is decreasing at the rate of 6 in./sec. At what rate is the volume of the cylinder changing when the radius is 20 in. and the height is 16 in.

Answer 1

`\approx \bold{6544\ in^3/sec}`

Explanation:

Given:

Rate of change of radius of cylinder:

`(dr)/(dt) = +7\ in/sec`

(This is increasing rate so positive)

Rate of change of height of cylinder:

`(dh)/(dt) = -6\ in/sec`

(This is decreasing rate so negative)

To find:

Rate of change of volume when r = 20 inches and h = 16 inches.

Solution:

First of all, let us have a look at the formula for Volume:

`V = \pi r^2h`

Differentiating it w.r.to 't':

`(dV)/(dt) = (d)/(dt)(\pi r^2h)`

Let us have a look at the formula:

`1.\ (d)/(dx) (C.f(x)) = C(d(f(x)))/(dx) \ \ \ (\text{C is a constant})\\2.\ (d)/(dx) (f(x).g(x)) = f(x)(d)/(dx) (g(x))+g(x)(d)/(dx) (f(x))`

`3.\ (dx^n)/(dx) = nx^(n-1)`

Applying the two formula for the above differentiation:

`\Rightarrow (dV)/(dt) = \pi(d)/(dt)( r^2h)\\\Rightarrow (dV)/(dt) = \pi h(d )/(dt)( r^2)+\pi r^2(dh )/(dt)\\\Rightarrow (dV)/(dt) = \pi h* 2r (dr )/(dt)+\pi r^2(dh )/(dt)`

Now, putting the values:

`\Rightarrow (dV)/(dt) = \pi * 16* 2* 20 * 7+\pi* 20^2* (-6)\\\Rightarrow (dV)/(dt) = 22 * 16* 2* 20 +3.14* 400* (-6)\\\Rightarrow (dV)/(dt) = 14080 -7536\\\Rightarrow (dV)/(dt) \approx \bold{6544\ in^3/sec}`

So, the answer is:
`\approx \bold{6544\ in^3/sec}`