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

Answer:

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

Step-by-step explanation:

Given:

Rate of change of radius of cylinder:

$\dfrac{dr}{dt} = +7\ in/sec$

(This is increasing rate so positive)

Rate of change of height of cylinder:

$\dfrac{dh}{dt} = -6\ in/sec$

(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':

$\dfrac{dV}{dt} = \dfrac{d}{dt}(\pi r^2h)$

Let us have a look at the formula:

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

$3.\ \dfrac{dx^n}{dx} = nx^{n-1}$

Applying the two formula for the above differentiation:

$\Rightarrow \dfrac{dV}{dt} = \pi\dfrac{d}{dt}( r^2h)\\\Rightarrow \dfrac{dV}{dt} = \pi h\dfrac{d }{dt}( r^2)+\pi r^2\dfrac{dh }{dt}\\\Rightarrow \dfrac{dV}{dt} = \pi h\times 2r \dfrac{dr }{dt}+\pi r^2\dfrac{dh }{dt}$

Now, putting the values:

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

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