Question

A cylinder gets taller at a rate of $3$ inches per second, but the radius shrinks at a rate of $1$ inch per second. How fast is the volume of the cylinder changing when the height is $20$ inches and the radius is $10$ inches?

Answer 1

Given that the cylinder gets taller at a rate of $3$ inches per second i.e., $\dfrac{{dh}}{{dt}} = 3$ inches per second and also given, the radius shrinks at a rate of $1$ inch per second i.e., $\dfrac{{dr}}{{dt}} = - 1$ inch per second. And we know that the volume of a cylinder is $V = \pi {r^2}h$, where $r$ is the radius of the cylinder and $h$ is the height of the cylinder. We will differentiate $V$ w.r.t to $t$ and then we will substitute the values of $\dfrac{{dh}}{{dt}}$ and $\dfrac{{dr}}{{dt}}$. We will calculate the required rate of change of the volume by putting $h$ as $20$ inches and $r$ as $10$ inches.

Complete step by step answer:
Given that the cylinder gets taller at a rate of $3$ inches per second i.e., $\dfrac{{dh}}{{dt}} = 3$ inches per second and also given, the radius shrinks at a rate of $1$ inch per second i.e., $\dfrac{{dr}}{{dt}} = - 1$ inch per second. Now, as we know that volume of a cylinder is given by,
$\pi {r^2}h$
where $r$ is the radius of the cylinder and $h$ is the height of the cylinder i.e., $V = \pi {r^2}h$.

Differentiating both the sides of $V = \pi {r^2}h$ with respect to $t$, we get
$\Rightarrow \dfrac{d}{{dt}}\left( V \right) = \dfrac{d}{{dt}}\left( {\pi {r^2}h} \right)$
Since, $\pi$ is a constant. So, we get
$\Rightarrow \dfrac{{dV}}{{dt}} = \pi \dfrac{d}{{dt}}\left( {{r^2}h} \right)$
From the product rule of differentiation, we know $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$. Using this, we get
$\Rightarrow \dfrac{{dV}}{{dt}} = \pi \left[ {{r^2}\dfrac{d}{{dt}}\left( h \right) + h\dfrac{d}{{dt}}\left( {{r^2}} \right)} \right]$

On differentiating, we get
$\Rightarrow \dfrac{{dV}}{{dt}} = \pi \left[ {{r^2}\dfrac{{dh}}{{dt}} + \left( {2r} \right)h\dfrac{{dr}}{{dt}}} \right]$
On rewriting, we get
$\Rightarrow \dfrac{{dV}}{{dt}} = \pi {r^2}\dfrac{{dh}}{{dt}} + 2\pi rh\dfrac{{dr}}{{dt}}$
Using the given values of $\dfrac{{dh}}{{dt}}$ and $\dfrac{{dr}}{{dt}}$, we get
$\Rightarrow \dfrac{{dV}}{{dt}} = \pi {r^2} \times \left( 3 \right) + 2\pi rh \times \left( { - 1} \right)$

Now, we have to find how fast the volume of the cylinder changes when the height is $20$ inches and the radius is $10$ inches. Therefore, we get
$\Rightarrow \dfrac{{dV}}{{dt}} = \pi \times {\left( {10} \right)^2} \times \left( 3 \right) + 2\pi \times \left( {10} \right) \times \left( {20} \right) \times \left( { - 1} \right)$
On simplifying, we get
$\Rightarrow \dfrac{{dV}}{{dt}} = 300\pi - 400\pi$
$\therefore \dfrac{{dV}}{{dt}} = 100\pi$

Therefore, the volume of the cylinder changes at a rate of $100\pi {\text{ }}\dfrac{{inche{s^3}}}{s}$ when the height is $20$ inches and the radius is $10$ inches.

Note: When a value $y$ varies with $x$ such that it satisfies $y = f(x)$, then the rate of change of $y$with respect to $x$ is $f'(x) = \dfrac{{dy}}{{dx}}$. If two variable $x$ and $y$ vary with respect to another variable $t$ such that $x = f(t)$ and $y = g(t)$, then using chain rule we can define $\dfrac{{dy}}{{dx}} = \dfrac{{\left( {\dfrac{{dy}}{{dt}}} \right)}}{{\left( {\dfrac{{dx}}{{dt}}} \right)}}$, if $\dfrac{{dx}}{{dt}} \ne 0$.