Related rates

Example 1: A stone falls into a pond and makes a circular ripple. The radius grows at $2$ cm/s. How fast is the area inside the ripple growing when the radius is $10$ cm?

Solution:

Let $r$ be the radius and $A$ the area.

We know: $\dfrac{dr}{dt}=2$ cm/s.

We want: $\dfrac{dA}{dt}$ when $r=10$ cm.

The equation: $A=\pi r^2$.

Differentiate with respect to $t$:

$\dfrac{dA}{dt}=2\pi r\cdot\dfrac{dr}{dt}$

Now put in $r=10$ and $\dfrac{dr}{dt}=2$:

$\begin{align*}&\dfrac{dA}{dt}\\&=2\pi(10)(2)\\&=40\pi\approx 125.7\ \text{cm}^2\text{/s}\end{align*}$

The area grows faster as the ripple gets bigger, even though the radius grows at a steady rate.

Example 2: Air is pumped into a spherical balloon at $100$ cm$^3$/s. How fast is the radius growing when the radius is $25$ cm?

Solution:

Let $r$ be the radius and $V$ the volume.

We know: $\dfrac{dV}{dt}=100$ cm$^3$/s.

We want: $\dfrac{dr}{dt}$ when $r=25$ cm.

The volume of a sphere is

$V=\dfrac{4}{3}\pi r^3$.

Differentiate with respect to $t$:

$\begin{align*}&\dfrac{dV}{dt}\\&=\dfrac{4}{3}\pi\cdot 3r^2\cdot\dfrac{dr}{dt}\\&=4\pi r^2\cdot\dfrac{dr}{dt}\end{align*}$

Put in the numbers:

$100=4\pi(25)^2\cdot\dfrac{dr}{dt}$

$\begin{align*}&\dfrac{dr}{dt}\\&=\dfrac{100}{2500\pi}\\&=\dfrac{1}{25\pi}\approx 0.0127\ \text{cm/s}\end{align*}$

The radius grows slowly, because a big balloon needs a lot of air to grow a little.

Example 3: A $10$-foot ladder leans against a wall. The bottom of the ladder slides away from the wall at $1$ ft/s. How fast is the top sliding down the wall when the bottom is $6$ feet from the wall?

Solution:

Let $x$ be the distance from the wall to the bottom of the ladder. Let $y$ be the height of the top.

We know: $\dfrac{dx}{dt}=1$ ft/s.

We want: $\dfrac{dy}{dt}$ when $x=6$ ft.

The equation, by the Pythagorean theorem:

$x^2+y^2=100$

Differentiate with respect to $t$:

$2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$

We also need $y$ at this moment.

When $x=6$:

$y=\sqrt{100-36}=8$ ft

Now put in the numbers:

$\begin{align*}2(6)(1)+2(8)\dfrac{dy}{dt}&=0\\16\dfrac{dy}{dt}&=-12\\\dfrac{dy}{dt}&=-\dfrac{3}{4}\ \text{ft/s}\end{align*}$

The rate is negative because $y$ is getting smaller. The top slides down the wall at $\dfrac{3}{4}$ ft/s.

Example 4: Two cars leave an intersection. Car A drives north at $60$ mi/h. Car B drives east at $80$ mi/h. How fast is the distance between the cars growing when car A is $3$ miles north and car B is $4$ miles east of the intersection?

Solution:

Let $y$ be the distance of car A from the intersection, $x$ the distance of car B, and $z$ the distance between the cars.

We know: $\dfrac{dy}{dt}=60$ mi/h and

$\dfrac{dx}{dt}=80$ mi/h.

We want: $\dfrac{dz}{dt}$ when $x=4$ and $y=3$.

The equation: $z^2=x^2+y^2$.

Differentiate with respect to $t$, and divide both sides by $2$:

$z\dfrac{dz}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$

At this moment, $z=\sqrt{4^2+3^2}=5$ miles. Put in the numbers:

$\begin{align*}5\dfrac{dz}{dt}&=4(80)+3(60)\\5\dfrac{dz}{dt}&=500\\\dfrac{dz}{dt}&=100\ \text{mi/h}\end{align*}$

The cars are moving apart at $100$ mi/h.

Example 5: A tank is shaped like a cone with its point at the bottom. The tank is $10$ m tall, and the radius of its top is $4$ m. Water is pumped in at $2$ m$^3$/min. How fast is the water level rising when the water is $5$ m deep?

Solution:

Let $h$ be the depth of the water and $r$ the radius of the water surface.

We know: $\dfrac{dV}{dt}=2$ m$^3$/min.

We want: $\dfrac{dh}{dt}$ when $h=5$ m.

The volume of a cone is

$V=\dfrac{1}{3}\pi r^2h$.

This has two changing variables, $r$ and $h$. The water forms a smaller cone with the same shape as the tank. By similar triangles,

$\dfrac{r}{h}=\dfrac{4}{10}$,   so   $r=\dfrac{2h}{5}$.

Replace $r$ in the volume formula:

$\begin{align*}&V\\&=\dfrac{1}{3}\pi\left(\dfrac{2h}{5}\right)^2h\\&=\dfrac{4\pi}{75}h^3\end{align*}$

Differentiate with respect to $t$:

$\begin{align*}&\dfrac{dV}{dt}\\&=\dfrac{4\pi}{75}\cdot 3h^2\cdot\dfrac{dh}{dt}\\&=\dfrac{4\pi}{25}h^2\cdot\dfrac{dh}{dt}\end{align*}$

Put in the numbers, $h=5$ and $\dfrac{dV}{dt}=2$:

$\begin{align*}2&=\dfrac{4\pi}{25}(25)\dfrac{dh}{dt}\\2&=4\pi\dfrac{dh}{dt}\\\dfrac{dh}{dt}&=\dfrac{1}{2\pi}\approx 0.159\ \text{m/min}\end{align*}$

The water level rises at about $0.16$ m per minute.