Introduction to Related Rates

When a pebble drops into a pond, the ripple’s radius grows, and so does the area inside it. The two rates are connected: if you know how fast the radius is growing, you can work out how fast the area is growing. Related rates problems use derivatives with respect to time to link rates like these. This page covers the core skill; related rates problems puts it to work on harder setups.

Key ideas

Everything is a function of time

In a related rates problem, the changing quantities (radius, area, side length, volume) all depend on time $t$, even if the formula doesn’t show $t$. So $r$ really means $r(t)$, and its rate of change is $\dfrac{dr}{dt}$.

Differentiate both sides with respect to t

Take the equation that links the quantities and differentiate both sides with respect to $t$. This is implicit differentiation, with $t$ as the variable. Every time you differentiate a quantity that depends on $t$, the chain rule attaches its rate:

$$\frac{d}{dt}\left[r^2\right] = 2r\,\frac{dr}{dt}, \qquad \frac{d}{dt}\left[s^3\right] = 3s^2\,\frac{ds}{dt}, \qquad \frac{d}{dt}\left[xy\right] = x\frac{dy}{dt} + y\frac{dx}{dt}$$

Constants stay constants: $\dfrac{d}{dt}[25] = 0$ and $\dfrac{d}{dt}[\pi r^2] = 2\pi r\,\dfrac{dr}{dt}$.

The basic steps

1. Write an equation that links the quantities.
2. Differentiate both sides with respect to $t$.
3. Substitute the values you know at the moment in question.
4. Solve for the unknown rate, and give units.

Substitute after you differentiate. If you plug in $r = 4$ first, $r^2$ becomes the constant $16$, and its derivative becomes $0$, which loses the information you need.

Signs and units

A rate is positive if the quantity is increasing and negative if it’s decreasing. A balloon losing air has $\dfrac{dV}{dt} \lt 0$. Units follow the quantity: area per second is cm²/s, volume per minute is m³/min.

As the radius grows at rate $\dfrac{dr}{dt}$, the area grows at rate $\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$.

Worked examples

Example 1: Differentiating an equation

The variables $x$ and $y$ both depend on $t$, and $x^2 + y^2 = 25$. Find $\dfrac{dy}{dt}$ when $x = 3$, $y = 4$, and $\dfrac{dx}{dt} = 2$.

Solution. Differentiate both sides with respect to $t$:

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

Substitute:

$$2(3)(2) + 2(4)\frac{dy}{dt} = 0 \quad\Rightarrow\quad 8\frac{dy}{dt} = -12 \quad\Rightarrow\quad \frac{dy}{dt} = -\frac{3}{2}$$

Example 2: A ripple

The radius of a circular ripple increases at $0.5$ m/s. How fast is the area inside it increasing when the radius is $4$ m?

Solution. Know: $\dfrac{dr}{dt} = 0.5$ m/s. Want: $\dfrac{dA}{dt}$ when $r = 4$.

$$A = \pi r^2 \quad\Rightarrow\quad \frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}$$ $$\frac{dA}{dt} = 2\pi(4)(0.5) = 4\pi \approx 12.566$$

The area is increasing at $4\pi \approx 12.566$ m² per second.

Example 3: A growing square

Each side of a square grows at $3$ cm/s. How fast is the area growing when the side is $10$ cm? How fast is the perimeter growing?

Solution. Let $s$ be the side length, so $\dfrac{ds}{dt} = 3$.

$$A = s^2 \quad\Rightarrow\quad \frac{dA}{dt} = 2s\frac{ds}{dt} = 2(10)(3) = 60$$ $$P = 4s \quad\Rightarrow\quad \frac{dP}{dt} = 4\frac{ds}{dt} = 12$$

The area grows at $60$ cm²/s and the perimeter at $12$ cm/s. Notice the perimeter’s rate doesn’t depend on $s$, but the area’s does: a bigger square gains more area for the same growth in side length.

Example 4: Inflating a balloon

Air is pumped into a spherical balloon at $100$ cm³/s. How fast is the radius increasing when the radius is $5$ cm?

Solution. Know: $\dfrac{dV}{dt} = 100$. Want: $\dfrac{dr}{dt}$ when $r = 5$.

$$V = \frac{4}{3}\pi r^3 \quad\Rightarrow\quad \frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}$$ $$100 = 4\pi(25)\frac{dr}{dt} \quad\Rightarrow\quad \frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \approx 0.318$$

The radius is increasing at $\dfrac{1}{\pi} \approx 0.318$ cm/s.

Common mistakes

Substituting before differentiating. If you put $r = 5$ into $V = \frac{4}{3}\pi r^3$ first, $V$ becomes a constant and you get $\dfrac{dV}{dt} = 0$. Differentiate the general equation, then substitute.

Forgetting the chain rule factor. $\dfrac{d}{dt}[r^2]$ is $2r\,\dfrac{dr}{dt}$, not just $2r$. Every variable that changes with time brings its own rate.

Getting the sign wrong. If something is shrinking, draining, or melting, its rate is negative. Put the negative sign in when you substitute.

Mixing up which rate is given. Write “Know:” and “Want:” with the derivative notation before you start, like $\dfrac{dV}{dt} = 100$ and $\dfrac{dr}{dt} = \,?$.

Leaving out units. A rate needs units, such as cm²/s for an area changing over time.

Practice

1. (Warm-up) The side $s$ of a square depends on time. Differentiate $A = s^2$ with respect to $t$.

Solution

$$\frac{dA}{dt} = 2s\,\frac{ds}{dt}$$

2. (Warm-up) The edge of a cube grows at $2$ cm/s. How fast is the volume increasing when the edge is $5$ cm?

Solution

$V = s^3$, so $\dfrac{dV}{dt} = 3s^2\dfrac{ds}{dt} = 3(25)(2) = 150$.

The volume is increasing at $150$ cm³/s.

3. (Warm-up) $y = x^2 + 3x$, where $x$ and $y$ depend on $t$. Find $\dfrac{dy}{dt}$ when $x = 2$ and $\dfrac{dx}{dt} = 4$.

Solution

$$\frac{dy}{dt} = (2x + 3)\frac{dx}{dt} = (4 + 3)(4) = 28$$

4. (Core) The area of a circle increases at $10$ cm²/s. How fast is the radius increasing when the radius is $2$ cm?

Solution

$$\frac{dA}{dt} = 2\pi r\frac{dr}{dt} \quad\Rightarrow\quad 10 = 2\pi(2)\frac{dr}{dt} \quad\Rightarrow\quad \frac{dr}{dt} = \frac{10}{4\pi} = \frac{5}{2\pi} \approx 0.796$$

The radius is increasing at about $0.796$ cm/s.

5. (Core) $xy = 12$, where $x$ and $y$ depend on $t$. Find $\dfrac{dy}{dt}$ when $x = 4$ and $\dfrac{dx}{dt} = 3$.

Solution

When $x = 4$, $y = \dfrac{12}{4} = 3$. Differentiate with the product rule:

$$x\frac{dy}{dt} + y\frac{dx}{dt} = 0 \quad\Rightarrow\quad 4\frac{dy}{dt} + 3(3) = 0 \quad\Rightarrow\quad \frac{dy}{dt} = -\frac{9}{4}$$

6. (Core) An ice cube melts so that its volume decreases at $6$ cm³/min. It stays a cube. How fast is its surface area changing when the edge is $3$ cm?

Solution

First find $\dfrac{ds}{dt}$ from $V = s^3$:

$$\frac{dV}{dt} = 3s^2\frac{ds}{dt} \quad\Rightarrow\quad -6 = 27\frac{ds}{dt} \quad\Rightarrow\quad \frac{ds}{dt} = -\frac{2}{9}$$

Then use $S = 6s^2$:

$$\frac{dS}{dt} = 12s\frac{ds}{dt} = 12(3)\left(-\frac{2}{9}\right) = -8$$

The surface area is decreasing at $8$ cm²/min.

7. (Core) The circumference of a circle increases at $6$ cm/s. How fast is the area increasing when the radius is $5$ cm?

Solution

$C = 2\pi r$, so $\dfrac{dC}{dt} = 2\pi\dfrac{dr}{dt}$, which gives $\dfrac{dr}{dt} = \dfrac{6}{2\pi} = \dfrac{3}{\pi}$.

$$\frac{dA}{dt} = 2\pi r\frac{dr}{dt} = 2\pi(5)\cdot\frac{3}{\pi} = 30$$

The area is increasing at $30$ cm²/s.

8. (Challenge) A rectangle’s length increases at $2$ cm/s while its width decreases at $1$ cm/s. When the length is $8$ cm and the width is $5$ cm:

• (a) Is the area increasing or decreasing, and how fast?
• (b) How fast is the diagonal changing? (3 decimal places)

Solution

(a) $A = LW$, so

$$\frac{dA}{dt} = L\frac{dW}{dt} + W\frac{dL}{dt} = 8(-1) + 5(2) = 2$$

The area is increasing at $2$ cm²/s.

(b) $D^2 = L^2 + W^2$, so $2D\dfrac{dD}{dt} = 2L\dfrac{dL}{dt} + 2W\dfrac{dW}{dt}$. Here $D = \sqrt{64 + 25} = \sqrt{89}$:

$$\sqrt{89}\,\frac{dD}{dt} = 8(2) + 5(-1) = 11 \quad\Rightarrow\quad \frac{dD}{dt} = \frac{11}{\sqrt{89}} \approx 1.166$$

The diagonal is increasing at about $1.166$ cm/s.

9. (Challenge) Air is pumped into a spherical balloon at $50$ cm³/s. How fast is its surface area increasing when the radius is $10$ cm? ($S = 4\pi r^2$)

Solution

From $V = \frac{4}{3}\pi r^3$:

$$50 = 4\pi(10)^2\frac{dr}{dt} \quad\Rightarrow\quad \frac{dr}{dt} = \frac{50}{400\pi} = \frac{1}{8\pi}$$

Then

$$\frac{dS}{dt} = 8\pi r\frac{dr}{dt} = 8\pi(10)\cdot\frac{1}{8\pi} = 10$$

The surface area is increasing at $10$ cm²/s.