Math Core

When several quantities are tied together by an equation, and all of them change over time, their rates of change are tied together too. Related rates problems use that connection: you know how fast one quantity changes and you want the rate of another that is harder to measure directly.

The key move: differentiate with respect to time

Suppose a circular ripple spreads on a pond. Its radius $r$ and area $A$ both depend on time $t$ , and at every instant

A = \pi r^2.

Differentiate both sides with respect to $t$ . Because $r$ is a function of $t$ , the chain rule applies, exactly as in implicit differentiation:

\frac{dA}{dt} = 2\pi r \frac{dr}{dt}.

This new equation links the two rates. If you know $r$ and $dr/dt$ at some instant, you know $dA/dt$ at that instant.

A strategy for related rates

Draw a picture and label every quantity that changes with a variable. Label constants with numbers.
Write down the rate you know and the rate you want, in $\dfrac{d}{dt}$ notation, with signs (decreasing quantities have negative rates).
Find an equation relating the variables: the Pythagorean theorem, a volume or area formula, similar triangles, or a trig ratio.
Differentiate implicitly with respect to $t$ .
Substitute the values at the instant in question, then solve for the unknown rate.
Answer in context with units.

Common mistake

Don't substitute the instantaneous values (like " $x = 5$ ") before you differentiate. A quantity that is changing must stay a variable until after step 4; otherwise its derivative becomes 0 and the rate you want disappears. Only true constants, like the length of a ladder, may be plugged in early.

Pythagorean problems

Right triangles appear whenever objects move along perpendicular lines: a ladder against a wall, two cars leaving an intersection, a kite on a string.

Worked example: A sliding ladder

A 13-foot ladder leans against a vertical wall. The bottom slides away from the wall at 2 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 5 feet from the wall?

The ladder, wall and ground form a right triangle with hypotenuse 13 ft.

Solution. Let $x$ be the distance from the wall to the bottom and $y$ the height of the top. We know $\dfrac{dx}{dt} = 2$ and want $\dfrac{dy}{dt}$ when $x = 5$ .

The relationship is $x^2 + y^2 = 169$ . Differentiating,

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

When $x = 5$ , $y = \sqrt{169 - 25} = 12$ . Substituting,

2(5)(2) + 2(12)\frac{dy}{dt} = 0 \quad\Longrightarrow\quad \frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6}.

The top of the ladder is sliding down at $\dfrac{5}{6}$ foot per second. The negative sign shows that $y$ is decreasing.

Worked example: Two cars

Car A leaves an intersection heading north at 60 mph. At the same moment, car B leaves heading east at 45 mph. How fast is the distance between the cars increasing 2 hours later?

Car A travels north, car B travels east, and D is the distance between them.

Solution. Let $y$ and $x$ be the distances of cars A and B from the intersection, and $D$ the distance between them. Then $x^2 + y^2 = D^2$ , and

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2D\frac{dD}{dt}.

After 2 hours, $y = 120$ , $x = 90$ and $D = \sqrt{90^2 + 120^2} = 150$ . Substituting and dividing by 2:

90(45) + 120(60) = 150\frac{dD}{dt} \quad\Longrightarrow\quad \frac{dD}{dt} = \frac{11250}{150} = 75.

The distance between the cars is increasing at 75 mph.

Volume problems and similar triangles

For a cone-shaped tank, the volume formula $V = \frac{1}{3}\pi r^2 h$ has two changing variables. Similar triangles let you eliminate one of them before differentiating.

Worked example: Filling a conical tank

A tank is shaped like an inverted cone with height 12 feet and top radius 4 feet. Water is pumped in at 2 cubic feet per minute. How fast is the water level rising when the water is 6 feet deep?

Water fills the cone to depth h, with surface radius r.

Solution. Let $h$ be the depth and $r$ the radius of the water's surface. By similar triangles, $\dfrac{r}{h} = \dfrac{4}{12}$ , so $r = \dfrac{h}{3}$ . Then

V = \frac{1}{3}\pi\left(\frac{h}{3}\right)^2 h = \frac{\pi h^3}{27}.

Differentiate: $\dfrac{dV}{dt} = \dfrac{\pi h^2}{9}\dfrac{dh}{dt}$ . With $\dfrac{dV}{dt} = 2$ and $h = 6$ :

2 = \frac{36\pi}{9}\frac{dh}{dt} = 4\pi\frac{dh}{dt} \quad\Longrightarrow\quad \frac{dh}{dt} = \frac{1}{2\pi} \approx 0.159.

The water level is rising at $\dfrac{1}{2\pi} \approx 0.159$ feet per minute.

Choosing the right equation

Most related rates problems use one of a small set of relationships. Recognizing which one fits is often the hardest step.

situation	typical equation
perpendicular motion, ladders, kites	Pythagorean theorem $x^2 + y^2 = z^2$
spreading circles, growing squares	area formulas, such as $A = \pi r^2$ or $A = s^2$
balloons, cubes, tanks	volume formulas, such as $V = \frac{4}{3}\pi r^3$ or $V = \frac{1}{3}\pi r^2 h$
shadows, cones	similar triangles (proportions)
angles of elevation	trig ratios, such as $\tan\theta = \dfrac{y}{x}$

When the equation involves an angle, remember that $\dfrac{d}{dt}\tan\theta = \sec^2\theta\,\dfrac{d\theta}{dt}$ , and angle rates must be in radians per unit of time.

Keep in mind that the rates you find are instantaneous. In the ladder example, the top of the ladder does not slide at $\frac{5}{6}$ ft/s for the whole motion; that is its speed only at the moment the bottom is 5 feet out. As the bottom moves farther away, the top falls faster and faster.

Tip

Check the sign of your answer against the story. Water pumped in should make $dh/dt$ positive; a leak should make it negative. A ladder sliding down has negative $dy/dt$ . If the sign disagrees with the picture, look for a sign error in the given rate.

Practice

Practice 1

The side of a square is increasing at 4 centimeters per second. How fast is the area increasing, in square centimeters per second, when the side is 7 centimeters?