What is the first step in a related rates problem?

Draw a picture of the situation at a general moment and give every changing quantity a letter. Then write down the rate you know and the rate you want as derivatives with respect to time.

Why do you differentiate with respect to time?

Because the rates in the problem (feet per second, liters per minute) are all changes over time. Differentiating the linking equation with respect to t turns a relation between quantities into a relation between their rates.

When do you plug in the numbers in related rates?

After differentiating. Constants that never change (a ladder's length, a tank's dimensions) can go in earlier, but changing quantities must stay as letters until the equation has been differentiated.

Why is my related rates answer negative?

A negative rate means that quantity is decreasing, like the top of a sliding ladder moving down. Check that the sign matches the situation, and state the answer as "decreasing at" a positive speed if the question asks how fast.

How are related rates and implicit differentiation connected?

Related rates is implicit differentiation with respect to time: every quantity in the equation is a hidden function of t, so the chain rule attaches a rate to each.

Related Rates Problems: Step-by-Step Method and Examples

When two quantities are linked by an equation, their rates of change are linked too. Pump air into a balloon and its radius grows; slide the bottom of a ladder away from a wall and its top slides down. A related rates problem gives you the rate at which one quantity changes and asks for the rate at which a linked quantity changes, at one particular moment.

All of these rates are derivatives with respect to time, $t$ . If $x$ is a distance in feet, $\frac{d x}{d t}$ is how fast it changes, in feet per second; if $V$ is a volume in cubic centimeters, $\frac{d V}{d t}$ is in cubic centimeters per second.

The details change from problem to problem, but the same six steps solve all of them:

Picture. Draw a diagram of the situation at a general moment, not just the special moment in the question.
Name. Give a letter to every quantity that changes. Write down the rate you know and the rate you want, as derivatives with respect to $t$ .
Relate. Write an equation linking the quantities that is true at every moment: from the Pythagorean theorem, an area or volume formula, similar triangles, or trigonometry.
Differentiate both sides with respect to $t$ . By the chain rule, each changing quantity brings its own rate: $\frac{d}{d t} x^{2} =$ $2 x \frac{d x}{d t}$ .
Substitute the values at the special moment, then solve for the rate you want.
Interpret. State the answer with units, and check its sign makes sense.

Step 4 is implicit differentiation, with $t$ in place of $x$ : every quantity that changes is treated as a hidden function of time.

How do you solve the sliding ladder problem?

Worked example: A sliding ladder

A $13$ ft ladder leans against a wall. Its bottom slides away from the wall at $3$ ft/s. How fast is the top sliding down when the bottom is $5$ ft from the wall?

Picture and name. Let $x$ be the distance from the wall to the bottom, and $y$ the height of the top. Known: $\frac{d x}{d t} = 3$ . Wanted: $\frac{d y}{d t}$ when $x = 5$ .
Relate. The wall, the ground and the ladder form a right triangle, so by the Pythagorean theorem $x^{2} + y^{2} = 169$ (since $13^{2} = 169$ ) at every moment.
Differentiate with respect to $t$ : $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} =$ $0$ . The ladder's length is constant, so the right side is $0$ .
Substitute. When $x = 5$ : $y^{2} = 169 - 25 = 144$ , so $y = 12$ . Then $2 (5) (3) + 2 (12) \frac{d y}{d t} =$ $0$ , which is $30 + 24 \frac{d y}{d t} =$ $0$ .
Solve: $24 \frac{d y}{d t} =$ $- 30$ , so $\frac{d y}{d t} =$ $- \frac{30}{24} =$ $- 1.25$ .
Interpret. The rate is negative because the height is decreasing.

Answer: The top slides down at $1.25$ ft/s.

The ladder at the moment in question: bottom

5

ft from the wall, top

12

ft up. The three sides satisfy

x^{2} + y^{2} = 13^{2}

at every moment.

Common mistake

Substituting too early. If you put $x = 5$ into the equation before differentiating, $x$ becomes a constant, its derivative becomes $0$ , and the rate you were given disappears. Only constants (like the ladder's length) can be substituted before step 4.

Worked example: An inflating balloon

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

Name. $V$ = volume, $r$ = radius. Known: $\frac{d V}{d t} = 100$ . Wanted: $\frac{d r}{d t}$ when $r = 5$ .
Relate. The volume of a sphere: $V = \frac{4}{3} π r^{3}$ .
Differentiate: $\frac{d V}{d t} =$ $\frac{4}{3} π \cdot 3 r^{2} \frac{d r}{d t} =$ $4 π r^{2} \frac{d r}{d t}$ .
Substitute: $100 =$ $4 π (5)^{2} \frac{d r}{d t} =$ $100 π \frac{d r}{d t}$ .
Solve: $\frac{d r}{d t} =$ $\frac{100}{100 π} =$ $\frac{1}{π} \approx 0.318$ .

Answer: The radius grows at $\frac{1}{π} \approx 0.32$ cm/s.

Notice that the radius grows more slowly as the balloon gets bigger: the same amount of air is spread over a larger surface.

Some problems have one quantity too many. In a cone-shaped tank, the volume depends on both the water's radius and its depth, but you're only told about one of them. Similar triangles (triangles with the same shape, so their sides are in the same ratio) link the two and remove the extra letter before you differentiate.

Worked example: Filling a cone-shaped tank

Water flows at $2$ m³/min into a tank shaped like an upside-down cone, $6$ m deep with a top radius of $3$ m. How fast is the water level rising when the water is $4$ m deep?

Name. $h$ = depth of the water, $r$ = radius of its surface, $V$ = volume. Known: $\frac{d V}{d t} = 2$ . Wanted: $\frac{d h}{d t}$ when $h = 4$ .
Relate. Volume of a cone: $V = \frac{1}{3} π r^{2} h$ .
Remove $r$ with similar triangles. The water forms a smaller cone with the same shape as the tank, so $\frac{r}{h} =$ $\frac{3}{6} =$ $\frac{1}{2}$ , giving $r = \frac{h}{2}$ .
Substitute: $V =$ $\frac{1}{3} π {(\frac{h}{2})}^{2} h =$ $\frac{1}{3} π \cdot \frac{h^{2}}{4} \cdot h =$ $\frac{π}{12} h^{3}$ .
Differentiate: $\frac{d V}{d t} =$ $\frac{π}{12} \cdot 3 h^{2} \frac{d h}{d t} =$ $\frac{π}{4} h^{2} \frac{d h}{d t}$ .
Substitute $h = 4$ : $2 =$ $\frac{π}{4} \cdot 16 \frac{d h}{d t} =$ $4 π \frac{d h}{d t}$ , so $\frac{d h}{d t} =$ $\frac{2}{4 π} =$ $\frac{1}{2 π} \approx 0.159$ .

Answer: The water level rises at $\frac{1}{2 π} \approx 0.16$ m/min.

Worked example: Two cars leaving an intersection

One car drives north from an intersection at $60$ mph, and another drives east at $45$ mph. How fast is the distance between them increasing when the first car is $3$ miles north and the second is $4$ miles east?

Name. $y$ = first car's distance north, $x$ = second car's distance east, $z$ = distance between them. Known: $\frac{d y}{d t} = 60$ , $\frac{d x}{d t} = 45$ . Wanted: $\frac{d z}{d t}$ .
Relate. The roads meet at a right angle: $z^{2} = x^{2} + y^{2}$ .
Differentiate: $2 z \frac{d z}{d t} =$ $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t}$ ; divide by $2$ : $z \frac{d z}{d t} =$ $x \frac{d x}{d t} + y \frac{d y}{d t}$ .
Substitute. At the moment, $z =$ $\sqrt{4^{2} + 3^{2}} =$ $\sqrt{25} =$ $5$ . So $5 \frac{d z}{d t} =$ $4 (45) + 3 (60) =$ $180 + 180 =$ $360$ .
Solve: $\frac{d z}{d t} =$ $\frac{360}{5} =$ $72$ .

Answer: The distance between the cars is increasing at $72$ mph.

Substituting before differentiating, which turns changing quantities into constants.
Wrong signs. A quantity that is decreasing has a negative rate: water draining out means $\frac{d V}{d t} < 0$ .
Forgetting a rate. When you differentiate, every changing quantity gets its own rate factor, including ones you aren't asked about.
Missing units. A rate's units are the quantity's units per unit of time: ft/s, cm³/s, m/min.

Tip

When a problem asks for the largest or smallest possible value of something, rather than a rate at one moment, it's an optimization problem instead.

Related rates: a step-by-step method with worked examples

How do you solve the sliding ladder problem?

Learn it step by step

Frequently asked questions

What is the first step in a related rates problem?

Why do you differentiate with respect to time?

When do you plug in the numbers in related rates?

Why is my related rates answer negative?

How are related rates and implicit differentiation connected?

What is a related rates problem?

What are the steps for solving related rates problems?

How do you solve the sliding ladder problem?

How do you solve related rates with volume?

How do you use similar triangles in related rates?

How do you solve related rates with distances between moving objects?

What are the most common mistakes in related rates?

Learn it step by step

Frequently asked questions

What is the first step in a related rates problem?

Why do you differentiate with respect to time?

When do you plug in the numbers in related rates?

Why is my related rates answer negative?

How are related rates and implicit differentiation connected?