Blog · Calculus guides · 26 October 2026 · 5 min read

Related Rates: Steps, Examples & Practice Problems

Q: What are related rates in calculus?

Problems where two or more quantities are linked by an equation and all change with time. Knowing one rate of change lets you find another, by differentiating the equation with respect to time.

Q: How do you know which equation to use?

It comes from the shape: Pythagoras for ladders and distances, the volume formula for balloons and tanks, similar triangles for shadows and cones. Draw the picture first; the equation is usually sitting in it.

Q: Why is my answer negative?

Because that quantity is decreasing. In the ladder problem, dy/dt = -5/6 means the top falls at 5/6 ft/s. If a question asks how fast it is "sliding down", some books want 5/6, but the derivative itself is negative.

Q: How do you do the shadow problem?

Similar triangles. For a 6 ft person walking away from a 15 ft lamp post, with x the person's distance and s the shadow's length, s/6 = (x+s)/15, so s = 2/3x. The shadow's tip is at x + s = 5/3x, so it moves 5/3 times as fast as the person.

Q: Where can I find related rates practice problems?

Right on this page: "Your turn" has three related rates practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, Diverge, which has no clock, is free and needs no account.

A related rates problem gives you how fast one quantity is changing and asks how fast another, linked one is changing at a particular moment. Write an equation connecting the quantities, differentiate both sides with respect to time $t$ , then substitute the values at that moment and solve for the unknown rate. For example, if a sphere's volume grows at $100 cm^{3} / s$ , then at radius 5 cm its radius grows at $\frac{1}{π} \approx 0.318$ cm/s.

The formula

\frac{d}{d t} [f (x)] = f^{'} (x) \frac{d x}{d t} for example \frac{d}{d t} [x^{2} + y^{2}] = 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t}

Every quantity in the problem is secretly a function of time, so differentiating with respect to $t$ makes each one cough up its own rate. It is implicit differentiation with a stopwatch.

When to use it

Use this method on any word problem that talks about how fast something is changing "when" something else has a particular value: a ladder sliding down a wall, a balloon inflating, water filling a tank, a shadow lengthening, two cars driving apart. The tell is a rate you know ( $\frac{d V}{d t} = 100$ ), a rate you want ( $\frac{d r}{d t}$ ), and a moment ( $r = 5$ ). If the question asks for the biggest or smallest value of something instead, that is optimization, a close relative with a different method.

How to do it, step by step

Draw it and name everything. Give a letter to every quantity that changes. Only write in numbers for things that never change, like the length of a ladder.
Write down the rates. Turn the words into derivatives: "the base slides away at 2 ft/s" is $\frac{d x}{d t} = 2$ . Write the one you want with a question mark.
Find an equation linking the quantities. Pythagoras, a volume formula, similar triangles. If it has an extra variable, use the geometry to remove it now.
Differentiate both sides with respect to $t$ . Every changing letter picks up its own $\frac{d}{d t}$ , by the chain rule.
Substitute the values at that moment, then solve. Only now do the "when" numbers go in. Finish with units and a sense check on the sign.

Worked examples

Example 1: Inflating a balloon

Easy

Air is pumped into a spherical balloon at $100 cm^{3} / s$ . How fast is the radius increasing when the radius is 5 cm?

Known: $\frac{d V}{d t} = 100$ . Wanted: $\frac{d r}{d t}$ when $r = 5$ . The link is the volume of a sphere.

V = \frac{4}{3} π r^{3}

Differentiate with respect to $t$ . The $r^{3}$ gives $3 r^{2}$ , times $\frac{d r}{d t}$ :

\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}

Now substitute $r = 5$ and $\frac{d V}{d t} = 100$ :

100 = 4 π (25) \frac{d r}{d t} ⟹ \frac{d r}{d t} = \frac{100}{100 π}

Note that $4 π r^{2}$ is the balloon's surface area: the new air is spread over the whole skin, so a bigger balloon grows more slowly.

Answer

\frac{d r}{d t} = \frac{1}{π} \approx 0.318 cm/s

Example 2: The sliding ladder

Medium

A 13 ft ladder leans against a wall. The base slides away from the wall at 2 ft/s. How fast is the top sliding down when the base is 5 ft from the wall?

Let $x$ be the distance from the wall to the base and $y$ the height of the top. The ladder stays 13 ft, so Pythagoras links them.

x^{2} + y^{2} = 169

Differentiate with respect to $t$ , and cancel the 2s:

2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 ⟹ x \frac{d x}{d t} + y \frac{d y}{d t} = 0

At this moment $x = 5$ , so $y = 169 - 25 = 12$ , and $\frac{d x}{d t} = 2$ :

5 (2) + 12 \frac{d y}{d t} = 0 ⟹ \frac{d y}{d t} = - \frac{10}{12}

The minus sign means $y$ is shrinking: the top is sliding down at $\frac{5}{6}$ ft/s, which is what the question asked.

Answer

\frac{d y}{d t} = - \frac{5}{6} \approx - 0.833 ft/s

Example 3: Filling a cone

Harder

Water flows at $2 m^{3} / min$ into an inverted cone with radius 3 m and height 6 m. How fast is the water level rising when the water is 4 m deep?

Let $h$ be the depth and $r$ the radius of the water's surface. The volume is $V = \frac{1}{3} π r^{2} h$ , but that has two changing variables and we only know about $h$ .

Similar triangles remove $r$ : the water is a smaller cone with the same shape as the tank, so $\frac{r}{h} = \frac{3}{6}$ , that is $r = \frac{h}{2}$ . Do this before differentiating.

V = \frac{1}{3} π (\frac{h}{2})^{2} h = \frac{π h ^{3}}{12}

\frac{d V}{d t} = \frac{π h ^{2}}{4} \frac{d h}{d t}

Substitute $h = 4$ and $\frac{d V}{d t} = 2$ :

2 = \frac{16 π}{4} \frac{d h}{d t} = 4 π \frac{d h}{d t}

The level rises more slowly as the cone widens, which is why a cocktail glass seems to take forever at the top and fill instantly at the bottom.

Answer

\frac{d h}{d t} = \frac{1}{2 π} \approx 0.159 m/min

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Air is pumped into a spherical balloon at $50 cm^{3} / s$ . How fast is the radius increasing, in cm/s, when the radius is 10 cm?

V = \frac{4}{3} π r^{3}, \frac{d V}{d t} = 50, r = 10, \frac{d r}{d t} = ?

Problem 2

Medium

A 10 ft ladder leans against a wall. The base slides away from the wall at 3 ft/s. How fast is the top moving, in ft/s, when the base is 6 ft from the wall? (Give a negative number.)

x^{2} + y^{2} = 100, \frac{d x}{d t} = 3, x = 6, \frac{d y}{d t} = ?

Problem 3

Harder

A 6 ft tall person walks away from an 18 ft lamppost at 4 ft/s. How fast, in ft/s, is the tip of their shadow moving?

\frac{s}{6} = \frac{x + s}{18}, \frac{d x}{d t} = 4, \frac{d}{d t} (x + s) = ?

Common mistakes

Substituting too early. If you put $x = 5$ into the equation before differentiating, $x$ becomes a constant and its rate vanishes. Differentiate first, substitute second, always.
Forgetting the $\frac{d}{d t}$ factors. $\frac{d}{d t} [r^{3}]$ is $3 r^{2} \frac{d r}{d t}$ , not $3 r^{2}$ . Every changing quantity carries its own rate.
Getting the sign of a rate wrong. A quantity that is shrinking has a negative rate. Water draining out means $\frac{d V}{d t} < 0$ .
Keeping too many variables. In the cone, differentiating $V = \frac{1}{3} π r^{2} h$ directly needs $\frac{d r}{d t}$ , which you do not know. Use the geometry to get down to one variable first.

Practice it now

No clock, three strikes

Related rates are Diverge-only, turning up around difficulty 7 and 8: a sliding ladder, a balloon, a filling cone and a lamppost shadow, much like the ones above. Give your answer as a decimal to three places, and keep the minus sign when something is shrinking.

Play Diverge See it on the formula sheet

FAQ

What are related rates in calculus?: Problems where two or more quantities are linked by an equation and all change with time. Knowing one rate of change lets you find another, by differentiating the equation with respect to time.
How do you know which equation to use?: It comes from the shape: Pythagoras for ladders and distances, the volume formula for balloons and tanks, similar triangles for shadows and cones. Draw the picture first; the equation is usually sitting in it.
Why is my answer negative?: Because that quantity is decreasing. In the ladder problem, $\frac{d y}{d t} = - \frac{5}{6}$ means the top falls at $\frac{5}{6}$ ft/s. If a question asks how fast it is "sliding down", some books want $\frac{5}{6}$ , but the derivative itself is negative.
How do you do the shadow problem?: Similar triangles. For a 6 ft person walking away from a 15 ft lamp post, with $x$ the person's distance and $s$ the shadow's length, $\frac{s}{6} = \frac{x + s}{15}$ , so $s = \frac{2}{3} x$ . The shadow's tip is at $x + s = \frac{5}{3} x$ , so it moves $\frac{5}{3}$ times as fast as the person.
Where can I find related rates practice problems?: Right on this page: "Your turn" has three related rates practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, Diverge, which has no clock, is free and needs no account.

Related Rates: Steps, Examples & Practice Problems

When to use it

\frac{d V}{d t} = 100

), a rate you want (

\frac{d r}{d t}

), and a moment (

r = 5

). If the question asks for the biggest or smallest value of something instead, that is optimization, a close relative with a different method.

How to do it, step by step

Draw it and name everything. Give a letter to every quantity that changes. Only write in numbers for things that never change, like the length of a ladder.

Write down the rates. Turn the words into derivatives: "the base slides away at 2 ft/s" is

\frac{d x}{d t} = 2

. Write the one you want with a question mark.

Find an equation linking the quantities. Pythagoras, a volume formula, similar triangles. If it has an extra variable, use the geometry to remove it now.

Differentiate both sides with respect to $t$ . Every changing letter picks up its own

\frac{d}{d t}

, by the chain rule.

Substitute the values at that moment, then solve. Only now do the "when" numbers go in. Finish with units and a sense check on the sign.

Worked examples

Example 1: Inflating a balloon

Easy

Air is pumped into a spherical balloon at $100 cm^{3} / s$ . How fast is the radius increasing when the radius is 5 cm?

Known: $\frac{d V}{d t} = 100$ . Wanted: $\frac{d r}{d t}$ when $r = 5$ . The link is the volume of a sphere.

V = \frac{4}{3} π r^{3}

Differentiate with respect to $t$ . The $r^{3}$ gives $3 r^{2}$ , times $\frac{d r}{d t}$ :

\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}

Now substitute $r = 5$ and $\frac{d V}{d t} = 100$ :

100 = 4 π (25) \frac{d r}{d t} ⟹ \frac{d r}{d t} = \frac{100}{100 π}

Note that $4 π r^{2}$ is the balloon's surface area: the new air is spread over the whole skin, so a bigger balloon grows more slowly.

Answer

\frac{d r}{d t} = \frac{1}{π} \approx 0.318 cm/s

Example 2: The sliding ladder

Medium

A 13 ft ladder leans against a wall. The base slides away from the wall at 2 ft/s. How fast is the top sliding down when the base is 5 ft from the wall?

Let $x$ be the distance from the wall to the base and $y$ the height of the top. The ladder stays 13 ft, so Pythagoras links them.

x^{2} + y^{2} = 169

Differentiate with respect to $t$ , and cancel the 2s:

2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 ⟹ x \frac{d x}{d t} + y \frac{d y}{d t} = 0

At this moment $x = 5$ , so $y = 169 - 25 = 12$ , and $\frac{d x}{d t} = 2$ :

5 (2) + 12 \frac{d y}{d t} = 0 ⟹ \frac{d y}{d t} = - \frac{10}{12}

The minus sign means $y$ is shrinking: the top is sliding down at $\frac{5}{6}$ ft/s, which is what the question asked.

Answer

\frac{d y}{d t} = - \frac{5}{6} \approx - 0.833 ft/s

Example 3: Filling a cone

Harder

Water flows at $2 m^{3} / min$ into an inverted cone with radius 3 m and height 6 m. How fast is the water level rising when the water is 4 m deep?

Let $h$ be the depth and $r$ the radius of the water's surface. The volume is $V = \frac{1}{3} π r^{2} h$ , but that has two changing variables and we only know about $h$ .

Similar triangles remove $r$ : the water is a smaller cone with the same shape as the tank, so $\frac{r}{h} = \frac{3}{6}$ , that is $r = \frac{h}{2}$ . Do this before differentiating.

V = \frac{1}{3} π (\frac{h}{2})^{2} h = \frac{π h ^{3}}{12}

\frac{d V}{d t} = \frac{π h ^{2}}{4} \frac{d h}{d t}

Substitute $h = 4$ and $\frac{d V}{d t} = 2$ :

2 = \frac{16 π}{4} \frac{d h}{d t} = 4 π \frac{d h}{d t}

The level rises more slowly as the cone widens, which is why a cocktail glass seems to take forever at the top and fill instantly at the bottom.

Answer

\frac{d h}{d t} = \frac{1}{2 π} \approx 0.159 m/min

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Air is pumped into a spherical balloon at $50 cm^{3} / s$ . How fast is the radius increasing, in cm/s, when the radius is 10 cm?

V = \frac{4}{3} π r^{3}, \frac{d V}{d t} = 50, r = 10, \frac{d r}{d t} = ?

Problem 2

Medium

A 10 ft ladder leans against a wall. The base slides away from the wall at 3 ft/s. How fast is the top moving, in ft/s, when the base is 6 ft from the wall? (Give a negative number.)

x^{2} + y^{2} = 100, \frac{d x}{d t} = 3, x = 6, \frac{d y}{d t} = ?

Problem 3

Harder

A 6 ft tall person walks away from an 18 ft lamppost at 4 ft/s. How fast, in ft/s, is the tip of their shadow moving?

\frac{s}{6} = \frac{x + s}{18}, \frac{d x}{d t} = 4, \frac{d}{d t} (x + s) = ?

Common mistakes

Substituting too early. If you put

x = 5

into the equation before differentiating,

x

becomes a constant and its rate vanishes. Differentiate first, substitute second, always.

Forgetting the $\frac{d}{d t}$ factors.

\frac{d}{d t} [r^{3}]

3 r^{2} \frac{d r}{d t}

, not

3 r^{2}

. Every changing quantity carries its own rate.

Getting the sign of a rate wrong. A quantity that is shrinking has a negative rate. Water draining out means

\frac{d V}{d t} < 0

Keeping too many variables. In the cone, differentiating

V = \frac{1}{3} π r^{2} h

directly needs

\frac{d r}{d t}

, which you do not know. Use the geometry to get down to one variable first.

Practice it now

No clock, three strikes

Play Diverge See it on the formula sheet

FAQ

What are related rates in calculus?

Problems where two or more quantities are linked by an equation and all change with time. Knowing one rate of change lets you find another, by differentiating the equation with respect to time.

How do you know which equation to use?

It comes from the shape: Pythagoras for ladders and distances, the volume formula for balloons and tanks, similar triangles for shadows and cones. Draw the picture first; the equation is usually sitting in it.

Why is my answer negative?

Because that quantity is decreasing. In the ladder problem,

\frac{d y}{d t} = - \frac{5}{6}

means the top falls at

\frac{5}{6}

ft/s. If a question asks how fast it is "sliding down", some books want

\frac{5}{6}

, but the derivative itself is negative.

How do you do the shadow problem?

Similar triangles. For a 6 ft person walking away from a 15 ft lamp post, with

x

the person's distance and

s

the shadow's length,

\frac{s}{6} = \frac{x + s}{15}

, so

s = \frac{2}{3} x

. The shadow's tip is at

x + s = \frac{5}{3} x

, so it moves

\frac{5}{3}

times as fast as the person.

Where can I find related rates practice problems?

Right on this page: "Your turn" has three related rates practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, Diverge, which has no clock, is free and needs no account.

Related Rates: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: Inflating a balloon

Example 2: The sliding ladder

Example 3: Filling a cone

Your turn

Common mistakes

FAQ

Keep going

Related Rates: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: Inflating a balloon

Example 2: The sliding ladder

Example 3: Filling a cone

Your turn

Common mistakes

FAQ

Keep going