4.3 Related rates

Achievable AP Calculus AB

4. Contextual uses

Our AP Calculus AB course is currently in development and is a work-in-progress.

Related rates

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What you’ll learn:

How to translate real-world changing quantities into variables and rates
Use geometric or trigonometric relationships to connect variables
Solve classic related rates problems involving ladders, balloons, cones, and angles

Related rates problems involve quantities that change over time. In many real-world situations, several variables change together, and related rates lets you calculate how the rate of change of one quantity affects another.

Step-by-step strategy

1. Define variables and draw a diagram if needed.

Rewrite the given information and the quantity you’re asked to find in mathematical terms. Choose variable names that are easy to track.

For example, if you’re told “the height is increasing at a rate of $2$ inches per second,” write

$\frac{d h}{d t} = 2 in/s$

where $h$ is height and $t$ is time.

AP tip:

When you define variables and when you check your final answer, make sure the sign makes sense.

If a variable increases over time, its rate of change is positive.
If a variable decreases over time, its rate of change is negative.

2. Find an equation that relates the given and target variables. There may be more than one equation involved in the problem.

If the relationship isn’t given directly, it’s often a geometric or trigonometric one, such as:

a similarity proportion
the Pythagorean theorem
an area or volume formula
a trigonometric function

3. Differentiate with respect to time using implicit differentiation.

Treat any variable that changes over time as a function of $t$ . For example, if the distance between two objects is $x$ and increases over time, then it’s really $x (t)$ . When you differentiate, the chain rule tells you to multiply by $\frac{d x}{d t}$ .

4. Substitute known values and solve for the unknown rate.

::: sidenote When to substitute

If a quantity stays constant throughout the problem, you can replace its variable with the constant value in step 2, since the rate of change of a constant is $0$ .

For quantities that change, wait to plug in numerical values until after you differentiate. :::

Let’s start with a classic:

A 10-foot ladder is leaning against a wall. The bottom of the ladder slides away from the wall at a rate of 3 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 8 feet from the wall?

Solution

1. Define variables and diagram

Given:

$z = 10$ ft (ladder length, constant)
$\frac{d x}{d t} = + 3$ ft/s

Important: This is positive because $x$ increases as the ladder slides away from the wall.

Find:

$\frac{d y}{d t}$ when $x = 8$ ft

2. Relate variables

To relate $x$ , $y$ , and $z$ , use the Pythagorean theorem for the right triangle.

$x^{2} + y^{2} = z^{2}$

Because the ladder length doesn’t change, replace $z$ with $10$ :

$x^{2} + y^{2} = 10$

3. Differentiate with respect to time

Differentiate both sides with respect to $t$ :

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

4. Substitute known variables

$x = 8$ ft
$\frac{d x}{d t} = 3$ ft/s
To find $y$ , use the Pythagorean theorem

$x^{2} + y^{2} = z^{2}$

$(8)^{2} + y^{2} = (10)^{2}$

$y = 6$

(Use the positive value because $y$ is a length.)

Now substitute into the differentiated equation:

$2 (8) (3) + 2 (6) \cdot \frac{d y}{d t} = 2 (10) (0)$

$48 + 12 \cdot \frac{d y}{d t} = 0$

$12 \cdot \frac{d y}{d t} = - 48$

$= - 4$

Notice the negative sign: $y$ is decreasing, which matches the picture (the top slides down).

So the top of the ladder is sliding down at $4 ft/s$ .

Sometimes more than one equation is involved.

Air is pumped into a balloon at a rate of $2$ ft $^{3}$ per minute. At what rate is the surface area increasing (in ft $^{2}$ per minute) when the volume is $\frac{1}{6} π$ ft $^{3}$ ?

Solution

(spoiler)

1. Define variables and/or diagram

Given:

$\frac{d V}{d t} = 2$ ft $^{3}$ /min

Important: The units tell you this is the rate of change of volume.

Find:

$\frac{d A}{d t}$ (rate of change of surface area) when$V= \dfrac16 \pi \text^3$

2. Relate variables

For a sphere,

$V = \frac{4}{3} π r^{3}$

$A = 4 π r^{2}$

You could combine these into a single equation relating $A$ and $V$ , but you don’t have to. It’s fine to differentiate both equations and connect the rates afterward.

3. Differentiate with respect to time

Differentiate the volume equation:

$\frac{d V}{d t} = \frac{4}{3} π (3 r^{2} \cdot \frac{d r}{d t})$

$= 4 π r^{2} \cdot \frac{d r}{d t}$

Differentiate the area equation:

$\frac{d A}{d t} = 4 π (2 r \cdot \frac{d r}{d t})$

$= 8 π r \cdot \frac{d r}{d t}$

We don’t know $\frac{d r}{d t}$ , so solve for it using the first derivative:

Since

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

divide both sides by $4 π r^{2}$ :

$\frac{d r}{d t} = \frac{1}{4 π r ^{2}} \cdot \frac{d V}{d t}$

Substitute into $\frac{d A}{d t}$ :

$\frac{d A}{d t} = 8 π r \cdot \frac{d r}{d t}$

$= 8 π r \cdot \frac{1}{4 π r ^{2}} \cdot \frac{d V}{d t}$

$\frac{d A}{d t} = \frac{2}{r} \cdot \frac{d V}{d t}$

Now you only need $r$ at the instant when $V = \frac{1}{6} π$ .

4. Substitute known values

Use the volume formula to find $r$ at that instant:

$\frac{4}{3} π r^{3} = \frac{1}{6} π$

$r^{3} = \frac{1}{6} π \cdot \frac{3}{4 π}$

$r^{3} = \frac{1}{8}$

$r = \frac{1}{2}$

Then

$\frac{d A}{d t} = \frac{2}{\frac{1}{2}} \cdot 2$

$= 8$

So the surface area of the balloon is increasing at a rate of $8 ft^{2} /min .$

In some problems, the relationship between quantities isn’t obvious at first. These often use similarity and proportional relationships.

Water drains from a cone at 2 $π$ cubic centimeters per second. The cone has a height of 12 cm and a base radius of 6 cm. How fast is the water level dropping when the height is 4 cm?

Solution

(spoiler)

1. Define variables and diagram

Given:

$\frac{d V}{d t} = - 2 π$ cm $^{3}$ /s
- Negative because the volume of water is decreasing over time.
The full cone has radius 6 cm and height 12 cm.

Find:

$\frac{d h}{d t}$ when $h = 4$ cm

2. Relate variables

The volume of a cone is

$V = \frac{1}{3} π r^{2} h$

Here, $r$ and $h$ refer to the radius and height of the water (the smaller cone). Both change as the water drains.

If we differentiate $V = \frac{1}{3} π r^{2} h$ directly, we’d need the product rule and we’d end up with both $\frac{d r}{d t}$ and $\frac{d h}{d t}$ . Since we aren’t given $\frac{d r}{d t}$ , it’s better to rewrite $V$ in terms of $h$ only.

The full cone and the smaller cone are similar, so the ratio $\frac{r}{h}$ stays constant:

$\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$

$r = \frac{h}{2}$

Substitute $r = \frac{h}{2}$ into the volume formula:

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

$V = \frac{1}{12} π h^{3}$

3. Differentiate with respect to time

$\frac{d V}{d t} = \frac{1}{12} π \cdot 3 h^{2} \cdot \frac{d h}{d t}$

$= \frac{1}{4} π h^{2} \cdot \frac{d h}{d t}$

4. Substitute known values

At $h = 4$ cm and $\frac{d V}{d t} = - 2 π$ :

$- 2 π = \frac{1}{4} π (4)^{2} \cdot \frac{d h}{d t}$

$- 2 π = 4 π \cdot \frac{d h}{d t}$

$\frac{d h}{d t} = - \frac{1}{2}$

The negative sign matches the situation: the water level is dropping at a rate of $0.5 cm/s$ .

Related rates problems can also involve trigonometry. ::: sidenote Trigonometry

For problems with trigonometry, $θ$ is always in radians for consistency. :::

A camera is mounted on a 10-meter pole and is pointed at a car moving toward it. The car is traveling along a straight road at a speed of 20 meters per second.

How fast is the angle of elevation from the car to the camera changing when the car is 10 meters away from the pole?

Solution

(spoiler)

1. Define variables and diagram

Given:

Height of pole $y = 10$ m
Car’s motion:$\dfrac = -20$ m/s
- Negative because the distance $x$ decreases as the car moves toward the pole.

Find:

$\frac{d θ}{d t}$ when $x = 10$ meters.

2. Relate variables

$x$ , $y$ , and $θ$ are related by a trigonometric ratio:

$tan (θ) = \frac{y}{x}$

Because the pole height is constant, replace $y$ with $10$ :

$tan (θ) = \frac{10}{x}$

3. Differentiate with respect to time

Differentiate both sides with respect to $t$ :

$sec^{2} (θ) \cdot \frac{d θ}{d t} = (- \frac{10}{x ^{2}}) \cdot \frac{d x}{d t}$

4. Substitute known values

You don’t need to compute $θ$ directly. When $x = 10$ m, the hypotenuse is $102$ m (by the Pythagorean theorem), so

$sec^{2} (θ) = (\frac{10 2}{10})^{2} = 2$

Now substitute $x = 10$ and $\frac{d x}{d t} = - 20$ :

$2 \cdot \frac{d θ}{d t} = (- \frac{10}{1 0 ^{2}}) \cdot (- 20)$

$\frac{d θ}{d t} = 1$

The angle is increasing at a rate of $1 radian/s$ , which matches the situation: as the car gets closer, the angle of elevation increases.

Alternatively, you could relate the variables using $cot (θ) = \frac{x}{y}$ in step 2. Try solving it that way to confirm you get the same result.

Step-by-step strategy for related rates

Define variables; use diagrams and assign clear variable names
Relate variables with geometric/trigonometric equations
Differentiate with respect to time using implicit differentiation
Substitute known values after differentiating; solve for unknown rate

Ladder against a wall problem

Use Pythagorean theorem: $x^{2} + y^{2} = z^{2}$ (with $z$ constant)
Differentiate: $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$
Substitute values to solve for $\frac{d y}{d t}$ ; negative sign means variable is decreasing

Balloon (sphere) problem

Volume: $V = \frac{4}{3} π r^{3}$ ; Surface area: $A = 4 π r^{2}$
Differentiate both equations; relate $\frac{d A}{d t}$ and $\frac{d V}{d t}$ via $\frac{d r}{d t}$
Substitute $r$ found from $V$ to compute $\frac{d A}{d t}$

Cone draining problem

Volume: $V = \frac{1}{3} π r^{2} h$ ; relate $r$ and $h$ via similarity: $r = \frac{h}{2}$
Substitute to get $V = \frac{1}{12} π h^{3}$ ; differentiate: $\frac{d V}{d t} = \frac{1}{4} π h^{2} \frac{d h}{d t}$
Plug in known values to solve for $\frac{d h}{d t}$ ; negative rate means height is decreasing

Trigonometric related rates (angle of elevation)

Relate variables: $tan (θ) = \frac{y}{x}$ (with $y$ constant)
Differentiate: $sec^{2} (θ) \frac{d θ}{d t} = - \frac{y}{x ^{2}} \frac{d x}{d t}$
Use Pythagorean theorem to find $sec^{2} (θ)$ at given $x$ ; solve for $\frac{d θ}{d t}$

General tips

Positive rate: variable increasing; negative rate: variable decreasing
For constants, substitute before differentiating; for changing quantities, substitute after differentiating
Use similarity or proportionality to reduce variables when possible

Related rates

What you’ll learn:

How to translate real-world changing quantities into variables and rates
Use geometric or trigonometric relationships to connect variables
Solve classic related rates problems involving ladders, balloons, cones, and angles

Step-by-step strategy

1. Define variables and draw a diagram if needed.

Rewrite the given information and the quantity you’re asked to find in mathematical terms. Choose variable names that are easy to track.

For example, if you’re told “the height is increasing at a rate of $2$ inches per second,” write

$\frac{d h}{d t} = 2 in/s$

where $h$ is height and $t$ is time.

AP tip:

When you define variables and when you check your final answer, make sure the sign makes sense.

If a variable increases over time, its rate of change is positive.
If a variable decreases over time, its rate of change is negative.

2. Find an equation that relates the given and target variables. There may be more than one equation involved in the problem.

If the relationship isn’t given directly, it’s often a geometric or trigonometric one, such as:

a similarity proportion
the Pythagorean theorem
an area or volume formula
a trigonometric function

3. Differentiate with respect to time using implicit differentiation.

Treat any variable that changes over time as a function of $t$ . For example, if the distance between two objects is $x$ and increases over time, then it’s really $x (t)$ . When you differentiate, the chain rule tells you to multiply by $\frac{d x}{d t}$ .

4. Substitute known values and solve for the unknown rate.

::: sidenote When to substitute

If a quantity stays constant throughout the problem, you can replace its variable with the constant value in step 2, since the rate of change of a constant is $0$ .

For quantities that change, wait to plug in numerical values until after you differentiate. :::

Let’s start with a classic:

A 10-foot ladder is leaning against a wall. The bottom of the ladder slides away from the wall at a rate of 3 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 8 feet from the wall?

Solution

1. Define variables and diagram

Given:

$z = 10$ ft (ladder length, constant)
$\frac{d x}{d t} = + 3$ ft/s

Important: This is positive because $x$ increases as the ladder slides away from the wall.

Find:

$\frac{d y}{d t}$ when $x = 8$ ft

2. Relate variables

To relate $x$ , $y$ , and $z$ , use the Pythagorean theorem for the right triangle.

$x^{2} + y^{2} = z^{2}$

Because the ladder length doesn’t change, replace $z$ with $10$ :

$x^{2} + y^{2} = 10$

3. Differentiate with respect to time

Differentiate both sides with respect to $t$ :

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

4. Substitute known variables

$x = 8$ ft
$\frac{d x}{d t} = 3$ ft/s
To find $y$ , use the Pythagorean theorem

$x^{2} + y^{2} = z^{2}$

$(8)^{2} + y^{2} = (10)^{2}$

$y = 6$

(Use the positive value because $y$ is a length.)

Now substitute into the differentiated equation:

$2 (8) (3) + 2 (6) \cdot \frac{d y}{d t} = 2 (10) (0)$

$48 + 12 \cdot \frac{d y}{d t} = 0$

$12 \cdot \frac{d y}{d t} = - 48$

$= - 4$

Notice the negative sign: $y$ is decreasing, which matches the picture (the top slides down).

So the top of the ladder is sliding down at $4 ft/s$ .

Sometimes more than one equation is involved.

Air is pumped into a balloon at a rate of $2$ ft $^{3}$ per minute. At what rate is the surface area increasing (in ft $^{2}$ per minute) when the volume is $\frac{1}{6} π$ ft $^{3}$ ?

Solution

(spoiler)

1. Define variables and/or diagram

Given:

$\frac{d V}{d t} = 2$ ft $^{3}$ /min

Important: The units tell you this is the rate of change of volume.

Find:

$\frac{d A}{d t}$ (rate of change of surface area) when$V= \dfrac{1}{6} \pi \text{ ft}^3$

2. Relate variables

For a sphere,

$V = \frac{4}{3} π r^{3}$

$A = 4 π r^{2}$

You could combine these into a single equation relating $A$ and $V$ , but you don’t have to. It’s fine to differentiate both equations and connect the rates afterward.

3. Differentiate with respect to time

Differentiate the volume equation:

$\frac{d V}{d t} = \frac{4}{3} π (3 r^{2} \cdot \frac{d r}{d t})$

$= 4 π r^{2} \cdot \frac{d r}{d t}$

Differentiate the area equation:

$\frac{d A}{d t} = 4 π (2 r \cdot \frac{d r}{d t})$

$= 8 π r \cdot \frac{d r}{d t}$

We don’t know $\frac{d r}{d t}$ , so solve for it using the first derivative:

Since

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

divide both sides by $4 π r^{2}$ :

$\frac{d r}{d t} = \frac{1}{4 π r ^{2}} \cdot \frac{d V}{d t}$

Substitute into $\frac{d A}{d t}$ :

$\frac{d A}{d t} = 8 π r \cdot \frac{d r}{d t}$

$= 8 π r \cdot \frac{1}{4 π r ^{2}} \cdot \frac{d V}{d t}$

$\frac{d A}{d t} = \frac{2}{r} \cdot \frac{d V}{d t}$

Now you only need $r$ at the instant when $V = \frac{1}{6} π$ .

4. Substitute known values

Use the volume formula to find $r$ at that instant:

$\frac{4}{3} π r^{3} = \frac{1}{6} π$

$r^{3} = \frac{1}{6} π \cdot \frac{3}{4 π}$

$r^{3} = \frac{1}{8}$

$r = \frac{1}{2}$

Then

$\frac{d A}{d t} = \frac{2}{\frac{1}{2}} \cdot 2$

$= 8$

So the surface area of the balloon is increasing at a rate of $8 ft^{2} /min .$

In some problems, the relationship between quantities isn’t obvious at first. These often use similarity and proportional relationships.

Water drains from a cone at 2 $π$ cubic centimeters per second. The cone has a height of 12 cm and a base radius of 6 cm. How fast is the water level dropping when the height is 4 cm?

Solution

(spoiler)

1. Define variables and diagram

Given:

$\frac{d V}{d t} = - 2 π$ cm $^{3}$ /s
- Negative because the volume of water is decreasing over time.
The full cone has radius 6 cm and height 12 cm.

Find:

$\frac{d h}{d t}$ when $h = 4$ cm

2. Relate variables

The volume of a cone is

$V = \frac{1}{3} π r^{2} h$

Here, $r$ and $h$ refer to the radius and height of the water (the smaller cone). Both change as the water drains.

The full cone and the smaller cone are similar, so the ratio $\frac{r}{h}$ stays constant:

$\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$

$r = \frac{h}{2}$

Substitute $r = \frac{h}{2}$ into the volume formula:

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

$V = \frac{1}{12} π h^{3}$

3. Differentiate with respect to time

$\frac{d V}{d t} = \frac{1}{12} π \cdot 3 h^{2} \cdot \frac{d h}{d t}$

$= \frac{1}{4} π h^{2} \cdot \frac{d h}{d t}$

4. Substitute known values

At $h = 4$ cm and $\frac{d V}{d t} = - 2 π$ :

$- 2 π = \frac{1}{4} π (4)^{2} \cdot \frac{d h}{d t}$

$- 2 π = 4 π \cdot \frac{d h}{d t}$

$\frac{d h}{d t} = - \frac{1}{2}$

The negative sign matches the situation: the water level is dropping at a rate of $0.5 cm/s$ .

Related rates problems can also involve trigonometry. ::: sidenote Trigonometry

For problems with trigonometry, $θ$ is always in radians for consistency. :::

A camera is mounted on a 10-meter pole and is pointed at a car moving toward it. The car is traveling along a straight road at a speed of 20 meters per second.

How fast is the angle of elevation from the car to the camera changing when the car is 10 meters away from the pole?

Solution

(spoiler)

1. Define variables and diagram

Given:

Height of pole $y = 10$ m
Car’s motion:$\dfrac{dx}{dt} = -20$ m/s
- Negative because the distance $x$ decreases as the car moves toward the pole.

Find:

$\frac{d θ}{d t}$ when $x = 10$ meters.

2. Relate variables

$x$ , $y$ , and $θ$ are related by a trigonometric ratio:

$tan (θ) = \frac{y}{x}$

Because the pole height is constant, replace $y$ with $10$ :

$tan (θ) = \frac{10}{x}$

3. Differentiate with respect to time

Differentiate both sides with respect to $t$ :

$sec^{2} (θ) \cdot \frac{d θ}{d t} = (- \frac{10}{x ^{2}}) \cdot \frac{d x}{d t}$

4. Substitute known values

You don’t need to compute $θ$ directly. When $x = 10$ m, the hypotenuse is $102$ m (by the Pythagorean theorem), so

$sec^{2} (θ) = (\frac{10 2}{10})^{2} = 2$

Now substitute $x = 10$ and $\frac{d x}{d t} = - 20$ :

$2 \cdot \frac{d θ}{d t} = (- \frac{10}{1 0 ^{2}}) \cdot (- 20)$

$\frac{d θ}{d t} = 1$

The angle is increasing at a rate of $1 radian/s$ , which matches the situation: as the car gets closer, the angle of elevation increases.

Alternatively, you could relate the variables using $cot (θ) = \frac{x}{y}$ in step 2. Try solving it that way to confirm you get the same result.

Achievable AP Calculus AB

4. Contextual uses

Our AP Calculus AB course is currently in development and is a work-in-progress.

Related rates

10 min read

Font

Discuss

Feedback

What you’ll learn:

How to translate real-world changing quantities into variables and rates
Use geometric or trigonometric relationships to connect variables
Solve classic related rates problems involving ladders, balloons, cones, and angles

Step-by-step strategy

1. Define variables and draw a diagram if needed.

Rewrite the given information and the quantity you’re asked to find in mathematical terms. Choose variable names that are easy to track.

For example, if you’re told “the height is increasing at a rate of $2$ inches per second,” write

$\frac{d h}{d t} = 2 in/s$

where $h$ is height and $t$ is time.

AP tip:

When you define variables and when you check your final answer, make sure the sign makes sense.

If a variable increases over time, its rate of change is positive.
If a variable decreases over time, its rate of change is negative.

2. Find an equation that relates the given and target variables. There may be more than one equation involved in the problem.

If the relationship isn’t given directly, it’s often a geometric or trigonometric one, such as:

a similarity proportion
the Pythagorean theorem
an area or volume formula
a trigonometric function

3. Differentiate with respect to time using implicit differentiation.

Treat any variable that changes over time as a function of $t$ . For example, if the distance between two objects is $x$ and increases over time, then it’s really $x (t)$ . When you differentiate, the chain rule tells you to multiply by $\frac{d x}{d t}$ .

4. Substitute known values and solve for the unknown rate.

::: sidenote When to substitute

If a quantity stays constant throughout the problem, you can replace its variable with the constant value in step 2, since the rate of change of a constant is $0$ .

For quantities that change, wait to plug in numerical values until after you differentiate. :::

Let’s start with a classic:

A 10-foot ladder is leaning against a wall. The bottom of the ladder slides away from the wall at a rate of 3 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 8 feet from the wall?

Solution

1. Define variables and diagram

Given:

$z = 10$ ft (ladder length, constant)
$\frac{d x}{d t} = + 3$ ft/s

Important: This is positive because $x$ increases as the ladder slides away from the wall.

Find:

$\frac{d y}{d t}$ when $x = 8$ ft

2. Relate variables

To relate $x$ , $y$ , and $z$ , use the Pythagorean theorem for the right triangle.

$x^{2} + y^{2} = z^{2}$

Because the ladder length doesn’t change, replace $z$ with $10$ :

$x^{2} + y^{2} = 10$

3. Differentiate with respect to time

Differentiate both sides with respect to $t$ :

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

4. Substitute known variables

$x = 8$ ft
$\frac{d x}{d t} = 3$ ft/s
To find $y$ , use the Pythagorean theorem

$x^{2} + y^{2} = z^{2}$

$(8)^{2} + y^{2} = (10)^{2}$

$y = 6$

(Use the positive value because $y$ is a length.)

Now substitute into the differentiated equation:

$2 (8) (3) + 2 (6) \cdot \frac{d y}{d t} = 2 (10) (0)$

$48 + 12 \cdot \frac{d y}{d t} = 0$

$12 \cdot \frac{d y}{d t} = - 48$

$= - 4$

Notice the negative sign: $y$ is decreasing, which matches the picture (the top slides down).

So the top of the ladder is sliding down at $4 ft/s$ .

Sometimes more than one equation is involved.

Air is pumped into a balloon at a rate of $2$ ft $^{3}$ per minute. At what rate is the surface area increasing (in ft $^{2}$ per minute) when the volume is $\frac{1}{6} π$ ft $^{3}$ ?

Solution

(spoiler)

1. Define variables and/or diagram

Given:

$\frac{d V}{d t} = 2$ ft $^{3}$ /min

Important: The units tell you this is the rate of change of volume.

Find:

$\frac{d A}{d t}$ (rate of change of surface area) when$V= \dfrac16 \pi \text^3$

2. Relate variables

For a sphere,

$V = \frac{4}{3} π r^{3}$

$A = 4 π r^{2}$

You could combine these into a single equation relating $A$ and $V$ , but you don’t have to. It’s fine to differentiate both equations and connect the rates afterward.

3. Differentiate with respect to time

Differentiate the volume equation:

$\frac{d V}{d t} = \frac{4}{3} π (3 r^{2} \cdot \frac{d r}{d t})$

$= 4 π r^{2} \cdot \frac{d r}{d t}$

Differentiate the area equation:

$\frac{d A}{d t} = 4 π (2 r \cdot \frac{d r}{d t})$

$= 8 π r \cdot \frac{d r}{d t}$

We don’t know $\frac{d r}{d t}$ , so solve for it using the first derivative:

Since

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

divide both sides by $4 π r^{2}$ :

$\frac{d r}{d t} = \frac{1}{4 π r ^{2}} \cdot \frac{d V}{d t}$

Substitute into $\frac{d A}{d t}$ :

$\frac{d A}{d t} = 8 π r \cdot \frac{d r}{d t}$

$= 8 π r \cdot \frac{1}{4 π r ^{2}} \cdot \frac{d V}{d t}$

$\frac{d A}{d t} = \frac{2}{r} \cdot \frac{d V}{d t}$

Now you only need $r$ at the instant when $V = \frac{1}{6} π$ .

4. Substitute known values

Use the volume formula to find $r$ at that instant:

$\frac{4}{3} π r^{3} = \frac{1}{6} π$

$r^{3} = \frac{1}{6} π \cdot \frac{3}{4 π}$

$r^{3} = \frac{1}{8}$

$r = \frac{1}{2}$

Then

$\frac{d A}{d t} = \frac{2}{\frac{1}{2}} \cdot 2$

$= 8$

So the surface area of the balloon is increasing at a rate of $8 ft^{2} /min .$

In some problems, the relationship between quantities isn’t obvious at first. These often use similarity and proportional relationships.

Water drains from a cone at 2 $π$ cubic centimeters per second. The cone has a height of 12 cm and a base radius of 6 cm. How fast is the water level dropping when the height is 4 cm?

Solution

(spoiler)

1. Define variables and diagram

Given:

$\frac{d V}{d t} = - 2 π$ cm $^{3}$ /s
- Negative because the volume of water is decreasing over time.
The full cone has radius 6 cm and height 12 cm.

Find:

$\frac{d h}{d t}$ when $h = 4$ cm

2. Relate variables

The volume of a cone is

$V = \frac{1}{3} π r^{2} h$

Here, $r$ and $h$ refer to the radius and height of the water (the smaller cone). Both change as the water drains.

The full cone and the smaller cone are similar, so the ratio $\frac{r}{h}$ stays constant:

$\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$

$r = \frac{h}{2}$

Substitute $r = \frac{h}{2}$ into the volume formula:

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

$V = \frac{1}{12} π h^{3}$

3. Differentiate with respect to time

$\frac{d V}{d t} = \frac{1}{12} π \cdot 3 h^{2} \cdot \frac{d h}{d t}$

$= \frac{1}{4} π h^{2} \cdot \frac{d h}{d t}$

4. Substitute known values

At $h = 4$ cm and $\frac{d V}{d t} = - 2 π$ :

$- 2 π = \frac{1}{4} π (4)^{2} \cdot \frac{d h}{d t}$

$- 2 π = 4 π \cdot \frac{d h}{d t}$

$\frac{d h}{d t} = - \frac{1}{2}$

The negative sign matches the situation: the water level is dropping at a rate of $0.5 cm/s$ .

Related rates problems can also involve trigonometry. ::: sidenote Trigonometry

For problems with trigonometry, $θ$ is always in radians for consistency. :::

A camera is mounted on a 10-meter pole and is pointed at a car moving toward it. The car is traveling along a straight road at a speed of 20 meters per second.

How fast is the angle of elevation from the car to the camera changing when the car is 10 meters away from the pole?

Solution

(spoiler)

1. Define variables and diagram

Given:

Height of pole $y = 10$ m
Car’s motion:$\dfrac = -20$ m/s
- Negative because the distance $x$ decreases as the car moves toward the pole.

Find:

$\frac{d θ}{d t}$ when $x = 10$ meters.

2. Relate variables

$x$ , $y$ , and $θ$ are related by a trigonometric ratio:

$tan (θ) = \frac{y}{x}$

Because the pole height is constant, replace $y$ with $10$ :

$tan (θ) = \frac{10}{x}$

3. Differentiate with respect to time

Differentiate both sides with respect to $t$ :

$sec^{2} (θ) \cdot \frac{d θ}{d t} = (- \frac{10}{x ^{2}}) \cdot \frac{d x}{d t}$

4. Substitute known values

You don’t need to compute $θ$ directly. When $x = 10$ m, the hypotenuse is $102$ m (by the Pythagorean theorem), so

$sec^{2} (θ) = (\frac{10 2}{10})^{2} = 2$

Now substitute $x = 10$ and $\frac{d x}{d t} = - 20$ :

$2 \cdot \frac{d θ}{d t} = (- \frac{10}{1 0 ^{2}}) \cdot (- 20)$

$\frac{d θ}{d t} = 1$

The angle is increasing at a rate of $1 radian/s$ , which matches the situation: as the car gets closer, the angle of elevation increases.

Alternatively, you could relate the variables using $cot (θ) = \frac{x}{y}$ in step 2. Try solving it that way to confirm you get the same result.

Step-by-step strategy for related rates

Define variables; use diagrams and assign clear variable names
Relate variables with geometric/trigonometric equations
Differentiate with respect to time using implicit differentiation
Substitute known values after differentiating; solve for unknown rate

Ladder against a wall problem

Use Pythagorean theorem: $x^{2} + y^{2} = z^{2}$ (with $z$ constant)
Differentiate: $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$
Substitute values to solve for $\frac{d y}{d t}$ ; negative sign means variable is decreasing

Balloon (sphere) problem

Volume: $V = \frac{4}{3} π r^{3}$ ; Surface area: $A = 4 π r^{2}$
Differentiate both equations; relate $\frac{d A}{d t}$ and $\frac{d V}{d t}$ via $\frac{d r}{d t}$
Substitute $r$ found from $V$ to compute $\frac{d A}{d t}$

Cone draining problem

Volume: $V = \frac{1}{3} π r^{2} h$ ; relate $r$ and $h$ via similarity: $r = \frac{h}{2}$
Substitute to get $V = \frac{1}{12} π h^{3}$ ; differentiate: $\frac{d V}{d t} = \frac{1}{4} π h^{2} \frac{d h}{d t}$
Plug in known values to solve for $\frac{d h}{d t}$ ; negative rate means height is decreasing

Trigonometric related rates (angle of elevation)

Relate variables: $tan (θ) = \frac{y}{x}$ (with $y$ constant)
Differentiate: $sec^{2} (θ) \frac{d θ}{d t} = - \frac{y}{x ^{2}} \frac{d x}{d t}$
Use Pythagorean theorem to find $sec^{2} (θ)$ at given $x$ ; solve for $\frac{d θ}{d t}$

General tips

Positive rate: variable increasing; negative rate: variable decreasing
For constants, substitute before differentiating; for changing quantities, substitute after differentiating
Use similarity or proportionality to reduce variables when possible

Related rates

What you’ll learn:

How to translate real-world changing quantities into variables and rates
Use geometric or trigonometric relationships to connect variables
Solve classic related rates problems involving ladders, balloons, cones, and angles

Step-by-step strategy

1. Define variables and draw a diagram if needed.

Rewrite the given information and the quantity you’re asked to find in mathematical terms. Choose variable names that are easy to track.

For example, if you’re told “the height is increasing at a rate of $2$ inches per second,” write

$\frac{d h}{d t} = 2 in/s$

where $h$ is height and $t$ is time.

AP tip:

When you define variables and when you check your final answer, make sure the sign makes sense.

If a variable increases over time, its rate of change is positive.
If a variable decreases over time, its rate of change is negative.

2. Find an equation that relates the given and target variables. There may be more than one equation involved in the problem.

If the relationship isn’t given directly, it’s often a geometric or trigonometric one, such as:

a similarity proportion
the Pythagorean theorem
an area or volume formula
a trigonometric function

3. Differentiate with respect to time using implicit differentiation.

Treat any variable that changes over time as a function of $t$ . For example, if the distance between two objects is $x$ and increases over time, then it’s really $x (t)$ . When you differentiate, the chain rule tells you to multiply by $\frac{d x}{d t}$ .

4. Substitute known values and solve for the unknown rate.

::: sidenote When to substitute

If a quantity stays constant throughout the problem, you can replace its variable with the constant value in step 2, since the rate of change of a constant is $0$ .

For quantities that change, wait to plug in numerical values until after you differentiate. :::

Let’s start with a classic:

A 10-foot ladder is leaning against a wall. The bottom of the ladder slides away from the wall at a rate of 3 feet per second. How fast is the top of the ladder sliding down the wall when the bottom is 8 feet from the wall?

Solution

1. Define variables and diagram

Given:

$z = 10$ ft (ladder length, constant)
$\frac{d x}{d t} = + 3$ ft/s

Important: This is positive because $x$ increases as the ladder slides away from the wall.

Find:

$\frac{d y}{d t}$ when $x = 8$ ft

2. Relate variables

To relate $x$ , $y$ , and $z$ , use the Pythagorean theorem for the right triangle.

$x^{2} + y^{2} = z^{2}$

Because the ladder length doesn’t change, replace $z$ with $10$ :

$x^{2} + y^{2} = 10$

3. Differentiate with respect to time

Differentiate both sides with respect to $t$ :

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

4. Substitute known variables

$x = 8$ ft
$\frac{d x}{d t} = 3$ ft/s
To find $y$ , use the Pythagorean theorem

$x^{2} + y^{2} = z^{2}$

$(8)^{2} + y^{2} = (10)^{2}$

$y = 6$

(Use the positive value because $y$ is a length.)

Now substitute into the differentiated equation:

$2 (8) (3) + 2 (6) \cdot \frac{d y}{d t} = 2 (10) (0)$

$48 + 12 \cdot \frac{d y}{d t} = 0$

$12 \cdot \frac{d y}{d t} = - 48$

$= - 4$

Notice the negative sign: $y$ is decreasing, which matches the picture (the top slides down).

So the top of the ladder is sliding down at $4 ft/s$ .

Sometimes more than one equation is involved.

Air is pumped into a balloon at a rate of $2$ ft $^{3}$ per minute. At what rate is the surface area increasing (in ft $^{2}$ per minute) when the volume is $\frac{1}{6} π$ ft $^{3}$ ?

Solution

(spoiler)

1. Define variables and/or diagram

Given:

$\frac{d V}{d t} = 2$ ft $^{3}$ /min

Important: The units tell you this is the rate of change of volume.

Find:

$\frac{d A}{d t}$ (rate of change of surface area) when$V= \dfrac{1}{6} \pi \text{ ft}^3$

2. Relate variables

For a sphere,

$V = \frac{4}{3} π r^{3}$

$A = 4 π r^{2}$

You could combine these into a single equation relating $A$ and $V$ , but you don’t have to. It’s fine to differentiate both equations and connect the rates afterward.

3. Differentiate with respect to time

Differentiate the volume equation:

$\frac{d V}{d t} = \frac{4}{3} π (3 r^{2} \cdot \frac{d r}{d t})$

$= 4 π r^{2} \cdot \frac{d r}{d t}$

Differentiate the area equation:

$\frac{d A}{d t} = 4 π (2 r \cdot \frac{d r}{d t})$

$= 8 π r \cdot \frac{d r}{d t}$

We don’t know $\frac{d r}{d t}$ , so solve for it using the first derivative:

Since

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

divide both sides by $4 π r^{2}$ :

$\frac{d r}{d t} = \frac{1}{4 π r ^{2}} \cdot \frac{d V}{d t}$

Substitute into $\frac{d A}{d t}$ :

$\frac{d A}{d t} = 8 π r \cdot \frac{d r}{d t}$

$= 8 π r \cdot \frac{1}{4 π r ^{2}} \cdot \frac{d V}{d t}$

$\frac{d A}{d t} = \frac{2}{r} \cdot \frac{d V}{d t}$

Now you only need $r$ at the instant when $V = \frac{1}{6} π$ .

4. Substitute known values

Use the volume formula to find $r$ at that instant:

$\frac{4}{3} π r^{3} = \frac{1}{6} π$

$r^{3} = \frac{1}{6} π \cdot \frac{3}{4 π}$

$r^{3} = \frac{1}{8}$

$r = \frac{1}{2}$

Then

$\frac{d A}{d t} = \frac{2}{\frac{1}{2}} \cdot 2$

$= 8$

So the surface area of the balloon is increasing at a rate of $8 ft^{2} /min .$

In some problems, the relationship between quantities isn’t obvious at first. These often use similarity and proportional relationships.

Water drains from a cone at 2 $π$ cubic centimeters per second. The cone has a height of 12 cm and a base radius of 6 cm. How fast is the water level dropping when the height is 4 cm?

Solution

(spoiler)

1. Define variables and diagram

Given:

$\frac{d V}{d t} = - 2 π$ cm $^{3}$ /s
- Negative because the volume of water is decreasing over time.
The full cone has radius 6 cm and height 12 cm.

Find:

$\frac{d h}{d t}$ when $h = 4$ cm

2. Relate variables

The volume of a cone is

$V = \frac{1}{3} π r^{2} h$

Here, $r$ and $h$ refer to the radius and height of the water (the smaller cone). Both change as the water drains.

The full cone and the smaller cone are similar, so the ratio $\frac{r}{h}$ stays constant:

$\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$

$r = \frac{h}{2}$

Substitute $r = \frac{h}{2}$ into the volume formula:

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

$V = \frac{1}{12} π h^{3}$

3. Differentiate with respect to time

$\frac{d V}{d t} = \frac{1}{12} π \cdot 3 h^{2} \cdot \frac{d h}{d t}$

$= \frac{1}{4} π h^{2} \cdot \frac{d h}{d t}$

4. Substitute known values

At $h = 4$ cm and $\frac{d V}{d t} = - 2 π$ :

$- 2 π = \frac{1}{4} π (4)^{2} \cdot \frac{d h}{d t}$

$- 2 π = 4 π \cdot \frac{d h}{d t}$

$\frac{d h}{d t} = - \frac{1}{2}$

The negative sign matches the situation: the water level is dropping at a rate of $0.5 cm/s$ .

Related rates problems can also involve trigonometry. ::: sidenote Trigonometry

For problems with trigonometry, $θ$ is always in radians for consistency. :::

A camera is mounted on a 10-meter pole and is pointed at a car moving toward it. The car is traveling along a straight road at a speed of 20 meters per second.

How fast is the angle of elevation from the car to the camera changing when the car is 10 meters away from the pole?

Solution

(spoiler)

1. Define variables and diagram

Given:

Height of pole $y = 10$ m
Car’s motion:$\dfrac{dx}{dt} = -20$ m/s
- Negative because the distance $x$ decreases as the car moves toward the pole.

Find:

$\frac{d θ}{d t}$ when $x = 10$ meters.

2. Relate variables

$x$ , $y$ , and $θ$ are related by a trigonometric ratio:

$tan (θ) = \frac{y}{x}$

Because the pole height is constant, replace $y$ with $10$ :

$tan (θ) = \frac{10}{x}$

3. Differentiate with respect to time

Differentiate both sides with respect to $t$ :

$sec^{2} (θ) \cdot \frac{d θ}{d t} = (- \frac{10}{x ^{2}}) \cdot \frac{d x}{d t}$

4. Substitute known values

You don’t need to compute $θ$ directly. When $x = 10$ m, the hypotenuse is $102$ m (by the Pythagorean theorem), so

$sec^{2} (θ) = (\frac{10 2}{10})^{2} = 2$

Now substitute $x = 10$ and $\frac{d x}{d t} = - 20$ :

$2 \cdot \frac{d θ}{d t} = (- \frac{10}{1 0 ^{2}}) \cdot (- 20)$

$\frac{d θ}{d t} = 1$

The angle is increasing at a rate of $1 radian/s$ , which matches the situation: as the car gets closer, the angle of elevation increases.

Alternatively, you could relate the variables using $cot (θ) = \frac{x}{y}$ in step 2. Try solving it that way to confirm you get the same result.