Related rates | Contextual uses | Achievable AP Calculus AB

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. Since real-world scenarios often involve multiple variables, these types of problems allow us to calculate how one changing quantity affects another.

Step-by-step strategy

1. Define variables and draw a diagram if needed.

Rewrite the given information and target in mathematical terms. Choose variables that are easy to keep track of.

For example, if given “the height is increasing at a rate of $2$ inches per second,” express this as

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

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

AP tip:

When defining variables or checking your final answer, check if the sign makes sense. If the value of a variable increases over time, then its rate of change is positive. If the value decreases, then 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 not given, this is often a geometric formula (like a similarity proportion, the Pythagorean theorem, area, or volume) or 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 actually a function $x (t)$ and upon differentiating, the chain rule must be used to immediately multiply the derivative by $\frac{d x}{d t}$ .

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

Sidenote

When to substitute

If a quantity remains constant throughout the problem, the variable can be replaced with the value right at step 2 since the rate of change of that quantity would just be $0$ anyway. For changing quantities, only plug in the values after differentiating.

Let’s start with a classic:

1. 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: It’s positive because the value of $x$ 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 this right triangle.

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

Because the length of the ladder doesn’t change over time, $z$ can be replaced with $10$ . Then the equation to work with is

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

3. Differentiate with respect to time

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

4. Substitute known variables

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

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

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

$y = 6$

(Positive version only, because $y$ is a length)

Then plugging in the values,

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

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

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

$= - 4$

Notice the negative sign! This means that $y$ is decreasing, which makes sense since the ladder is sliding down.

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

Sometimes more than one equation is involved.

2. 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: Look at the units! Since the rate of air being pumped is specified as $2$ ft $^{3}$ /min, it means this quantity is the rate of change of the volume.

Find:

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

2. Relate variables

For a sphere,

$V = \frac{4}{3} π r^{3}$

$A = 4 π r^{2}$

While we could combine these into a single equation that relates $A$ and $V$ directly, it’s not necessary to do so at this step. It’s perfectly fine to have more than one equation to differentiate. Manipulating the equations after that step might be easier.

3. Differentiate with respect to time

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

$= 4 π r^{2} \cdot \frac{d r}{d t}$

and

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

$= 8 π r \cdot \frac{d r}{d t}$

Since we don’t have $\frac{d r}{d t}$ , it’s at this step that we can relate $\frac{d A}{d t}$ and $\frac{d V}{d t}$ .

Since

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

then dividing both sides by $4 π r^{2}$ ,

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

Substituting 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 we only need $r$ to find $\frac{d A}{d t}$ .

4. Substitute known values

While $r$ is not given, we do know that the volume at the target instant is $\frac{1}{6} π ft^{3}$ . So the radius at that instant can be found using the volume formula.

$\frac{4}{3} π r^{3} = \frac{1}{6} π$

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

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

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

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 quantities to relate are not immediately obvious. These questions often involve similarity and proportional relationships.

3. 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 in the cone 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

$V = \frac{1}{3} π r^{2} h$

$r$ and $h$ here refer to the height of the water and the radius of the smaller cone formed by the water. Both of these change as the water drains, so differentiating directly would require the product rule and leave us with both $\frac{d r}{d t}$ and $\frac{d h}{d t}$ , and we don’t have the former. To make the process easier, we can rewrite the formula in terms of $h$ to relate $V$ and $h$ directly.

The entire cone and the smaller cone formed by the water are similar, so we can set up a similarity proportion to relate the changing radius and height of the water:

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

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

Then the volume in terms of the height is

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

$V = \frac{1}{12} π h^{3}$

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

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

$- 2 π = 4 π \cdot \frac{d h}{d t}$

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

The sign is consistent with the fact that the water is draining, and the 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:
$\frac{d x}{d t} = - 20$ m/s
- Negative because distance $x$ is decreasing as car moves toward pole

Find:

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

2. Relate variables

$x, y,$ and $θ$ are related with trig function

$tan (θ) = \frac{y}{x}$

Because the height of the pole remains constant over time, $y$ can be replaced with $10$ .

$tan (θ) = \frac{10}{x}$

3. Differentiate with respect to time

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

4. Substitute known values

We don’t actually need to calculate $θ$ for this problem - at the instant when $x = 10$ m, the hypotenuse of the triangle (distance between car and camera) is is $102$ m by the Pythagorean theorem which means

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

Then substituting the known values,

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

$\frac{d θ}{d t} = 1$

The angle is increasing at a rate of $1 radian/s$ . The sign makes sense since the car moves toward the pole.

Alternatively, the trig function $cot (θ) = \frac{x}{y}$ could have also been used to relate the variables for step 2. Try to solve the problem that way to confirm the answer!

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 target in mathematical terms. Choose variables that are easy to keep track of.

For example, if given “the height is increasing at a rate of $2$ inches per second,” express this as

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

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

AP tip:

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

If not given, this is often a geometric formula (like a similarity proportion, the Pythagorean theorem, area, or volume) or 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 actually a function $x (t)$ and upon differentiating, the chain rule must be used to immediately multiply the derivative by $\frac{d x}{d t}$ .

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

Sidenote

When to substitute

Let’s start with a classic:

1. 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: It’s positive because the value of $x$ 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 this right triangle.

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

Because the length of the ladder doesn’t change over time, $z$ can be replaced with $10$ . Then the equation to work with is

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

3. Differentiate with respect to time

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

4. Substitute known variables

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

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

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

$y = 6$

(Positive version only, because $y$ is a length)

Then plugging in the values,

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

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

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

$= - 4$

Notice the negative sign! This means that $y$ is decreasing, which makes sense since the ladder is sliding down.

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

Sometimes more than one equation is involved.

2. 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: Look at the units! Since the rate of air being pumped is specified as $2$ ft $^{3}$ /min, it means this quantity is the rate of change of the volume.

Find:

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

2. Relate variables

For a sphere,

$V = \frac{4}{3} π r^{3}$

$A = 4 π r^{2}$

3. Differentiate with respect to time

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

$= 4 π r^{2} \cdot \frac{d r}{d t}$

and

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

$= 8 π r \cdot \frac{d r}{d t}$

Since we don’t have $\frac{d r}{d t}$ , it’s at this step that we can relate $\frac{d A}{d t}$ and $\frac{d V}{d t}$ .

Since

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

then dividing both sides by $4 π r^{2}$ ,

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

Substituting 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 we only need $r$ to find $\frac{d A}{d t}$ .

4. Substitute known values

While $r$ is not given, we do know that the volume at the target instant is $\frac{1}{6} π ft^{3}$ . So the radius at that instant can be found using the volume formula.

$\frac{4}{3} π r^{3} = \frac{1}{6} π$

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

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

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

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 quantities to relate are not immediately obvious. These questions often involve similarity and proportional relationships.

3. 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 in the cone 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

$V = \frac{1}{3} π r^{2} h$

The entire cone and the smaller cone formed by the water are similar, so we can set up a similarity proportion to relate the changing radius and height of the water:

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

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

Then the volume in terms of the height is

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

$V = \frac{1}{12} π h^{3}$

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

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

$- 2 π = 4 π \cdot \frac{d h}{d t}$

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

The sign is consistent with the fact that the water is draining, and the 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:
$\frac{d x}{d t} = - 20$ m/s
- Negative because distance $x$ is decreasing as car moves toward pole

Find:

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

2. Relate variables

$x, y,$ and $θ$ are related with trig function

$tan (θ) = \frac{y}{x}$

Because the height of the pole remains constant over time, $y$ can be replaced with $10$ .

$tan (θ) = \frac{10}{x}$

3. Differentiate with respect to time

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

4. Substitute known values

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

Then substituting the known values,

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

$\frac{d θ}{d t} = 1$

The angle is increasing at a rate of $1 radian/s$ . The sign makes sense since the car moves toward the pole.

Alternatively, the trig function $cot (θ) = \frac{x}{y}$ could have also been used to relate the variables for step 2. Try to solve the problem that way to confirm the answer!