# Related Rates Practice Problems with Answers and Solutions

> 10 related rates practice problems with answers and full solutions: balloons, ladders, cones, shadows, cars and angles, each with a clear setup.

Source: https://practicecalculus.com/practice-problems/related-rates
Updated: 2026-11-06

$$ x^2 + y^2 = z^2 \;\Rightarrow\; 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} $$

## At a glance

- **Problems:** 10 word problems, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 4
- **You need:** [Implicit differentiation](/practice-problems/implicit-differentiation) and basic geometry formulas
- **Main idea:** Write an equation that links the quantities, differentiate it with respect to t, then plug in the numbers
- **Practice with new numbers:** [Related rates lesson](/learn/calculus-1/applications-of-derivatives/related-rates)

## Key takeaways

- Differentiate the equation first and plug in the numbers for that one instant after. Plugging in too early turns a changing quantity into a constant.
- Give rates a sign: a quantity that is shrinking has a negative rate.
- For cones and shadows, use similar triangles to remove the extra variable before you differentiate.

These **related rates practice problems** each describe a situation where two or more quantities change together. You know how fast one changes and need the rate of another. Work through each setup before you check. The answers come right after the list, then the full solutions.

Every problem follows the same five steps:

1. Draw a picture and name the changing quantities.
2. Write the known rate and the rate you want, with units and signs.
3. Write an equation that links the quantities.
4. Differentiate both sides with respect to $t$.
5. Substitute the values at the given instant and solve.

For more on the method, read the [related rates guide](/blog/related-rates).

## Problems

**Geometry formulas**

1. **Ripple.** A stone drops into a pond and makes a circular ripple. The radius grows at $3$ cm/s. How fast is the area inside the ripple growing when the radius is $10$ cm?
2. **Balloon.** Air is pumped into a spherical balloon at $100$ cm³/s. How fast is the radius increasing when the radius is $5$ cm?
3. **Growing cube.** The edges of a cube grow at $0.5$ cm/s. How fast is the volume growing when each edge is $4$ cm long?

**Right triangles**

4. **Sliding ladder.** A $10$ ft ladder leans against a wall. The bottom is pulled away from the wall at $2$ ft/s. How fast is the top sliding down the wall when the bottom is $6$ ft from the wall?
5. **Ladder angle.** In problem 4, how fast is the angle between the ladder and the ground changing at that same moment?
6. **Two cars.** Car A leaves an intersection heading north at $60$ mph. At the same time car B leaves heading east at $45$ mph. How fast is the distance between them growing after $2$ hours?
7. **Balloon and observer.** A hot air balloon rises straight up at $4$ m/s. An observer stands $300$ m from the launch point. How fast is the angle of elevation from the observer to the balloon changing when the balloon is $400$ m high?

**Similar triangles and curves**

8. **Filling a cone.** A water tank is a cone with the point down. It is $12$ ft tall and $4$ ft in radius at the top. Water flows in at $9$ ft³/min. How fast is the water level rising when the water is $6$ ft deep?
9. **Shadow.** A $6$ ft tall person walks away from a $15$ ft lamp post at $5$ ft/s. How fast is the length of their shadow growing?
10. **Moving point.** A point moves along the curve $y = x^2$ so that its $x$-coordinate increases at $3$ units/s. How fast is its distance from the origin changing when the point is at $(2, 4)$?

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $60\pi \approx 188.5$ cm²/s |
| 2 | $\dfrac{1}{\pi} \approx 0.318$ cm/s |
| 3 | $24$ cm³/s |
| 4 | The top slides down at $1.5$ ft/s ($\frac{dy}{dt} = -1.5$) |
| 5 | $-\dfrac{1}{4}$ rad/s (the angle shrinks at $0.25$ rad/s) |
| 6 | $75$ mph |
| 7 | $0.0048$ rad/s |
| 8 | $\dfrac{9}{4\pi} \approx 0.716$ ft/min |
| 9 | $\dfrac{10}{3} \approx 3.33$ ft/s |
| 10 | $\dfrac{27}{\sqrt{5}} \approx 12.07$ units/s |

## Step-by-step solutions

**1.** Known: $\frac{dr}{dt} = 3$. Want: $\frac{dA}{dt}$ when $r = 10$. The area of a circle is $A = \pi r^2$. Differentiate with respect to $t$:

$$\frac{dA}{dt} = 2\pi r\frac{dr}{dt} = 2\pi(10)(3) = 60\pi \approx 188.5 \text{ cm}^2/\text{s}$$

**2.** Known: $\frac{dV}{dt} = 100$. Want: $\frac{dr}{dt}$ when $r = 5$. The volume is $V = \frac{4}{3}\pi r^3$:

$$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} \quad\Rightarrow\quad 100 = 4\pi(25)\frac{dr}{dt} \quad\Rightarrow\quad \frac{dr}{dt} = \frac{1}{\pi} \approx 0.318 \text{ cm/s}$$

**3.** Known: $\frac{ds}{dt} = 0.5$. Want: $\frac{dV}{dt}$ when $s = 4$. The volume is $V = s^3$:

$$\frac{dV}{dt} = 3s^2\frac{ds}{dt} = 3(16)(0.5) = 24 \text{ cm}^3/\text{s}$$

**4.** Let $x$ be the distance from the wall to the bottom and $y$ the height of the top. Known: $\frac{dx}{dt} = 2$. Want: $\frac{dy}{dt}$ when $x = 6$. The ladder length is fixed:

$$x^2 + y^2 = 100$$

When $x = 6$, $y = \sqrt{100 - 36} = 8$. Differentiate:

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \quad\Rightarrow\quad \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{6}{8}(2) = -1.5 \text{ ft/s}$$

The minus sign means the top is moving down.

**5.** Let $\theta$ be the angle at the ground. Then $\cos\theta = \frac{x}{10}$. Differentiate:

$$-\sin\theta\frac{d\theta}{dt} = \frac{1}{10}\frac{dx}{dt}$$

At this moment $\sin\theta = \frac{y}{10} = \frac{8}{10}$, so

$$-\frac{8}{10}\frac{d\theta}{dt} = \frac{2}{10} \quad\Rightarrow\quad \frac{d\theta}{dt} = -\frac{1}{4} \text{ rad/s}$$

The angle gets smaller as the ladder slides down.

**6.** Let $y$ be car A's distance north, $x$ car B's distance east and $z$ the distance between them. After $2$ hours, $y = 120$, $x = 90$ and $z = \sqrt{90^2 + 120^2} = 150$. From $x^2 + y^2 = z^2$:

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \quad\Rightarrow\quad \frac{dz}{dt} = \frac{90(45) + 120(60)}{150} = \frac{11250}{150} = 75 \text{ mph}$$

**7.** Let $h$ be the height and $\theta$ the angle of elevation. Known: $\frac{dh}{dt} = 4$. The observer is $300$ m away, so $\tan\theta = \frac{h}{300}$. Differentiate:

$$\sec^2\theta\frac{d\theta}{dt} = \frac{1}{300}\frac{dh}{dt}$$

When $h = 400$, the line of sight is $\sqrt{300^2 + 400^2} = 500$ m, so $\cos\theta = \frac{3}{5}$ and $\sec^2\theta = \frac{25}{9}$:

$$\frac{d\theta}{dt} = \frac{9}{25}\cdot\frac{4}{300} = 0.0048 \text{ rad/s}$$

**8.** Let $h$ be the water depth and $r$ the radius of the water surface. Known: $\frac{dV}{dt} = 9$. Want: $\frac{dh}{dt}$ when $h = 6$. By similar triangles, $\frac{r}{h} = \frac{4}{12}$, so $r = \frac{h}{3}$. Substitute into the cone volume to remove $r$:

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

Differentiate and substitute $h = 6$:

$$\frac{dV}{dt} = \frac{\pi h^2}{9}\frac{dh}{dt} \quad\Rightarrow\quad 9 = \frac{36\pi}{9}\frac{dh}{dt} = 4\pi\frac{dh}{dt} \quad\Rightarrow\quad \frac{dh}{dt} = \frac{9}{4\pi} \approx 0.716 \text{ ft/min}$$

**9.** Let $x$ be the person's distance from the post and $s$ the shadow length. Known: $\frac{dx}{dt} = 5$. The lamp, the person and the tip of the shadow form two similar triangles:

$$\frac{15}{x + s} = \frac{6}{s} \quad\Rightarrow\quad 15s = 6x + 6s \quad\Rightarrow\quad s = \frac{2}{3}x$$

Differentiate:

$$\frac{ds}{dt} = \frac{2}{3}\cdot 5 = \frac{10}{3} \approx 3.33 \text{ ft/s}$$

The answer does not depend on where the person is. The tip of the shadow moves at $5 + \frac{10}{3} = \frac{25}{3}$ ft/s.

**10.** Let $D$ be the distance to the origin. Since $y = x^2$:

$$D^2 = x^2 + y^2 = x^2 + x^4$$

Differentiate with respect to $t$:

$$2D\frac{dD}{dt} = (2x + 4x^3)\frac{dx}{dt}$$

At $(2, 4)$, $D = \sqrt{4 + 16} = \sqrt{20}$, and $\frac{dx}{dt} = 3$:

$$\frac{dD}{dt} = \frac{(4 + 32)(3)}{2\sqrt{20}} = \frac{108}{2\sqrt{20}} = \frac{27}{\sqrt{5}} \approx 12.07 \text{ units/s}$$

## Common mistakes on these problems

- **Plugging in numbers too early.** In problem 4, writing $36 + y^2 = 100$ before differentiating makes $x$ a constant and the rate disappears.
- **Ignoring the sign.** In problem 4 the answer is negative. Saying "the top moves at 1.5 ft/s" is fine, but you should know which way.
- **Keeping both $r$ and $h$ in the cone.** In problem 8, use similar triangles to write $r$ in terms of $h$ before differentiating.
- **Using degrees.** In problems 5 and 7 the rates come out in radians per second, because the trig derivatives only work in radians.

## Keep practicing

These 10 problems use fixed numbers. For unlimited practice with new numbers and instant feedback on each step, try the [related rates lesson](/learn/calculus-1/applications-of-derivatives/related-rates), then the mixed [applications practice](/learn/calculus-1/applications-of-derivatives/applications-practice). The other big word problem type is optimization: practice it with the [optimization practice problems](/practice-problems/optimization).

## Frequently asked questions

### How do you solve related rates problems?

Draw a picture and name every quantity that changes. Write down the rate you know and the rate you want. Find an equation linking the quantities, differentiate both sides with respect to time $t$, then substitute the values at the instant in question and solve.

### Why can't you plug in numbers before differentiating?

Because the numbers only hold at one instant. If you replace a changing quantity with a number before differentiating, its derivative becomes $0$ and you lose the rate. Only fixed quantities, like the length of a ladder, can go in early.

### Why is the answer to the ladder problem negative?

The top of the ladder is moving down, so its height $y$ is decreasing and $\frac{dy}{dt} < 0$. In words, the answer is "sliding down at 1.5 ft/s", and the negative sign is what tells you the direction.