25 Related Rates Problems to Practice On (With Step-by-Step Solutions)

Every related rates problem hides the same move: write an equation connecting two changing quantities, then differentiate both sides with respect to time. The scenery changes, ladders, cones, shadows, a rocket launch, but the mechanics don't.

Twenty-five problems below, sorted from a straight plug-into-a-formula warm-up to multi-step setups that stack two skills at once. Work each one yourself before opening the solution. That's the only way the pattern actually sticks.

Section 1: Direct-Formula Rates

Start where there's no setup work at all: a known formula, one differentiation, done. This is the raw chain-rule move you'll lean on everywhere else, so get it automatic before anything gets added on top.

Problem 1

The radius of a circle is increasing at $3\text{ cm/s}$ . How fast is the area increasing when $r = 5\text{ cm}$ ?

Step 1: Differentiate the area formula with respect to time. $A = \pi r^2 \quad \Rightarrow \quad \frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

Step 2: Plug in the values. $\frac{dA}{dt} = 2\pi(5)(3) = 30\pi \approx 94.2 \text{ cm}^2/\text{s}$

Answer30π ≈ 94.2 cm²/s

Problem 2

The side of a square is growing at $4\text{ cm/s}$ . How fast is the area increasing when the side is $10\text{ cm}$ ?

Step 1: Differentiate the area formula. $A = s^2 \quad \Rightarrow \quad \frac{dA}{dt} = 2s \frac{ds}{dt}$

Step 2: Plug in the values. $\frac{dA}{dt} = 2(10)(4) = 80 \text{ cm}^2/\text{s}$

Answer80 cm²/s

Problem 3

The side of a cube is increasing at $2\text{ cm/s}$ . How fast is the volume increasing when the side is $6\text{ cm}$ ?

Step 1: Differentiate the volume formula. $V = s^3 \quad \Rightarrow \quad \frac{dV}{dt} = 3s^2 \frac{ds}{dt}$

Step 2: Plug in the values. $\frac{dV}{dt} = 3(6)^2(2) = 216 \text{ cm}^3/\text{s}$

Answer216 cm³/s

Problem 4

The radius of a circular oil slick is shrinking at $0.5\text{ m/s}$ . How fast is the area decreasing when $r = 20\text{ m}$ ?

Step 1: Differentiate the area formula. $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

Step 2: Plug in the values, using a negative rate since the radius is shrinking. $\frac{dA}{dt} = 2\pi(20)(-0.5) = -20\pi \approx -62.8 \text{ m}^2/\text{s}$

AnswerShrinking at 20π ≈ 62.8 m²/s

Problem 5

Air is pumped into a spherical balloon, and its radius grows at $0.2\text{ cm/s}$ . How fast is the volume increasing when $r = 10\text{ cm}$ ?

Step 1: Differentiate the volume formula. $V = \frac{4}{3}\pi r^3 \quad \Rightarrow \quad \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

Step 2: Plug in the values. $\frac{dV}{dt} = 4\pi(10)^2(0.2) = 80\pi \approx 251.3 \text{ cm}^3/\text{s}$

Answer80π ≈ 251.3 cm³/s

Section 2: Right-Triangle Setups

Now you have to build the equation yourself before you can differentiate it. Most of these come from the Pythagorean theorem, and the setup is the real challenge. The calculus at the end is quick.

Problem 6

A $10\text{-ft}$ ladder leans against a wall. The bottom slides away from the wall at $1\text{ ft/s}$ . How fast is the top sliding down the wall when the bottom is $6\text{ ft}$ from the wall?

Step 1: Set up the constraint and differentiate. With $x$ the base distance and $y$ the height, $x^2 + y^2 = 10^2 \quad \Rightarrow \quad 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

Step 2: Find $y$ when $x = 6$ . Since $6^2 + y^2 = 100$ , $y = 8$ .

Step 3: Solve for $\frac{dy}{dt}$ . $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{6}{8}(1) = -0.75 \text{ ft/s}$

AnswerThe top slides down at 0.75 ft/s

Problem 7

A $13\text{-ft}$ ladder leans against a wall. The bottom slides away from the wall at $2\text{ ft/s}$ . How fast is the top sliding down when the bottom is $5\text{ ft}$ from the wall?

Step 1: Find $y$ when $x = 5$ . From $x^2+y^2=13^2$ , $y = 12$ .

Step 2: Solve for $\frac{dy}{dt}$ . $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{5}{12}(2) = -\frac{5}{6} \approx -0.83 \text{ ft/s}$

AnswerThe top slides down at ≈0.83 ft/s

Problem 8

Two cars leave the same intersection at the same time. Car A drives east at $30\text{ mph}$ , car B drives north at $40\text{ mph}$ . How fast is the distance between them growing after $2$ hours?

Step 1: Find the positions at $t = 2\text{ hr}$ . $x = 30(2) = 60\text{ mi}$ , $y = 40(2) = 80\text{ mi}$ .

Step 2: Set up the distance constraint and differentiate. $s^2 = x^2 + y^2 \quad \Rightarrow \quad 2s\frac{ds}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$

Step 3: Find $s$ and solve. $s = \sqrt{60^2+80^2} = 100\text{ mi}$ . $\frac{ds}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{s} = \frac{60(30) + 80(40)}{100} = 50 \text{ mph}$

Answer50 mph

Problem 9

A boat is pulled toward a dock by a rope through a pulley $4\text{ ft}$ above the water. The rope is reeled in at $2\text{ ft/s}$ . How fast is the boat approaching the dock when $5\text{ ft}$ of rope is out?

Step 1: Set up the constraint. With $L$ the rope length and $x$ the horizontal distance to the dock, $L^2 = x^2 + 4^2$

Step 2: Find $x$ when $L = 5$ . $x^2 = 25 - 16 = 9$ , so $x = 3\text{ ft}$ .

Step 3: Differentiate and solve. The rope is being reeled in, so $\frac{dL}{dt} = -2$ . $2L\frac{dL}{dt} = 2x\frac{dx}{dt} \quad \Rightarrow \quad \frac{dx}{dt} = \frac{L}{x}\frac{dL}{dt} = \frac{5}{3}(-2) = -\frac{10}{3} \approx -3.33 \text{ ft/s}$

AnswerThe boat approaches at ≈3.33 ft/s

Problem 10

A $25\text{-ft}$ ladder is pushed toward a wall so the bottom moves toward the wall at $2\text{ ft/s}$ . How fast is the top rising when the bottom is $7\text{ ft}$ from the wall?

Step 1: Find $y$ when $x = 7$ . From $x^2+y^2=25^2$ , $y = 24$ .

Step 2: Solve for $\frac{dy}{dt}$ . The bottom moves toward the wall, so $\frac{dx}{dt} = -2$ . $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{7}{24}(-2) = \frac{7}{12} \approx 0.58 \text{ ft/s}$

AnswerThe top rises at ≈0.58 ft/s

Section 3: Volumes with a Hidden Constraint

Cones, piles, and troughs don't give you two independent variables. Radius and height are locked together by geometry, so substitute that relationship in before you differentiate, or the derivative won't simplify.

Problem 11

Water is pumped into a cylindrical tank of radius $3\text{ m}$ at $2\text{ m}^3/\text{min}$ . How fast is the water level rising?

Step 1: Differentiate the volume formula. The radius is constant, so $V = \pi r^2 h \quad \Rightarrow \quad \frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$

Step 2: Solve. $\frac{dh}{dt} = \frac{2}{\pi(3)^2} = \frac{2}{9\pi} \approx 0.071 \text{ m/min}$

Answer≈0.071 m/min

Problem 12

Water drains from a conical tank, vertex down, with radius $4\text{ ft}$ and height $10\text{ ft}$ at the top, at $5\text{ ft}^3/\text{min}$ . How fast is the water level falling when $h = 6\text{ ft}$ ?

Step 1: Relate $r$ and $h$ with similar triangles. $\frac{r}{h} = \frac{4}{10} = \frac{2}{5}$ , so $r = \frac{2h}{5}$ .

Step 2: Substitute into the volume formula and differentiate. $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{2h}{5}\right)^2 h = \frac{4\pi}{75}h^3 \quad \Rightarrow \quad \frac{dV}{dt} = \frac{4\pi h^2}{25}\frac{dh}{dt}$

Step 3: Plug in $h = 6$ and $\frac{dV}{dt} = -5$ . $-5 = \frac{4\pi(36)}{25}\frac{dh}{dt} = \frac{144\pi}{25}\frac{dh}{dt} \quad \Rightarrow \quad \frac{dh}{dt} = -\frac{125}{144\pi} \approx -0.276 \text{ ft/min}$

AnswerThe level falls at ≈0.276 ft/min

Problem 13

Sand poured onto a conical pile at $10\text{ ft}^3/\text{min}$ always keeps its height equal to its radius. How fast is the height increasing when $h = 5\text{ ft}$ ?

Step 1: Substitute $r = h$ into the volume formula. $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi h^3 \quad \Rightarrow \quad \frac{dV}{dt} = \pi h^2 \frac{dh}{dt}$

Step 2: Plug in $h = 5$ and $\frac{dV}{dt} = 10$ . $10 = \pi(25)\frac{dh}{dt} \quad \Rightarrow \quad \frac{dh}{dt} = \frac{2}{5\pi} \approx 0.127 \text{ ft/min}$

Answer≈0.127 ft/min

Problem 14

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

Step 1: Differentiate the volume formula. $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

Step 2: Plug in $r = 5$ and $\frac{dV}{dt} = 100$ . $100 = 4\pi(25)\frac{dr}{dt} = 100\pi\frac{dr}{dt} \quad \Rightarrow \quad \frac{dr}{dt} = \frac{1}{\pi} \approx 0.318 \text{ cm/s}$

Answer≈0.318 cm/s

Problem 15

A trough $10\text{ ft}$ long has a triangular cross-section: $2\text{ ft}$ wide at the top, $3\text{ ft}$ deep, vertex at the bottom. Water flows in at $5\text{ ft}^3/\text{min}$ . How fast is the water level rising when $h = 1\text{ ft}$ ?

Step 1: Relate the surface width $w$ to the depth $h$ with similar triangles. $\frac{w}{h} = \frac{2}{3}$ , so $w = \frac{2h}{3}$ .

Step 2: Write the volume and differentiate. Cross-sectional area is $\frac{1}{2}wh = \frac{h^2}{3}$ , so $V = 10\left(\frac{h^2}{3}\right) \quad \Rightarrow \quad \frac{dV}{dt} = \frac{20h}{3}\frac{dh}{dt}$

Step 3: Plug in $h = 1$ and $\frac{dV}{dt} = 5$ . $5 = \frac{20}{3}\frac{dh}{dt} \quad \Rightarrow \quad \frac{dh}{dt} = 0.75 \text{ ft/min}$

Answer0.75 ft/min

Section 4: Angles and Trig Relationships

Distances turn into angles here, so sine, cosine, and tangent do the connecting instead of the Pythagorean theorem. Watch for $\sec^2\theta$ : it shows up constantly once tangent enters the picture.

Problem 16

A $6\text{-ft}$ person walks away from a $15\text{-ft}$ lamppost at $5\text{ ft/s}$ . How fast is the tip of their shadow moving?

Step 1: Set up the similar triangles. With $x$ the person's distance from the pole and $s$ the shadow length, $\frac{15}{x+s} = \frac{6}{s} \quad \Rightarrow \quad 9s = 6x \quad \Rightarrow \quad s = \frac{2}{3}x$

Step 2: Differentiate. $\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt} = \frac{2}{3}(5) = \frac{10}{3} \text{ ft/s}$

Step 3: Add the person's speed to get the tip's speed. The tip is at distance $x+s$ from the pole. $\frac{d(x+s)}{dt} = 5 + \frac{10}{3} = \frac{25}{3} \approx 8.33 \text{ ft/s}$

Answer≈8.33 ft/s

Problem 17

A hot air balloon rises at $5\text{ ft/s}$ from a point $200\text{ ft}$ from an observer. How fast is the angle of elevation changing when the balloon is $150\text{ ft}$ high?

Step 1: Set up the trig relationship and differentiate. With $h$ the height and $x = 200$ fixed, $\tan\theta = \frac{h}{200} \quad \Rightarrow \quad \sec^2\theta\frac{d\theta}{dt} = \frac{1}{200}\frac{dh}{dt}$

Step 2: Find $\sec^2\theta$ at $h = 150$ . The hypotenuse is $\sqrt{200^2+150^2}=250$ , so $\sec\theta = \frac{250}{200}=1.25$ and $\sec^2\theta = 1.5625$ .

Step 3: Solve. $\frac{d\theta}{dt} = \frac{\frac{1}{200}(5)}{1.5625} = \frac{0.025}{1.5625} = 0.016 \text{ rad/s}$

Answer0.016 rad/s

Problem 18

For the ladder in Problem 6 (the $10\text{-ft}$ ladder, bottom sliding away at $1\text{ ft/s}$ ), how fast is the angle between the ladder and the ground changing when the bottom is $6\text{ ft}$ from the wall?

Step 1: Write $x$ in terms of $\theta$ and differentiate. With $\theta$ the angle between the ladder and the ground, $x = 10\cos\theta$ , so $\frac{dx}{dt} = -10\sin\theta\frac{d\theta}{dt}$

Step 2: Find $\sin\theta$ at $x = 6, y = 8$ . $\sin\theta = \frac{y}{10} = 0.8$ .

Step 3: Solve. $1 = -10(0.8)\frac{d\theta}{dt} = -8\frac{d\theta}{dt} \quad \Rightarrow \quad \frac{d\theta}{dt} = -0.125 \text{ rad/s}$

AnswerThe angle shrinks at 0.125 rad/s

Problem 19

A kite flies at a constant height of $100\text{ ft}$ , blown away horizontally at $8\text{ ft/s}$ . How fast is the angle of elevation decreasing when $260\text{ ft}$ of string is out?

Step 1: Find the horizontal distance $x$ . From $x^2 + 100^2 = 260^2$ , $x = 240\text{ ft}$ .

Step 2: Set up the trig relationship and differentiate. $\tan\theta = \frac{100}{x} \quad \Rightarrow \quad \sec^2\theta\frac{d\theta}{dt} = -\frac{100}{x^2}\frac{dx}{dt}$

Step 3: Find $\sec^2\theta$ . $\tan\theta = \frac{100}{240} = \frac{5}{12}$ , so $\sec^2\theta = 1 + \frac{25}{144} = \frac{169}{144}$ .

Step 4: Solve. $\frac{d\theta}{dt} = \frac{-\frac{100}{57600}(8)}{\frac{169}{144}} = -\frac{2}{169} \approx -0.0118 \text{ rad/s}$

AnswerDecreasing at ≈0.0118 rad/s

Problem 20

A rocket launches vertically $3000\text{ ft}$ from an observer, rising at $880\text{ ft/s}$ . How fast is the angle of elevation changing when the rocket is $4000\text{ ft}$ high?

Step 1: Set up the trig relationship and differentiate. $\tan\theta = \frac{h}{3000} \quad \Rightarrow \quad \sec^2\theta\frac{d\theta}{dt} = \frac{1}{3000}\frac{dh}{dt}$

Step 2: Find $\sec^2\theta$ at $h = 4000$ . $\tan\theta = \frac{4000}{3000} = \frac{4}{3}$ , so $\sec^2\theta = 1 + \frac{16}{9} = \frac{25}{9}$ .

Step 3: Solve. $\frac{d\theta}{dt} = \frac{\frac{880}{3000}}{\frac{25}{9}} = \frac{66}{625} \approx 0.106 \text{ rad/s}$

Answer≈0.106 rad/s

Section 5: Multi-Step and Combined Setups

These stack two skills at once: a unit conversion buried inside a trig setup, a product rule sitting next to an implicit differentiation, a physical law standing in for the equation you'd normally have to derive yourself. This is what the earlier sections were building toward.

Problem 21

A plane flies at a constant altitude of $3\text{ mi}$ and a constant speed of $600\text{ mph}$ , passing directly over a radar station. How fast is the angle of elevation changing, in radians per minute, when the plane is $4\text{ mi}$ of horizontal distance past the station?

Step 1: Convert the speed to miles per minute. $600 \text{ mph} = \frac{600}{60} = 10 \text{ mi/min}$

Step 2: Set up the trig relationship and differentiate. With $x$ the horizontal distance, $\tan\theta = \frac{3}{x} \quad \Rightarrow \quad \sec^2\theta\frac{d\theta}{dt} = -\frac{3}{x^2}\frac{dx}{dt}$

Step 3: Find $\sec^2\theta$ at $x = 4$ . $\sec^2\theta = 1 + \frac{9}{x^2} = \frac{x^2+9}{x^2} = \frac{25}{16}$ .

Step 4: Solve. $\frac{d\theta}{dt} = \frac{-\frac{3}{16}(10)}{\frac{25}{16}} = -\frac{3(10)}{25} = -1.2 \text{ rad/min}$

Answer-1.2 rad/min

Problem 22

Two resistors are wired in parallel: $R_1 = 20\ \Omega$ increasing at $3\ \Omega/\text{s}$ , and $R_2 = 30\ \Omega$ increasing at $2\ \Omega/\text{s}$ . Using $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ , how fast is the combined resistance $R$ changing?

Step 1: Find $R$ at the given instant. $\frac{1}{R} = \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12} \quad \Rightarrow \quad R = 12\ \Omega$

Step 2: Differentiate the constraint. $-\frac{1}{R^2}\frac{dR}{dt} = -\frac{1}{R_1^2}\frac{dR_1}{dt} - \frac{1}{R_2^2}\frac{dR_2}{dt}$

Step 3: Solve for $\frac{dR}{dt}$ . $\frac{dR}{dt} = R^2\left(\frac{1}{R_1^2}\frac{dR_1}{dt} + \frac{1}{R_2^2}\frac{dR_2}{dt}\right) = 144\left(\frac{3}{400} + \frac{2}{900}\right) = 144\left(\frac{7}{720}\right) = 1.4\ \Omega/\text{s}$

Answer1.4 Ω/s

Problem 23

Two sides of a triangle are $a = 4\text{ m}$ and $b = 5\text{ m}$ , growing at $2\text{ m/s}$ and $3\text{ m/s}$ respectively. The included angle stays fixed at $60°$ . How fast is the area growing?

Step 1: Write the area formula and differentiate with the product rule. Since $\theta$ is constant, $A = \frac{1}{2}ab\sin\theta \quad \Rightarrow \quad \frac{dA}{dt} = \frac{1}{2}\sin\theta\left(a\frac{db}{dt} + b\frac{da}{dt}\right)$

Step 2: Plug in the values. $\sin 60° = \frac{\sqrt{3}}{2}$ . $\frac{dA}{dt} = \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)\left(4(3) + 5(2)\right) = \frac{\sqrt{3}}{4}(22) = \frac{11\sqrt{3}}{2} \approx 9.53 \text{ m}^2/\text{s}$

Answer11√3/2 ≈ 9.53 m²/s

Problem 24

A gas obeys Boyle's Law, $PV = 800$ (constant, in appropriate units). If the pressure is increasing at $2\text{ kPa/min}$ , how fast is the volume changing when $P = 40\text{ kPa}$ and $V = 20\text{ L}$ ?

Step 1: Differentiate the constraint. $PV = 800$ is constant, so $P\frac{dV}{dt} + V\frac{dP}{dt} = 0$

Step 2: Solve for $\frac{dV}{dt}$ . $\frac{dV}{dt} = -\frac{V}{P}\frac{dP}{dt} = -\frac{20}{40}(2) = -1 \text{ L/min}$

AnswerShrinking at 1 L/min

Problem 25

Two sides of a triangle are $a = 7\text{ cm}$ and $b = 4\text{ cm}$ , growing at $2\text{ cm/s}$ and $3\text{ cm/s}$ . The included angle is fixed at $60°$ . Using the law of cosines, how fast is the third side $c$ changing?

Step 1: Find $c$ at the given instant. With $\cos 60° = 0.5$ , $c^2 = a^2 + b^2 - 2ab\cos\theta = 49 + 16 - 28 = 37 \quad \Rightarrow \quad c = \sqrt{37}$

Step 2: Differentiate the law of cosines. Since $\theta$ is constant, the $\cos\theta$ term only needs the product rule on $ab$ . $2c\frac{dc}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt} - 2\cos\theta\left(a\frac{db}{dt} + b\frac{da}{dt}\right)$

Step 3: Plug in the values. $2c\frac{dc}{dt} = 2(7)(2) + 2(4)(3) - 2(0.5)\left(7(3)+4(2)\right) = 28 + 24 - 29 = 23$

Step 4: Solve. $\frac{dc}{dt} = \frac{23}{2\sqrt{37}} \approx 1.89 \text{ cm/s}$

Answer≈1.89 cm/s

Every angle problem here is differentiated in radians, since $\frac{d}{d\theta}\sin\theta$ , $\cos\theta$ , and $\tan\theta$ only take that clean form when $\theta$ is measured in radians. If you need degrees for a final report, convert last, never before differentiating.