Day 10

Related Rates

Related rates problems connect two or more changing quantities through a single equation — you're given how fast one quantity changes and asked to find how fast another one is changing at the same instant, using implicit differentiation with respect to time.

Core Theorem

General strategy:
(1) Draw a picture and label all quantities.
(2) List known and unknown rates.
(3) Write an equation relating the variables — do NOT substitute values that change until after differentiating.
(4) Differentiate both sides with respect to

t

.
(5) Substitute the known values at the specific instant.
(6) Solve for the desired rate.

Watch the TikTok ExplanationRelated Rates: Spheres→Watch the TikTok ExplanationRelated Rates: Triangles/Ladders→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Air is pumped into a spherical balloon at a rate of

100\ \text{cm}^3/\text{sec}

. Find the rate at which the radius is increasing when the diameter is

50\ \text{cm}

How to think about it

Everything in the problem is a function of time. Write one equation relating the quantities, differentiate the whole thing with respect to

t

(chain rule on every variable), then plug in the instant last.

Key idea: Start from $V = \frac{4}{3}\pi r^3$ , differentiate with respect to $t$ , then substitute $r=25$ (half the diameter) and $\frac{dV}{dt}=100$ .

Full solution

1. Step 1: Write the Volume Equation

V = \frac{4}{3}\pi r^3

2. Step 2: Differentiate w.r.t. Time

\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

3. Step 3: Substitute Known Values

D=50

r=25

. So

100 = 4\pi(25)^2\frac{dr}{dt} = 2500\pi\frac{dr}{dt}

4. Step 4: Solve

\frac{dr}{dt} = \frac{100}{2500\pi} = \frac{1}{25\pi}\ \text{cm/sec} \approx 0.0127\ \text{cm/sec}

Worked Example 02Medium

5\text{m}

ladder leans against a wall. The base slides away from the wall at

1\ \text{m/sec}

. How fast is the top of the ladder descending when the base is

3\text{m}

from the wall?

How to think about it

Key idea: Let $x$ be the base's distance from the wall and $y$ the height of the top. Use the Pythagorean relationship $x^2+y^2=25$ .

Full solution

1. Step 1: Set Up the Equation

x^2 + y^2 = 25

(the ladder's length is fixed at 5m).

2. Step 2: Differentiate w.r.t. Time

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies x\frac{dx}{dt} + y\frac{dy}{dt} = 0

3. Step 3: Find y When x = 3

3^2+y^2=25 \implies y^2=16 \implies y=4

4. Step 4: Substitute and Solve

3(1) + 4\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{3}{4}\ \text{m/sec}

. The negative sign confirms the top is descending, at

\frac{3}{4}

m/sec.

Step-by-Step SOP

1
Draw and Label
Sketch the situation and label every quantity, marking which ones are changing and which are fixed.
2
Write the Governing Equation
Find a single equation relating all the changing quantities — geometric formulas, the Pythagorean theorem, and trig ratios are the most common sources.
3
Differentiate, Then Substitute
Differentiate both sides with respect to $t$ first. Only after that should you plug in the specific numerical values for this instant.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

At an intersection, a police car is

0.6

miles north of the intersection, driving south at

60\ \text{mph}

. A speeding car is

0.8

miles east of the intersection, driving east. Radar shows the distance between the two cars is increasing at

20\ \text{mph}

. Find the speeding car's actual speed.

Need a hint?

Place the intersection at the origin. Let

x

be the speeding car's distance east and

y

the police car's distance north, with

z=\sqrt{x^2+y^2}

the distance between them. Careful with the sign of

\frac{dy}{dt}

— the police car is APPROACHING the intersection.

Show solution

1. Step 1: Set Up the Distance Equation

z^2 = x^2+y^2

, where

x=0.8

y=0.6

\frac{dy}{dt}=-60

(decreasing, since the police car moves toward the intersection), and

\frac{dz}{dt}=20

2. Step 2: Find z at This Instant

z = \sqrt{0.8^2+0.6^2} = \sqrt{0.64+0.36} = \sqrt{1} = 1

3. Step 3: Differentiate w.r.t. Time

2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \implies z\frac{dz}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}

4. Step 4: Substitute and Solve

(1)(20) = (0.8)\frac{dx}{dt} + (0.6)(-60) \implies 20 = 0.8\frac{dx}{dt} - 36 \implies 0.8\frac{dx}{dt} = 56 \implies \frac{dx}{dt} = 70\ \text{mph}

Practice 02AB/BC Standard

An inverted conical tank has height

4\text{m}

and top radius

2\text{m}

. Water is poured in at

2\ \text{m}^3/\text{min}

. Find how fast the water level is rising when the depth is

3\text{m}

, and find the acceleration of the water level at that instant.

Need a hint?

Use similar triangles to write

r

in terms of

h

first, so the volume formula only has one variable. For the acceleration, differentiate the rate equation a second time.

Show solution

1. Step 1: Relate r and h with Similar Triangles

Since the tank has radius 2 at height 4,

\frac{r}{h} = \frac{2}{4} = \frac{1}{2} \implies r = \frac{h}{2}

2. Step 2: Write Volume in Terms of h Only

V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}

3. Step 3: Differentiate for dh/dt

\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}

. At

h=3

2 = \frac{9\pi}{4}\frac{dh}{dt} \implies \frac{dh}{dt} = \frac{8}{9\pi}\ \text{m/min}

4. Step 4: Differentiate Again for the Acceleration

Since

\frac{dV}{dt}=2

is constant, differentiating

2 = \frac{\pi h^2}{4}\frac{dh}{dt}

again gives

0 = \frac{\pi}{4}\left[2h\left(\frac{dh}{dt}\right)^2 + h^2\frac{d^2h}{dt^2}\right]

, so

\frac{d^2h}{dt^2} = -\frac{2\left(\frac{dh}{dt}\right)^2}{h}

5. Step 5: Substitute

h=3

\frac{dh}{dt}=\frac{8}{9\pi}

\frac{d^2h}{dt^2} = -\frac{2\left(\frac{8}{9\pi}\right)^2}{3} = -\frac{128}{243\pi^2}\ \text{m/min}^2

— the rate of rise is actually slowing down over time, even though the tank is filled at a constant volume rate.

Practice 03Hard

A person walks along a straight road at

1.5\ \text{m/s}

. A spotlight

6\text{m}

from the road stays trained on them. Find how fast the spotlight is rotating when the person is

8\text{m}

from the point on the road closest to the spotlight.

Need a hint?

Let

\theta

be the angle between the spotlight's perpendicular beam and the beam aimed at the person, and

x

their distance along the road. Use

\tan\theta = \frac{x}{6}

Show solution

1. Step 1: Set Up the Angle Equation

\tan\theta = \frac{x}{6}

, where

\frac{dx}{dt}=1.5

2. Step 2: Differentiate w.r.t. Time

\sec^2\theta \cdot \frac{d\theta}{dt} = \frac{1}{6}\frac{dx}{dt}

3. Step 3: Find sec²θ When x = 8

The hypotenuse is

\sqrt{6^2+8^2} = \sqrt{100}=10

, so

\cos\theta = \frac{6}{10}=\frac{3}{5}

and

\sec^2\theta = \left(\frac{5}{3}\right)^2 = \frac{25}{9}

4. Step 4: Substitute and Solve

\frac{25}{9}\frac{d\theta}{dt} = \frac{1}{6}(1.5) = \frac{1}{4} \implies \frac{d\theta}{dt} = \frac{1}{4}\cdot\frac{9}{25} = \frac{9}{100} = 0.09\ \text{rad/sec}

Practice 04Medium

A person

2\text{m}

tall walks away from a

6\text{m}

lamppost at

1.2\ \text{m/s}

. Find the rate at which their shadow's length is growing, and the rate at which the tip of the shadow is moving.

Need a hint?

Let

x

be the person's distance from the pole and

s

the shadow's length. Use similar triangles between the pole/full distance triangle and the person/shadow triangle.

Show solution

1. Step 1: Set Up the Similar Triangles

\frac{6}{x+s} = \frac{2}{s} \implies 6s = 2x+2s \implies 4s = 2x \implies s = \frac{x}{2}

2. Step 2: Differentiate for ds/dt

\frac{ds}{dt} = \frac{1}{2}\frac{dx}{dt} = \frac{1}{2}(1.2) = 0.6\ \text{m/s}

— notice this rate is constant, independent of how far the person has walked.

3. Step 3: Find the Tip's Speed

The shadow tip's position is

x+s

, so its speed is

\frac{d(x+s)}{dt} = \frac{dx}{dt}+\frac{ds}{dt} = 1.2+0.6 = 1.8\ \text{m/s}

Common Pitfalls

⚠
Never Substitute a Changing Value Before DifferentiatingPlugging in a specific number for a variable that's still changing (like $x=3$ ) before you differentiate destroys that variable's rate of change — always differentiate the general equation first, substitute last.
⚠
Watch the Sign of Given RatesA quantity that's decreasing (like the police car's distance from the intersection) has a NEGATIVE rate — don't automatically treat every given rate as positive.
⚠
Reduce to One Variable When PossibleIf two variables are linked by a fixed ratio (like the cone's radius and height, or the ladder's constant length), substitute that relationship in BEFORE differentiating so you only have to track one changing variable.

Exponential Growth & Decay

Whenever a quantity's rate of change is proportional to the quantity itself — population, radioactive mass, temperature difference, money under continuous interest — it follows the exact same exponential model. Recognizing this pattern turns a wide variety of word problems into the same three-step solve.

Core Theorem

\frac{dy}{dt} = ky

for a constant

k

, then

y = y_0 e^{kt}

, where

y_0 = y(0)

k>0

gives exponential growth;

k<0

gives exponential decay.
Half-life

T

satisfies

\frac{1}{2} = e^{kT}

, i.e.

k = -\frac{\ln 2}{T}

.
Newton's Law of Cooling:

H(t) = H_s + (H_0-H_s)e^{-kt}

, where

H_s

is the surrounding temperature and

H_0

is the initial temperature.

Watch the TikTok ExplanationExponential Growth and Decay→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

World population was

2.56

billion in 1950 and

3.04

billion in 1960. Assuming exponential growth, find the relative growth rate, then estimate the population in 1993 (43 years after 1950).

How to think about it

Any 'grows/decays proportionally' situation is

y = y_0 e^{kt}

. Use one data point to solve for

k

first; only then answer the actual question.

Key idea: Let $t=0$ correspond to 1950. Use the 1960 data point ( $t=10$ ) to solve for $k$ first.

Full solution

1. Step 1: Set Up the Model

P(t) = 2.56e^{kt}

(billions), with

t

in years since 1950.

2. Step 2: Solve for k Using the 1960 Data

P(10) = 3.04 = 2.56e^{10k} \implies e^{10k} = \frac{3.04}{2.56} = 1.1875 \implies k = \frac{\ln(1.1875)}{10} \approx 0.0172\ \text{per year}

3. Step 3: Estimate the 1993 Population (t = 43)

P(43) = 2.56e^{43(0.0172)} = 2.56e^{0.739} \approx 2.56(2.094) \approx 5.36

billion — reasonably close to the true 1993 figure of about

5.5

billion.

Worked Example 02Medium

Radium-226 has a half-life of

1590

years. A sample starts at

100\text{mg}

. Find the mass after

1000

years, and the time until only

30\text{mg}

remains.

How to think about it

Key idea: Use the half-life to find $k$ first: $\frac{1}{2} = e^{1590k}$ .

Full solution

1. Step 1: Find k from the Half-Life

k = -\frac{\ln 2}{1590} \approx -0.000436\ \text{per year}

2. Step 2: Write the Mass Function

m(t) = 100e^{-0.000436t}

3. Step 3: Evaluate at t = 1000

m(1000) = 100e^{-0.436} \approx 100(0.6466) \approx 64.7\text{mg}

4. Step 4: Solve for t When m(t) = 30

30 = 100e^{-0.000436t} \implies e^{-0.000436t} = 0.3 \implies t = \frac{\ln(0.3)}{-0.000436} \approx 2762\ \text{years}

Step-by-Step SOP

1
Identify y₀ and Set Up y = y₀e^(kt)
Find the initial value (at $t=0$ ) to use as $y_0$ ; if the problem gives a half-life or doubling time instead, convert that to $k$ directly.
2
Solve for k Using the Given Data Point
Substitute the known $(t,y)$ pair and isolate $k$ using logarithms.
3
Answer the Actual Question
Substitute back into the full formula to evaluate at a new time, or solve for $t$ given a target value of $y$ .

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

A cup of water at

22°C

is placed in a

7°C

refrigerator. After half an hour, its temperature is

16°C

. Find the temperature after one full hour, and the time needed to reach

10°C

Need a hint?

Use

H(t) = H_s + (H_0-H_s)e^{-kt}

with

H_s=7

H_0=22

, and the half-hour data point to solve for

k

Show solution

1. Step 1: Set Up the Cooling Equation

H(t) = 7 + 15e^{-kt}

2. Step 2: Solve for k Using H(0.5) = 16

16 = 7+15e^{-0.5k} \implies e^{-0.5k} = \frac{9}{15} = 0.6 \implies k = -2\ln(0.6) \approx 1.022\ \text{per hour}

3. Step 3: Find H(1)

Since

e^{-k} = (e^{-0.5k})^2 = 0.6^2 = 0.36

H(1) = 7+15(0.36) = 12.4°C

4. Step 4: Solve for t When H(t) = 10

10 = 7+15e^{-1.022t} \implies e^{-1.022t} = 0.2 \implies t = \frac{\ln(0.2)}{-1.022} \approx 1.575\ \text{hours} \approx 94.5\ \text{minutes}

Practice 02Easy

You invest

1000

dollars at a continuously compounded interest rate of

6\%

. Find the amount after 3 years.

Need a hint?

Continuous compounding is exactly the exponential growth model:

A(t) = A_0e^{rt}

Show solution

1. Step 1: Set Up the Formula

A(t) = 1000e^{0.06t}

2. Step 2: Evaluate at t = 3

A(3) = 1000e^{0.18} \approx 1000(1.1972) \approx 1197.22

dollars.

Practice 03Medium

A quantity

y

satisfies

y'(t) = ky(t)

with

y(0)=5

and

y(1)=15

. Find

k

and

y(2)

Need a hint?

Write

y(t)=5e^{kt}

, use

y(1)=15

to solve for

k

exactly (the answer will be a clean logarithm).

Show solution

1. Step 1: Set Up and Solve for k

y(t) = 5e^{kt}

y(1)=15 \implies 5e^k=15 \implies e^k=3 \implies k=\ln 3

2. Step 2: Find y(2)

y(2) = 5e^{2\ln 3} = 5(e^{\ln 3})^2 = 5(3^2) = 5(9) = 45

Common Pitfalls

⚠
Always Solve for k FirstEvery one of these problems follows the same order: use the given data point to pin down $k$ exactly, THEN answer whatever the question actually asks — trying to skip straight to the answer usually backfires.
⚠
Keep Units ConsistentIf $t$ is measured in years for one part of a problem, don't accidentally switch to months or hours later without converting — this is a very common careless error.

Linearization & Differentials

Near any point where a function is differentiable, its curve looks almost exactly like its tangent line — close enough to use that line as a fast approximation tool, and to estimate how errors in an input propagate into errors in the output.

Core Theorem

The linearization of

f

x=a

L(x) = f(a) + f'(a)(x-a)

, and

f(x) \approx L(x)

for

x

near

a

.
The differential

dy = f'(x)\,dx

approximates the actual change

\Delta y = f(x+\Delta x) - f(x)

when

\Delta x

(written

dx

) is small.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Find the linearization of

f(x) = \sqrt{3+x}

x=0

and at

x=1

How to think about it

The tangent line at

x=a

is the best straight-line stand-in for

f

near

a

. Build it once with

L(x) = f(a) + f'(a)(x-a)

, then evaluate it instead of the messy function.

Key idea: Compute $f(a)$ and $f'(a)$ at each point, then plug into $L(x)=f(a)+f'(a)(x-a)$ .

Full solution

1. Step 1: Find f'(x)

f'(x) = \frac{1}{2\sqrt{3+x}}

2. Step 2: Linearize at a = 0

f(0)=\sqrt{3}

f'(0)=\frac{1}{2\sqrt3}

L(x) = \sqrt{3} + \frac{1}{2\sqrt{3}}x

3. Step 3: Linearize at a = 1

f(1)=2

f'(1)=\frac{1}{4}

L(x) = 2 + \frac{1}{4}(x-1)

Worked Example 02Medium

Use linearization to estimate

\sqrt{3.98}

and

\sqrt{4.05}

How to think about it

Key idea: Pick $a=4$ since $\sqrt{4}=2$ is clean, and $3.98$ , $4.05$ are both very close to it.

Full solution

1. Step 1: Set Up the Linearization at a = 4

f(x)=\sqrt{x}

f'(x)=\frac{1}{2\sqrt{x}}

f(4)=2

f'(4)=\frac{1}{4}

L(x) = 2+\frac{1}{4}(x-4)

2. Step 2: Estimate √3.98

L(3.98) = 2+\frac{1}{4}(-0.02) = 2-0.005 = 1.995

3. Step 3: Estimate √4.05

L(4.05) = 2+\frac{1}{4}(0.05) = 2+0.0125 = 2.0125

Step-by-Step SOP

1
Choose a Nearby 'Nice' Point a
Pick a point close to the value you actually want, where $f(a)$ and $f'(a)$ are easy to compute exactly.
2
Build L(x) = f(a) + f'(a)(x-a)
Compute $f(a)$ and $f'(a)$ , then substitute into the linearization formula.
3
For Error Propagation, Use dy = f'(x)dx
Treat the measurement error as $dx$ , and the resulting output error is approximately $f'(x)\,dx$ .

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

For

f(x) = x^3+x^2-2x+1

, compare

\Delta y

and

dy

x

changes from

2

2.05

, and separately from

2

2.01

Need a hint?

Compute

dy=f'(2)\Delta x

for each case, then compute the actual

\Delta y = f(2+\Delta x)-f(2)

and compare.

Show solution

1. Step 1: Find f'(2)

f'(x)=3x^2+2x-2 \implies f'(2) = 12+4-2=14

. Also

f(2)=8+4-4+1=9

2. Step 2: Case Δx = 0.05

dy = 14(0.05)=0.7

f(2.05) = 2.05^3+2.05^2-2(2.05)+1 \approx 9.7176

, so

\Delta y \approx 0.7176

— close to

dy

, but not exact.

3. Step 3: Case Δx = 0.01

dy=14(0.01)=0.14

f(2.01) \approx 9.1407

, so

\Delta y \approx 0.1407

— much closer to

dy

than in the previous case.

4. Step 4: Conclude

\Delta x

shrinks,

dy

becomes a better and better approximation of the true change

\Delta y

— this is exactly the content of the linear approximation theorem.

Practice 02Medium

A sphere's radius is measured as

21\text{cm}

, with a possible measurement error of

0.05\text{cm}

. Estimate the resulting error in the computed volume.

Need a hint?

Use the differential

dV = 4\pi r^2\,dr

with

r=21

and

dr=0.05

Show solution

1. Step 1: Differentiate V(r)

V = \frac{4}{3}\pi r^3 \implies dV = 4\pi r^2\,dr

2. Step 2: Substitute

dV = 4\pi(21)^2(0.05) = 4\pi(441)(0.05) = 88.2\pi \approx 277.1\ \text{cm}^3

3. Step 3: Interpret

A tiny

0.05\text{cm}

error in the radius measurement could throw off the computed volume by roughly

277\ \text{cm}^3

Practice 03Easy

Use linearization to estimate

\sqrt[3]{8.1}

Need a hint?

Use

a=8

since

\sqrt[3]{8}=2

is clean.

Show solution

1. Step 1: Set Up the Linearization

f(x)=x^{1/3}

f'(x)=\frac{1}{3}x^{-2/3}

f(8)=2

f'(8)=\frac{1}{3}(8)^{-2/3}=\frac{1}{12}

L(x)=2+\frac{1}{12}(x-8)

2. Step 2: Estimate

L(8.1) = 2+\frac{1}{12}(0.1) \approx 2.0083

Common Pitfalls

⚠
Linearization Only Works Near a $L(x)$ is a good approximation only for $x$ close to $a$ — the farther away you estimate, the worse the approximation gets.
⚠
dx and Δx Are the Same ThingIn this context, $dx$ is just notation for the (small) change in $x$ — it's identical to $\Delta x$ , unlike $dy$ and $\Delta y$ , which are only approximately equal.
⚠
dy Approximates Δy, But Isn't ExactThe differential $dy$ is a linear estimate — it will always differ slightly from the true $\Delta y$ , and that gap shrinks as $\Delta x \to 0$ (see Example 3).

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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Worked Examples

1. Step 1: Write the Volume Equation

2. Step 2: Differentiate w.r.t. Time

3. Step 3: Substitute Known Values

4. Step 4: Solve

1. Step 1: Set Up the Equation

2. Step 2: Differentiate w.r.t. Time

3. Step 3: Find y When x = 3

4. Step 4: Substitute and Solve

Draw and Label

Write the Governing Equation

Differentiate, Then Substitute

Practice — Your Turn

1. Step 1: Set Up the Distance Equation

2. Step 2: Find z at This Instant

3. Step 3: Differentiate w.r.t. Time

4. Step 4: Substitute and Solve

1. Step 1: Relate r and h with Similar Triangles

2. Step 2: Write Volume in Terms of h Only

3. Step 3: Differentiate for dh/dt

4. Step 4: Differentiate Again for the Acceleration

5. Step 5: Substitute

1. Step 1: Set Up the Angle Equation

2. Step 2: Differentiate w.r.t. Time

3. Step 3: Find sec²θ When x = 8

4. Step 4: Substitute and Solve

1. Step 1: Set Up the Similar Triangles

2. Step 2: Differentiate for ds/dt

3. Step 3: Find the Tip's Speed

Worked Examples

1. Step 1: Set Up the Model

2. Step 2: Solve for k Using the 1960 Data

3. Step 3: Estimate the 1993 Population (t = 43)

1. Step 1: Find k from the Half-Life

2. Step 2: Write the Mass Function

3. Step 3: Evaluate at t = 1000

4. Step 4: Solve for t When m(t) = 30

Identify y₀ and Set Up y = y₀e^(kt)

Solve for k Using the Given Data Point

Answer the Actual Question

Practice — Your Turn

1. Step 1: Set Up the Cooling Equation

2. Step 2: Solve for k Using H(0.5) = 16

3. Step 3: Find H(1)

4. Step 4: Solve for t When H(t) = 10

1. Step 1: Set Up the Formula

2. Step 2: Evaluate at t = 3

1. Step 1: Set Up and Solve for k

2. Step 2: Find y(2)

Worked Examples

1. Step 1: Find f'(x)

2. Step 2: Linearize at a = 0

3. Step 3: Linearize at a = 1

1. Step 1: Set Up the Linearization at a = 4

2. Step 2: Estimate √3.98

3. Step 3: Estimate √4.05

Choose a Nearby 'Nice' Point a

Build L(x) = f(a) + f'(a)(x-a)

For Error Propagation, Use dy = f'(x)dx

Practice — Your Turn

1. Step 1: Find f'(2)

2. Step 2: Case Δx = 0.05

3. Step 3: Case Δx = 0.01

4. Step 4: Conclude

1. Step 1: Differentiate V(r)

2. Step 2: Substitute

3. Step 3: Interpret

1. Step 1: Set Up the Linearization

2. Step 2: Estimate

Gary Chang