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→

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 Exercises

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}

NEED A HINT?

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

SHOW DETAILED EXPLANATION

Step 1: Write the Volume Equation

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

Step 2: Differentiate w.r.t. Time

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

Step 3: Substitute Known Values

D=50

r=25

. So

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

Step 4: Solve

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

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?

NEED A HINT?

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

SHOW DETAILED EXPLANATION

Step 1: Set Up the Equation

x^2 + y^2 = 25

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

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

Step 3: Find y When x = 3

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

Example 03Hard

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 DETAILED EXPLANATION

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

Step 2: Find z at This Instant

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

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}

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}

Example 04AB/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 DETAILED EXPLANATION

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}

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}

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}

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}

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.

Example 05Hard

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 DETAILED EXPLANATION

Step 1: Set Up the Angle Equation

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

, where

\frac{dx}{dt}=1.5

Step 2: Differentiate w.r.t. Time

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

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}

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}

Example 06Medium

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 DETAILED EXPLANATION

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}

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.

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→

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 Exercises

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).

NEED A HINT?

Let

t=0

correspond to 1950. Use the 1960 data point (

t=10

) to solve for

k

first.

SHOW DETAILED EXPLANATION

Step 1: Set Up the Model

P(t) = 2.56e^{kt}

(billions), with

t

in years since 1950.

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}

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.

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.

NEED A HINT?

Use the half-life to find

k

first:

\frac{1}{2} = e^{1590k}

SHOW DETAILED EXPLANATION

Step 1: Find k from the Half-Life

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

Step 2: Write the Mass Function

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

Step 3: Evaluate at t = 1000

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

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}

Example 03Hard

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 DETAILED EXPLANATION

Step 1: Set Up the Cooling Equation

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

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}

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

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}

Example 04Easy

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 DETAILED EXPLANATION

Step 1: Set Up the Formula

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

Step 2: Evaluate at t = 3

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

dollars.

Example 05Medium

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 DETAILED EXPLANATION

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

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.

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 Exercises

Example 01Easy

Find the linearization of

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

x=0

and at

x=1

NEED A HINT?

Compute

f(a)

and

f'(a)

at each point, then plug into

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

SHOW DETAILED EXPLANATION

Step 1: Find f'(x)

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

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

Step 3: Linearize at a = 1

f(1)=2

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

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

Example 02Medium

Use linearization to estimate

\sqrt{3.98}

and

\sqrt{4.05}

NEED A HINT?

Pick

a=4

since

\sqrt{4}=2

is clean, and

3.98

4.05

are both very close to it.

SHOW DETAILED EXPLANATION

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)

Step 2: Estimate √3.98

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

Step 3: Estimate √4.05

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

Example 03Hard

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 DETAILED EXPLANATION

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

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.

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.

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.

Example 04Medium

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 DETAILED EXPLANATION

Step 1: Differentiate V(r)

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

Step 2: Substitute

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

Example 05Easy

Use linearization to estimate

\sqrt[3]{8.1}

NEED A HINT?

Use

a=8

since

\sqrt[3]{8}=2

is clean.

SHOW DETAILED EXPLANATION

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)

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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Draw and Label

Write the Governing Equation

Differentiate, Then Substitute

Practice Exercises

Step 1: Write the Volume Equation

Step 2: Differentiate w.r.t. Time

Step 3: Substitute Known Values

Step 4: Solve

Step 1: Set Up the Equation

Step 2: Differentiate w.r.t. Time

Step 3: Find y When x = 3

Step 4: Substitute and Solve

Step 1: Set Up the Distance Equation

Step 2: Find z at This Instant

Step 3: Differentiate w.r.t. Time

Step 4: Substitute and Solve

Step 1: Relate r and h with Similar Triangles

Step 2: Write Volume in Terms of h Only

Step 3: Differentiate for dh/dt

Step 4: Differentiate Again for the Acceleration

Step 5: Substitute

Step 1: Set Up the Angle Equation

Step 2: Differentiate w.r.t. Time

Step 3: Find sec²θ When x = 8

Step 4: Substitute and Solve

Step 1: Set Up the Similar Triangles

Step 2: Differentiate for ds/dt

Step 3: Find the Tip's Speed

Identify y₀ and Set Up y = y₀e^(kt)

Solve for k Using the Given Data Point

Answer the Actual Question

Practice Exercises

Step 1: Set Up the Model

Step 2: Solve for k Using the 1960 Data

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

Step 1: Find k from the Half-Life

Step 2: Write the Mass Function

Step 3: Evaluate at t = 1000

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

Step 1: Set Up the Cooling Equation

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

Step 3: Find H(1)

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

Step 1: Set Up the Formula

Step 2: Evaluate at t = 3

Step 1: Set Up and Solve for k

Step 2: Find y(2)

Choose a Nearby 'Nice' Point a

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

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

Practice Exercises

Step 1: Find f'(x)

Step 2: Linearize at a = 0

Step 3: Linearize at a = 1

Step 1: Set Up the Linearization at a = 4

Step 2: Estimate √3.98

Step 3: Estimate √4.05

Step 1: Find f'(2)

Step 2: Case Δx = 0.05

Step 3: Case Δx = 0.01

Step 4: Conclude

Step 1: Differentiate V(r)

Step 2: Substitute

Step 3: Interpret

Step 1: Set Up the Linearization

Step 2: Estimate

Gary Chang