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Grade 12 Calculus · Lesson 2.6120 min

Related Rates · Connect the Changes

Different quantities can change at the same instant. Build the model first, differentiate with respect to time, then use the instant data to find the rate you need.

Model changing quantities, differentiate with respect to time, and solve real-life rate problems using the teacher routine.

One clock: timeKnown → WantedSigns matterModel before calculusDifferentiate before substitutingUnits + direction

Key takeaway

Mr. Azzam’s core idea

1 · Model the situation

Name changing quantities, rates, signs, and the equation connecting them.

2 · Differentiate in time

Treat every changing quantity as a function of $t$ .

3 · Use the instant

Substitute values only after the rate equation exists, then solve with units.

Different quantities · one clock

Related Rates connects quantities changing with respect to the same variable: time. Lesson 2.5 supplied the implicit-differentiation idea; now the derivatives are rates such as $\frac{dx}{dt}$ , $\frac{dr}{dt}$ , and $\frac{dV}{dt}$ .

x=x(t),\quad y=y(t),\quad r=r(t),\quad V=V(t)

\frac{d}{dt}(r^2)=2r\frac{dr}{dt}

Your turn

If the radius

r

changes with time, differentiate

r^2

with respect to

t

2r

2t\frac{dr}{dt}

2r\frac{dr}{dt}

Common mistake. Do not differentiate a changing variable as if it were constant. If

r=r(t)

, the derivative of

r^2

must include

\frac{dr}{dt}

Mr. Azzam’s 10-step routine

STEP 1

Read & pull out information

Identify what changes and what formula can connect the quantities.

STEP 2

Sketch a diagram

Build a model whenever the geometry matters.

STEP 3

Write Known & Wanted

Name the given rate and exactly which rate is required.

STEP 4

Check the signs

Increasing is positive; decreasing is negative.

STEP 5

Constant or varying?

A fixed length stays constant; a changing length is a function of time.

STEP 6

Set up the equation

Relate only the quantities needed for the problem.

STEP 7

Differentiate in time

Differentiate with respect to

t

before inserting instant values.

STEP 8

Plug in known items

Now use the instantaneous values and signed rates.

STEP 9

Solve the wanted rate

Keep the requested derivative visible and isolate it.

STEP 10

State units + direction

Finish with a rate unit and interpret the sign when appropriate.

Worked example

Warm-up example · full question

Air is being pumped into a spherical balloon at a rate of 5 cm³/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.

Step 1 of 7

What is the known rate?

Step 1

What is the known rate?

$\frac{dV}{dt}=5$
$\frac{dr}{dt}=5$
$V=5$

Answer: $\frac{dV}{dt}=5$

Correct — air is entering at 5 cubic centimeters per minute.

Step 2

What rate are we asked to find?

$\frac{dr}{dt}$
$\frac{dV}{dt}$
$\frac{dd}{dt}$

Answer: $\frac{dr}{dt}$

Correct — the question asks how fast the radius increases.

Step 3

When the diameter is

20

cm, what is the radius?

$10$ cm
$20$ cm
$40$ cm

Answer: $10$ cm

Radius is half the diameter.

Step 4

Choose the model.

$V=\frac43\pi r^3$
$V=\pi r^2$
$V=2\pi r$

Answer: $V=\frac43\pi r^3$

Use the volume of a sphere.

Step 5

Differentiate with respect to

t

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

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

Chain Rule attaches the time rate of the radius.

Step 6

Substitute the instant data.

$5=4\pi(10)^2\frac{dr}{dt}$
$5=4\pi(20)^2\frac{dr}{dt}$
$10=4\pi(5)^2\frac{dr}{dt}$

Answer: $5=4\pi(10)^2\frac{dr}{dt}$

Now substitute the rate and instantaneous radius.

Step 7

Solve and choose the answer with correct units.

$\frac{dr}{dt}=\frac{1}{80\pi}\,\text{cm/min}$
$\frac{dr}{dt}=80\pi\,\text{cm/min}$
$\frac{dr}{dt}=\frac{1}{40\pi}\,\text{cm}^3/\text{min}$

Answer: $\frac{dr}{dt}=\frac{1}{80\pi}\,\text{cm/min}$

Excellent — the radius is a length, so its rate is cm/min.

\frac{dV}{dt}=5,\quad r=10,\quad V=\frac43\pi r^3

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

5=4\pi(10)^2\frac{dr}{dt}

\boxed{\frac{dr}{dt}=\frac{1}{80\pi}\ \text{cm/min}}

Interactive lab

Related Rates Simulator

Choose an instant, predict the direction, then calculate the rate. Enter your answer to four decimal places.

Horizontal distance x (ft)

$x^2+y^2=15^2$ ; $dx/dt=-0.25\text{ ft/s}$

Predict the sign of the top height rate.

Calculate the instantaneous rate (ft/s), to 4 decimal places.

Translate words into signed rates

ExplanationIncreasing

\frac{dq}{dt}>0

pumped in · expanding · moving away

ExplanationDecreasing

\frac{dq}{dt}<0

leaking · shrinking · moving toward

Your turn

Water is leaking from a tank at

2\,\text{ft}^3/\text{hr}

. How should the volume rate be written?

\frac{dh}{dt}=-2

\frac{dV}{dt}=-2

\frac{dV}{dt}=2

Common mistake. The word “rate” does not automatically mean positive. Decide the sign from the direction of change.

Model first · differentiate second

The diagram decides the equation. Right triangles use Pythagorean relationships, shadows and cones use similar triangles, and area or volume problems use the appropriate geometry formula.

ExplanationRight triangle

x^2+y^2=z^2

ExplanationSphere

V=\frac43\pi r^3

ExplanationCone

V=\frac13\pi r^2h

Your turn

In a related-rates problem, when should the instantaneous values be substituted?

After writing the model and differentiating with respect to

t

Before differentiatingBefore writing the equation

Key takeawayExam decision

If the model still contains two changing geometric variables that are related by similarity, simplify the model first when useful. Keep the variables alive until after differentiation.

Spherical balloon · volume drives radius

Interactive lab

Filling balloon

Play or move the time slider. Playback advances 1 min of model time per second.

Ready0 / 60 min

Elapsed time (min)

r = 10.000 cm; dr/dt = 0.00398 cm/min.

Added volume: +0.0 cm³; radius: +0.000 cm. Growth is subtle at 5 cm³/min; the drawing keeps the same scale.

With a constant volume inflow, what happens to the radius rate as r grows?

Worked exampleBefore calculating

Air enters, so $\frac{dV}{dt}>0$ . The radius should also increase. Predict the sign before solving.

Worked example

Warm-up example · full question

Air is being pumped into a spherical balloon at a rate of 5 cm³/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.

Step 1 of 7

What is the known rate?

Step 1

What is the known rate?

$\frac{dV}{dt}=5$
$\frac{dr}{dt}=5$
$V=5$

Answer: $\frac{dV}{dt}=5$

Correct — air is entering at 5 cubic centimeters per minute.

Step 2

What rate are we asked to find?

$\frac{dr}{dt}$
$\frac{dV}{dt}$
$\frac{dd}{dt}$

Answer: $\frac{dr}{dt}$

Correct — the question asks how fast the radius increases.

Step 3

When the diameter is

20

cm, what is the radius?

$10$ cm
$20$ cm
$40$ cm

Answer: $10$ cm

Radius is half the diameter.

Step 4

Choose the model.

$V=\frac43\pi r^3$
$V=\pi r^2$
$V=2\pi r$

Answer: $V=\frac43\pi r^3$

Use the volume of a sphere.

Step 5

Differentiate with respect to

t

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

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

Chain Rule attaches the time rate of the radius.

Step 6

Substitute the instant data.

$5=4\pi(10)^2\frac{dr}{dt}$
$5=4\pi(20)^2\frac{dr}{dt}$
$10=4\pi(5)^2\frac{dr}{dt}$

Answer: $5=4\pi(10)^2\frac{dr}{dt}$

Now substitute the rate and instantaneous radius.

Step 7

Solve and choose the answer with correct units.

$\frac{dr}{dt}=\frac{1}{80\pi}\,\text{cm/min}$
$\frac{dr}{dt}=80\pi\,\text{cm/min}$
$\frac{dr}{dt}=\frac{1}{40\pi}\,\text{cm}^3/\text{min}$

Answer: $\frac{dr}{dt}=\frac{1}{80\pi}\,\text{cm/min}$

Excellent — the radius is a length, so its rate is cm/min.

\frac{dV}{dt}=5,\quad r=10,\quad V=\frac43\pi r^3

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

5=4\pi(10)^2\frac{dr}{dt}

\boxed{\frac{dr}{dt}=\frac{1}{80\pi}\ \text{cm/min}}

Your turn

Air enters a sphere at

5\,\mathrm{cm}^3/\mathrm{min}

. At diameter

20\,\mathrm{cm}

, what is

dr/dt

\frac1{320\pi}\,\mathrm{cm/min}

5\,\mathrm{cm/min}

\frac1{80\pi}\,\mathrm{cm/min}

80\pi\,\mathrm{cm/min}

Ladder against a wall · one length fixed

Interactive lab

Moving ladder

Play or move the time slider. Playback advances 1 s of model time per second.

Ready0 / 12 s

Elapsed time (s)

x = 10.00 ft, y = 11.18 ft; dy/dt = 0.2236 ft/s.

As the base approaches the wall, how does the top move?

Worked example

Ladder · guided exam example · full question

A 15-foot ladder is resting against the wall. The bottom is initially 10 feet away from the wall and is being pushed towards the wall at a rate of 1/4 ft/sec. How fast is the top of the ladder moving up the wall 12 seconds after we start pushing?

Step 1 of 5

What are the known rate and wanted rate?

Your turn

A fixed ladder satisfies

x^2+y^2=L^2

, with

x,y>0

. If its bottom moves toward the wall, what is the sign of

dy/dt

NegativeZeroCannot determinePositive

Conical tank · geometry + similarity + rates

Interactive lab

Draining cone

Play or move the time slider. Playback advances 1 hr of model time per second.

Ready0 / 60 hr

Elapsed time (hr)

h = 12.000 ft; dh/dt = -0.0347 ft/hr.

With a constant leak, what happens to the magnitude of the depth rate as h falls?

Worked example

Conical tank · guided exam example · full question

A tank of water in the shape of a cone is leaking water at a constant rate of 2 ft³/hour. The base radius is 5 ft and the height is 14 ft. Find the depth rate when h=6 ft, then the radius rate at the same instant.

Step 1 of 5

Translate the leak.

Your turn

For a similar water cone with

r=\frac5{14}h

, how are the length rates related?

\frac{dr}{dt}=\frac5{14}\frac{dh}{dt}

\frac{dr}{dt}=\frac{14}5\frac{dh}{dt}

\frac{dr}{dt}=\frac{dh}{dt}

\frac{dr}{dt}=-2

Light and shadow · similar triangles

Interactive lab

Moving shadow

Play or move the time slider. Playback advances 1 s of model time per second.

Ready0 / 8 s

Elapsed time (s)

Shadow = 3.38 ft; tip = 7.38 ft from pole.

How does the shadow tip speed compare with the person’s 2 ft/s?

Worked example

Shadow · guided exam example · full question

A light is on top of a 12-ft pole and a 5-ft 6-in person walks away at 2 ft/sec. When the person is 25 ft from the pole, find the speed of the shadow tip from the pole and from the person.

Step 1 of 4

Convert the height and identify the walking rate.

Your turn

x

is the person's distance from the pole and

s

the shadow length, which rate measures the shadow tip moving away from the pole?

ds/dt

dx/dt+ds/dt

dx/dt-ds/dt

dx/dt

Mixed applications · complete the toolbox

These are not side notes. Each model is a genuine related-rates pattern: define the changing quantities, connect them, differentiate in time, then use the requested instant.

Worked example

Angle & distance · guided example · full question

A spotlight is 30 m from a straight wall. Its beam rotates so that the angle θ between the beam and the perpendicular to the wall increases at 0.04 rad/s. How fast is the light spot moving along the wall when θ=π/4?

Step 1 of 3

Which triangle relation connects the wall distance y and θ?

Worked example

Perpendicular motion · guided example · full question

Car A is 6 km east of an intersection and moves east at 50 km/h. Car B is 8 km north of the intersection and moves north at 40 km/h. How fast is the distance z between the cars changing at that instant?

Step 1 of 4

Choose the model.

Worked example

Parallel resistors · guided example · full question

Two variable resistors satisfy 1/R=1/R₁+1/R₂. At an instant R₁=6 Ω and R₂=3 Ω, with dR₁/dt=0.2 Ω/s and dR₂/dt=0.1 Ω/s. Find dR/dt.

Step 1 of 4

Which quantities are allowed to change?

Worked example

Delayed motion · guided example · full question

A car leaves an intersection traveling east at 40 km/h. Thirty minutes later, a second car leaves the same intersection traveling north at 30 km/h. How fast is the distance between them changing 1 hour after the second car starts?

Step 1 of 4

At the requested instant, how long has the eastbound car traveled?

Key takeawayExam strategy

Write Known, Wanted, and the Model. Check signs and elapsed times. Differentiate before substituting instant values. Finish with direction and units.

Different quantities · one clock

Your turn

Your turn · student questions

Mr. Azzam’s 10-step routine

Step 1

Step 2

Step 3

Step 4

Step 5

Step 6

Step 7

Translate words into signed rates

Your turn

Your turn · student questions

Model first · differentiate second

Your turn

Your turn · student questions

Spherical balloon · volume drives radius

Step 1

Step 2

Step 3

Step 4

Step 5

Step 6

Step 7

Your turn

Your turn · student questions

Ladder against a wall · one length fixed

Step 1

Step 2

Step 3

Step 4

Step 5

Your turn

Your turn · student questions

Conical tank · geometry + similarity + rates

Step 1

Step 2

Step 3

Step 4

Step 5

Your turn

Your turn · student questions

Light and shadow · similar triangles

Step 1

Step 2

Step 3

Step 4

Your turn

Your turn · student questions

Mixed applications · complete the toolbox

Step 1

Step 2

Step 3

Step 1

Step 2

Step 3

Step 4

Step 1

Step 2

Step 3

Step 4

Step 1

Step 2

Step 3

Step 4