Connected Rates of Change

Using the chain rule to relate rates of change in real-world problems

SPEC 7.6 • APPLICATIONS OF DIFFERENTIATION

What is a Rate of Change?

A rate of change measures how quickly a quantity changes with respect to time.

Definition

If a quantity

Q

varies with time

t

, then its rate of change is

\frac{d Q}{d t}

.

Common Rates

\frac{d V}{d t}

— rate of change of volume

\frac{d A}{d t}

— rate of change of area

\frac{d r}{d t}

— rate of change of radius

\frac{d h}{d t}

— rate of change of height

Key Insight

Positive rate = quantity increasing

Negative rate = quantity decreasing

e.g., melting ice has

\frac{d V}{d t} < 0

The Chain Rule for Connected Rates

When quantities are connected through a formula, their rates of change are also connected.

Core Formula

\frac{d A}{d t} = \frac{d A}{d B} \times \frac{d B}{d t}

This connects the rate of change of

A

to the rate of change of

B

through the relationship between

A

and

B

.

Extended Chain Rule

For three connected quantities:

\frac{d A}{d t} = \frac{d A}{d B} \times \frac{d B}{d C} \times \frac{d C}{d t}

Important

Notice how the "middle" differentials cancel out, just like fractions!

Problem-Solving Method

Step-by-Step Approach

1

Identify all variables and what they represent

2

Write the formula connecting the variables (e.g.,

V = \frac{4}{3} π r^{3}

)

3

Identify what's given (the rate you know)

4

Identify what's required (the rate you need)

5

Differentiate to find the connecting derivative

6

Apply the chain rule to link the rates

7

Substitute values at the specific instant

8

State your answer with correct units

Key Tip

Always substitute values AFTER differentiating, not before!

Worked Example — Expanding Circle01

A circular oil spill expands so that its radius increases at 2 m/s. Find the rate at which the area is increasing when the radius is 5 m.

1

Given:

\frac{d r}{d t} = 2

m/s. Find:

\frac{d A}{d t}

when

r = 5

m.

2

Formula:

A = π r^{2}

3

Differentiate:

\frac{d A}{d r} = 2 π r

4

Chain rule:

\frac{d A}{d t} = \frac{d A}{d r} \times \frac{d r}{d t} = 2 π r \times 2 = 4 π r

5

Substitute: When

r = 5

:

\frac{d A}{d t} = 4 π (5) = 20 π

AnswerArea is increasing at

20 π

m²/s (approximately 62.8 m²/s)

Worked Example — Inflating Sphere02

A spherical balloon is being inflated at a rate of 100 cm³/s. Find the rate at which the radius is increasing when the radius is 10 cm.

1

Given:

\frac{d V}{d t} = 100

cm³/s. Find:

\frac{d r}{d t}

when

r = 10

cm.

2

Formula:

V = \frac{4}{3} π r^{3}

3

Differentiate:

\frac{d V}{d r} = 4 π r^{2}

4

Chain rule:

\frac{d V}{d t} = \frac{d V}{d r} \times \frac{d r}{d t}

5

Rearrange:

\frac{d r}{d t} = \frac{d V / d t}{d V / d r} = \frac{100}{4 π r ^{2}}

6

Substitute: When

r = 10

:

\frac{d r}{d t} = \frac{100}{4 π ( 100 )} = \frac{100}{400 π} = \frac{1}{4 π}

AnswerRadius is increasing at

\frac{1}{4 π}

cm/s (approximately 0.0796 cm/s)

Key Geometric Formulas

These formulas appear frequently in connected rates problems:

ShapeVolumeSurface Area

Sphere

V = \frac{4}{3} π r^{3}

A = 4 π r^{2}

Cone

V = \frac{1}{3} π r^{2} h

A = π r l

(curved)

Cylinder

V = π r^{2} h

A = 2 π r h

(curved)

Circle

A = π r^{2}

C = 2 π r

Cube

V = x^{3}

A = 6 x^{2}

For Cones

When filling a cone, the ratio

\frac{r}{h}

is constant! Use similar triangles to express

r

in terms of

h

(or vice versa) to eliminate one variable.

Worked Example — Filling a Cone03

Water is poured into an inverted cone of base radius 6 cm and height 12 cm at a rate of 4 cm³/s. Find the rate at which the water level is rising when the depth is 3 cm.

1

Given:

\frac{d V}{d t} = 4

cm³/s. Find:

\frac{d h}{d t}

when

h = 3

cm.

2

Similar triangles:

\frac{r}{h} = \frac{6}{12} = \frac{1}{2}

, so

r = \frac{h}{2}

3

Volume in terms of h only:

V = \frac{1}{3} π r^{2} h = \frac{1}{3} π (\frac{h}{2})^{2} h = \frac{π h ^{3}}{12}

4

Differentiate:

\frac{d V}{d h} = \frac{3 π h ^{2}}{12} = \frac{π h ^{2}}{4}

5

Chain rule:

\frac{d V}{d t} = \frac{d V}{d h} \times \frac{d h}{d t}

, so

\frac{d h}{d t} = \frac{d V / d t}{d V / d h} = \frac{4}{π h ^{2} /4} = \frac{16}{π h ^{2}}

6

Substitute: When

h = 3

:

\frac{d h}{d t} = \frac{16}{π ( 9 )} = \frac{16}{9 π}

AnswerWater level is rising at

\frac{16}{9 π}

cm/s (approximately 0.566 cm/s)

The Cone Proportion Technique

In cone problems, the water forms a smaller cone similar to the container. The key relationship is:

Similar Triangles

\frac{r}{h} = \frac{R}{H}

(constant)where

R

and

H

are the container's radius and height.

Method

Use similar triangles to express

r

in terms of

h

(or

h

in terms of

r

)

Substitute into

V = \frac{1}{3} π r^{2} h

to get volume in terms of one variable only

Differentiate and apply the chain rule

Example

If container has

R = 6

,

H = 12

, then

\frac{r}{h} = \frac{6}{12} = \frac{1}{2}

, so

r = \frac{h}{2}

.

Volume becomes:

V = \frac{1}{3} π (\frac{h}{2})^{2} h = \frac{π h ^{3}}{12}

Worked Example — Expanding Cube04

The side of a cube is increasing at 0.5 cm/s. Find the rate at which the volume is increasing when the side length is 4 cm.

1

Given:

\frac{d x}{d t} = 0.5

cm/s. Find:

\frac{d V}{d t}

when

x = 4

cm.

2

Formula:

V = x^{3}

3

Differentiate:

\frac{d V}{d x} = 3 x^{2}

4

Chain rule:

\frac{d V}{d t} = \frac{d V}{d x} \times \frac{d x}{d t} = 3 x^{2} \times 0.5 = 1.5 x^{2}

5

Substitute: When

x = 4

:

\frac{d V}{d t} = 1.5 (16) = 24

AnswerVolume is increasing at 24 cm³/s

Worked Example — Cylinder (Two Rates)05

A cylinder has fixed radius 5 cm. Liquid is poured in at 20 cm³/s. Find the rate at which the height increases, and the rate at which the curved surface area increases.

1

Given:

r = 5

(constant),

\frac{d V}{d t} = 20

cm³/s.

2

Part (a) — Height:

V = π r^{2} h = 25 πh

(since

r = 5

)

3

\frac{d V}{d h} = 25 π

, so

\frac{d h}{d t} = \frac{20}{25 π} = \frac{4}{5 π}

cm/s

4

Part (b) — Curved surface area:

A = 2 π r h = 10 πh

5

\frac{d A}{d h} = 10 π

, so

\frac{d A}{d t} = \frac{d A}{d h} \times \frac{d h}{d t} = 10 π \times \frac{4}{5 π} = 8

AnswerHeight increases at

\frac{4}{5 π}

cm/s ≈ 0.255 cm/s; Surface area increases at 8 cm²/s

Practice

Question 1: A square has side length increasing at 3 cm/s. Find the rate of increase of area when the side is 8 cm.

1

Let side =

s

. Given:

\frac{d s}{d t} = 3

. Find:

\frac{d A}{d t}

when

s = 8

.

2

Area:

A = s^{2}

, so

\frac{d A}{d s} = 2 s

3

\frac{d A}{d t} = \frac{d A}{d s} \times \frac{d s}{d t} = 2 s \times 3 = 6 s

4

When

s = 8

:

\frac{d A}{d t} = 6 (8) = 48

cm²/s

Answer48 cm²/s

Practice

Question 2: A spherical balloon is deflating at 50 cm³/s. Find the rate at which the radius is decreasing when

r = 5

cm.

1

Given:

\frac{d V}{d t} = - 50

cm³/s (negative = deflating). Find:

\frac{d r}{d t}

when

r = 5

.

2

V = \frac{4}{3} π r^{3}

, so

\frac{d V}{d r} = 4 π r^{2}

3

\frac{d r}{d t} = \frac{d V / d t}{d V / d r} = \frac{- 50}{4 π r ^{2}}

4

When

r = 5

:

\frac{d r}{d t} = \frac{- 50}{4 π ( 25 )} = \frac{- 50}{100 π} = \frac{- 1}{2 π}

cm/s

AnswerRadius is decreasing at

\frac{1}{2 π}

cm/s ≈ 0.159 cm/s

Practice

Question 3: Water drains from a hemispherical bowl of radius 10 cm. When the water depth is

h

, the volume is

V = \frac{π h ^{2} ( 30 - h )}{3}

. Find the rate at which the water level is falling when

h = 5

cm, given the water drains at 25π cm³/s.

1

Given:

\frac{d V}{d t} = - 25 π

cm³/s (negative = draining). Find:

\frac{d h}{d t}

when

h = 5

.

2

Expand:

V = \frac{π ( 30 h ^{2} - h ^{3} )}{3} = 10 π h^{2} - \frac{π h ^{3}}{3}

3

Differentiate:

\frac{d V}{d h} = 20 πh - π h^{2} = πh (20 - h)

4

\frac{d h}{d t} = \frac{d V / d t}{d V / d h} = \frac{- 25 π}{πh ( 20 - h )} = \frac{- 25}{h ( 20 - h )}

5

When

h = 5

:

\frac{d h}{d t} = \frac{- 25}{5 ( 15 )} = \frac{- 25}{75} = \frac{- 1}{3}

cm/s

AnswerWater level is falling at

\frac{1}{3}

cm/s

Summary

Key Concepts

ConceptFormula/Method

Chain Rule for Rates

\frac{d A}{d t} = \frac{d A}{d B} \times \frac{d B}{d t}

Sphere Volume

V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d r} = 4 π r^{2}

Cone Proportion

\frac{r}{h} = \frac{R}{H}

(constant ratio)

Decreasing QuantitiesUse negative rates of change

Exam Skills

Always differentiate BEFORE substituting values

For cones, use similar triangles to eliminate one variable

Include correct units in your final answer

Use negative rates for decreasing quantities

Check your answer makes physical sense