Year 12 Maths Extension 1 (2027) Further applications of calculus

Related rates of change (chain rule; area/volume)

Q: How do you solve a related-rates problem?

Write the relating equation, differentiate with respect to t, substitute the known rate and values, then solve for the unknown rate.

Q: When do you differentiate implicitly?

When the variables are linked by a relation like x squared plus y squared equals c that is not solved for one variable.

20 practice questions 2 video lessons Theory + worked examples

Learn related rates of change for NSW Year 12 Mathematics Extension 1. When two quantities change together, the chain rule connects their rates so one can be found from the other.

You will learn to model a rate as a composition of functions, apply the chain rule to relate rates, and use given area, surface-area and volume formulas — solving real-world problems such as expanding circles and filling tanks in Extension 1.

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Related rates with the chain rule Watch
How to Solve Related Rates Problems Using the Chain Rule | Sphere Surface Area & Volume Explained Watch

Theory

Related rates connect the rates of change of two quantities linked by an equation, using the chain rule $\frac{d A}{d t} = \frac{d A}{d r} \cdot \frac{d r}{d t}$ . This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-05) covers area, surface-area, volume and implicit problems.

Related rates link the rates of change of two quantities that are connected by an equation, using the chain rule.

If a quantity $A$ depends on $r$ , and $r$ depends on time $t$ , then $\frac{d A}{d t} = \frac{d A}{d r} \cdot \frac{d r}{d t}$ . Write the equation relating the quantities, differentiate to get $\frac{d A}{d r}$ (or differentiate implicitly for a relation like $x^{2} + y^{2} = c$ ), then multiply by the known rate.

Often you must first reduce to one variable — for a cone, use a given ratio to write $V$ in terms of $h$ alone.

NESA link. Part of the Year 12 Further applications of calculus focus area, outcome ME1-12-05 ("applies calculus to solve problems involving polynomials, further rates of change, areas and volumes and differential equations") with MAO-WM-01.

The chain rule multiplies the linking rate by the known rate.

A sliding ladder:

x^{2} + y^{2} = 169

, differentiated implicitly in

t

\frac{d A}{d t} = \frac{d A}{d r} \cdot \frac{d r}{d t} .

dA/dt = (dA/dr)(dr/dt)

For an implicit relation, differentiate every term with respect to $t$ :

x^{2} + y^{2} = c \Rightarrow 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0.

d/dt of x^2 + y^2 = c gives 2x x' + 2y y' = 0

Reduce first. Write the quantity in one variable before differentiating (e.g. a cone with radius half its depth gives $V = \frac{1}{12} π h^{3}$ ). A negative rate means the quantity is decreasing.

How to solve a related-rates problem

Write the equation relating the quantities, reducing to one variable if needed.
Differentiate with respect to $t$ (implicitly if the relation mixes variables).
Substitute the known rate and the instant's values.
Solve for the unknown rate, and read the sign (negative means decreasing).

Example 1 — Expanding circle

A circle has area

A = π r^{2}

. If

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

cm/s, find

\frac{d A}{d t}

when

r = 4

cm.

Solution

$\frac{d A}{d r} = 2 π r$ .

\frac{d A}{d t}

=

2 π r \cdot \frac{d r}{d t} = 2 π (4) (3) = 24 π

dA/dt = 24 pi cm^2/s

$\frac{d A}{d t} = 24 π$ cm $^{2}$ /s.

Example 2 — Inflating sphere

A sphere has

V = \frac{4}{3} π r^{3}

. If

\frac{d r}{d t} = 0.2

cm/s, find

\frac{d V}{d t}

when

r = 5

cm.

Solution

$\frac{d V}{d r} = 4 π r^{2}$ .

\frac{d V}{d t}

=

4 π (25) (0.2) = 20 π

dV/dt = 20 pi cm^3/s

$\frac{d V}{d t} = 20 π$ cm $^{3}$ /s.

Example 3 — Sliding ladder

13

m ladder rests on a wall (

x^{2} + y^{2} = 169

). The base slides out at

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

m/s. Find

\frac{d y}{d t}

when

x = 5

Solution

Differentiate implicitly; at $x = 5$ , $y = 12$ .

$2 x \frac{d x}{d t} + 2 y \frac{d y}{d t}$	$=$	$0$
$10 (2) + 24 \frac{d y}{d t}$	$=$	$0 \Rightarrow \frac{d y}{d t} = - \frac{5}{6}$

dy/dt = -5/6 m/s (falling)

$\frac{d y}{d t} = - \frac{5}{6}$ m/s (falling).

Example 4 — Filling a cone

An inverted cone has radius half its depth, so

V = \frac{1}{12} π h^{3}

. Water is poured in at

6

^{3}

/s. Find

\frac{d h}{d t}

when

h = 2

Solution

$\frac{d V}{d h} = \frac{1}{4} π h^{2} = π$ at $h = 2$ .

6

=

π \cdot \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = \frac{6}{π}

dh/dt = 6/pi cm/s

$\frac{d h}{d t} = \frac{6}{π}$ cm/s.

Common pitfalls

Too many variables. Reduce to one variable first (use a given ratio), or you cannot differentiate cleanly.

Forgetting implicit differentiation. For a relation like

x^{2} + y^{2} = c

, differentiate every term with respect to

t

Substituting too early. Differentiate first, then substitute the instant's values.

Ignoring the sign. A negative rate means the quantity is decreasing.

Frequently asked questions

What are related rates?

They connect the rates of change of two quantities linked by an equation, via the chain rule $\frac{d A}{d t} = \frac{d A}{d r} \cdot \frac{d r}{d t}$ .