Related-rate problems | Year 12

01 / LEARN

Understand the idea.

Write a relation before differentiating

A related-rates problem connects quantities that vary with the same time. Define the changing dimensions and write their geometric or physical relation. Keep all changing quantities as variables while differentiating with respect to $t$ .

Apply the chain rule to each changing quantity, then substitute the measurements at the requested instant. For a sphere, $V = \frac{4}{3} π r^{3}$ gives $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ ; the radius rate is part of the derivative.

Read the sign and units

Choose positive directions and use signed rates consistently. A ladder foot moving away from a wall has increasing horizontal distance, while its top has decreasing height. The corresponding rates therefore have opposite signs in this setting.

Check units as well as arithmetic: a radius rate has length per time, while a volume rate has length cubed per time. If several dimensions change, a constraint such as similar triangles may let you express the model in a single changing variable.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

A sphere expands with $d r / d t = 0.2$ cm/s. Find $d V / d t$ when $r = 5$ cm.

Worked solution

1
Relate radius and volume.
$V = \frac{4}{3} π r^{3}$
A sphere volume depends on its changing radius.
2
Differentiate with respect to time.
$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
The chain rule includes the time rate of the radius.
3
Insert the measurements at the instant.
$\frac{d V}{d t} = 4 π {(5)}^{2} (0.2) = 20 π {cm}^{3} s^{- 1}$
Square centimetres times centimetres per second gives cubic centimetres per second; the positive sign means expanding volume.

Answer

$20 π$ cm $3$ /s

EXAMPLE 02

Worked example

A 10 m ladder has its foot 6 m from a wall, moving away at 0.5 m/s. Find the top height rate.

Worked solution

1
Define the changing distances.
$x^{2} + y^{2} = 100, y = \sqrt{100 - 6^{2}} = 8 m$
The fixed 10 metre ladder forms the hypotenuse, and the top height is positive.
2
Differentiate the constraint.
$2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$
Both x and y change with time while the ladder length remains constant.
3
Solve for the height rate.
$\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} = - \frac{6}{8} (\frac{1}{2}) = - \frac{3}{8} m s^{- 1}$
The positive horizontal rate produces a negative height rate.
4
Interpret the sign.
$\frac{d y}{d t} = - \frac{3}{8} m s^{- 1}$
The top descends at a speed of three eighths of a metre per second at this instant.

Answer

$- 3 / 8$ m/s

03 / YOUR TURN

Try it yourself.

Check your understanding

If $A = π r^{2}$ , $r = 4$ cm and $d r / d t = 3$ cm/s, find $d A / d t$ .

Show answer & reasoning

$24 π$ cm $2$ /s

Differentiate $A = π r^{2}$ to obtain $\frac{d A}{d t} = 2 π r \frac{d r}{d t}$ . At the stated instant this is $2 π (4) (3) = 24 π {cm}^{2} s^{- 1}$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

The side of a square grows at 2 cm/s. Find its area rate when the side is 7 cm.
2 marks
A cube edge grows at 0.4 cm/s. Find its volume rate at edge 3 cm.
2 marks
A circle area shrinks at $12 π$ cm $2$ /s. Find $\frac{d r}{d t}$ at $r = 3$ cm.
2 marks
A 13 m ladder has foot distance 5 m and foot rate 0.6 m/s away from the wall. Find the top rate.
2 marks
Water enters a cylinder of radius 2 m at 3 m $3$ /min. Find the depth rate.
2 marks
A spherical balloon radius is 6 cm when its volume grows at $72 π$ cm $3$ /s. Find its radius rate.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

The side of a square grows at 2 cm/s. Find its area rate when the side is 7 cm.
2 marks
$A = s^{2}$ gives $\dot{A} = 2 s \dot{s}$ .
At $s = 7$ , $\dot{A} = 2 (7) (2) = 28$ .
$28$ cm $2$ /s
A cube edge grows at 0.4 cm/s. Find its volume rate at edge 3 cm.
2 marks
$V = s^{3}$ gives $\dot{V} = 3 s^{2} \dot{s}$ .
$3 (9) (0.4) = 10.8$ .
$10.8$ cm $3$ /s
A circle area shrinks at $12 π$ cm $2$ /s. Find $\frac{d r}{d t}$ at $r = 3$ cm.
2 marks
$\dot{A} = 2 π r \dot{r}$ .
$- 12 π = 6 π \dot{r}$ , so $\dot{r} = - 2$ .
$- 2$ cm/s
A 13 m ladder has foot distance 5 m and foot rate 0.6 m/s away from the wall. Find the top rate.
2 marks
The height is $\sqrt{169 - 25} = 12$ m.
$\dot{y} = - \frac{x \dot{x}}{y} = - \frac{5 (0.6)}{12} = - 0.25$ .
$- 0.25$ m/s
Water enters a cylinder of radius 2 m at 3 m $3$ /min. Find the depth rate.
2 marks
$V = 4 π h$ , since the cylinder radius is fixed.
$\dot{V} = 4 π \dot{h}$ , giving $\dot{h} = \frac{3}{4 π}$ .
$\frac{3}{4 π}$ m/min
A spherical balloon radius is 6 cm when its volume grows at $72 π$ cm $3$ /s. Find its radius rate.
2 marks
$\dot{V} = 4 π r^{2} \dot{r}$ .
$72 π = 144 π \dot{r}$ , so $\dot{r} = \frac{1}{2}$ .
$\frac{1}{2}$ cm/s

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.1–6.2, with the approved school-level rates, differential-equation and motion applications.

Official curriculum source ↗

In this lesson

Understand the idea.

Write a relation before differentiating

Read the sign and units

See the method in action.

Worked example

Relate radius and volume.

Differentiate with respect to time.

Insert the measurements at the instant.

Worked example

Define the changing distances.

Differentiate the constraint.

Solve for the height rate.

Interpret the sign.

Try it yourself.

Check your understanding

Keep practising