What is the normal line?

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal -1m (provided m ≠ 0). Use the same point with this gradient.

Find the tangent to y = 1x at (1, 1). [3 marks]

The side of a square increases at 2\ cm s^-1. Find the rate of increase of the area when the side is 5\ cm. [3 marks]

SEAB Original-style practice: A spherical balloon is inflated so that its volume increases at 20\ cm^3\,s^-1. Find the rate at which the radius is increasing when the radius is 5\ cm. (Volume V = 43π r^3.)

dVdt = 20. Differentiate V = 43π r^3: dVdr = 4π r^2. By the chain rule, dVdt = dVdr·drdt, so drdt = dV/dtdV/dr = 204π r^2. At r = 5: drdt = 204π(25) = 20100π = 15π ≈ 0.0637\ cm s^-1. Markers reward linking the rates by the chain rule, computing dVdr, and the numerical rate with units.

SingaporeMathsSyllabus dot point

How do derivatives give tangent and normal lines and connect related rates of change?

Find equations of tangents and normals to curves, and solve connected rates of change problems using the chain rule

A focused answer to the H2 Mathematics outcome on tangents, normals and related rates. Finding tangent and normal equations, the perpendicular gradient relation, and linking rates of change through the chain rule.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to find the equations of tangents and normals to curves using the derivative as gradient, and to solve connected (related) rates of change problems by linking derivatives through the chain rule.

The answer

The tangent line

The derivative $\dfrac{dy}{dx}$ evaluated at a point gives the gradient of the tangent there. The tangent equation follows from the point-gradient form:

y - y_1 = m(x - x_1), \qquad m = \left.\frac{dy}{dx}\right|_{(x_1, y_1)}.

The normal line

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal $-\dfrac{1}{m}$ (provided $m \neq 0$ ). Use the same point with this gradient.

Connected rates of change

When two quantities are related and both change with time, their rates link through the chain rule:

\frac{dA}{dt} = \frac{dA}{dx}\cdot\frac{dx}{dt}.

The strategy: write the relationship between the quantities, differentiate it (often implicitly or via a known formula) to get the connecting derivative, and substitute the known rate to find the unknown one.

Setting up a related-rates problem

Identify the quantities and which rate is given and which is wanted.
Find an equation connecting the quantities.
Differentiate with respect to time using the chain rule.
Substitute the given values and rate, then solve.

Worked example

The radius of a circular oil slick increases at $0.5\ \text{m s}^{-1}$ . Find the rate at which the area is increasing when the radius is $10\ \text{m}$ .

Step 1: Identify the rates

Given $\dfrac{dr}{dt} = 0.5\ \text{m s}^{-1}$ ; want $\dfrac{dA}{dt}$ when $r = 10\ \text{m}$ .

Step 2: Write the connecting relationship

Area of a circle: $A = \pi r^2$ .

Step 3: Differentiate with respect to r

\frac{dA}{dr} = 2\pi r

Step 4: Apply the chain rule

\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt} = 2\pi r \times 0.5 = \pi r

Step 5: Substitute r = 10

\frac{dA}{dt} = \pi \times 10 = 10\pi \approx 31.4\ \text{m}^2\,\text{s}^{-1}

Examples in context

Example 1. Filling a cone. Water poured into a cone at a steady volume rate makes the depth rise faster when the cone is narrow; relating volume to depth and differentiating gives the changing depth rate, a standard related-rates application.

Example 2. Designing a road curve. The normal to a curve at a point gives the direction along which a banking force or a perpendicular barrier should be placed, so the negative-reciprocal gradient has a direct engineering reading.

Try this

Q1. Find the tangent to $y = \dfrac{1}{x}$ at $(1, 1)$ . [3 marks]

Cue. $\frac{dy}{dx} = -\frac{1}{x^2} = -1$ at $x = 1$ ; tangent $y - 1 = -1(x - 1)$ , so $y = -x + 2$ .

Q2. State the gradient of the normal if the tangent gradient is $\frac{2}{3}$ . [1 mark]

Cue. $-\frac{3}{2}$ .

Q3. The side of a square increases at $2\ \text{cm s}^{-1}$ . Find the rate of increase of the area when the side is $5\ \text{cm}$ . [3 marks]

Cue. $A = x^2$ , $\frac{dA}{dt} = 2x\frac{dx}{dt} = 2(5)(2) = 20\ \text{cm}^2\,\text{s}^{-1}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksFind the equation of the tangent and the normal to the curve

y = x^2 - 2x

at the point

(3, 3)

Show worked answer →

$\dfrac{dy}{dx} = 2x - 2$ . At $x = 3$ : gradient $= 4$ .

Tangent: $y - 3 = 4(x - 3)$ , that is $y = 4x - 9$ .

Normal gradient $= -\dfrac{1}{4}$ (negative reciprocal). Normal: $y - 3 = -\dfrac{1}{4}(x - 3)$ , that is $y = -\dfrac{1}{4}x + \dfrac{15}{4}$ .

Markers reward the gradient at the point, the tangent equation, the perpendicular gradient, and the normal equation.

Original5 marksA spherical balloon is inflated so that its volume increases at

20\ \text{cm}^3\,\text{s}^{-1}

. Find the rate at which the radius is increasing when the radius is

5\ \text{cm}

. (Volume

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

Show worked answer →

$\dfrac{dV}{dt} = 20$ . Differentiate $V = \tfrac{4}{3}\pi r^3$ : $\dfrac{dV}{dr} = 4\pi r^2$ .

By the chain rule, $\dfrac{dV}{dt} = \dfrac{dV}{dr}\cdot\dfrac{dr}{dt}$ , so $\dfrac{dr}{dt} = \dfrac{dV/dt}{dV/dr} = \dfrac{20}{4\pi r^2}$ .

At $r = 5$ : $\dfrac{dr}{dt} = \dfrac{20}{4\pi(25)} = \dfrac{20}{100\pi} = \dfrac{1}{5\pi} \approx 0.0637\ \text{cm s}^{-1}$ .

Markers reward linking the rates by the chain rule, computing $\dfrac{dV}{dr}$ , and the numerical rate with units.