Tangents, Normals and Rates of Change (Pure Mathematics 1 – 9709) — Mathematics

Introduction

Differentiation transforms a function into its gradient function — and this section is where that tool is put to direct use. In CAIE 9709 Paper 1, questions on tangents and normals, increasing/decreasing intervals, and rates of change are among the most consistently examined applications of calculus. Mastering this topic unlocks several marks across a range of question styles, from short "find the equation of the tangent" problems to multi-step connected rates of change scenarios involving related quantities.

Core Concept

Gradient at a Point

If $y = f(x)$ , then $\dfrac{dy}{dx}$ gives the gradient of the curve at any point. To find the gradient at a specific point $x = a$ , substitute $x = a$ into $\dfrac{dy}{dx}$ .

Tangents

The tangent at a point $P$ on a curve has the same gradient as the curve at $P$ . Using the gradient $m$ and the coordinates of $P$ , the equation is found via the straight-line formula.

Normals

The normal at point $P$ is perpendicular to the tangent at $P$ . If the tangent has gradient $m \neq 0$ , the normal has gradient $-\dfrac{1}{m}$ .

Increasing and Decreasing Functions

A function is increasing on an interval when $\dfrac{dy}{dx} > 0$ throughout that interval, and decreasing when $\dfrac{dy}{dx} < 0$ . Examiners frequently ask you to find these intervals by solving inequalities involving $\dfrac{dy}{dx}$ .

Rates of Change

The derivative $\dfrac{dy}{dx}$ represents the rate of change of $y$ with respect to $x$ . When quantities change with respect to time $t$ , we write $\dfrac{dy}{dt}$ , $\dfrac{dA}{dt}$ , etc.

Connected rates of change use the chain rule to link two rates:

\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}

More generally, for any chain of related variables:

\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}

This is the key technique when one quantity changes at a known rate and you must find the rate of change of a related quantity.

Key Formulae & Definitions

Gradient of curve at $x = a$ :

m = \left.\frac{dy}{dx}\right|_{x=a}

Equation of tangent at point $(x_1, y_1)$ with gradient $m$ :

y - y_1 = m(x - x_1)

Gradient of normal (perpendicular to tangent):

m_{\text{normal}} = -\frac{1}{m}

Equation of normal at $(x_1, y_1)$ :

y - y_1 = -\frac{1}{m}(x - x_1)

Increasing/decreasing conditions:

Condition	Behaviour of $f$
$\dfrac{dy}{dx} > 0$ on an interval	$f$ is increasing on that interval
$\dfrac{dy}{dx} < 0$ on an interval	$f$ is decreasing on that interval
$\dfrac{dy}{dx} = 0$ at a point	stationary point (neither increasing nor decreasing there)

Connected rates of change (chain rule):

\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}

Worked Examples

Example 1 — Tangent and Normal to a Curve

The curve $C$ has equation $y = x^3 - 4x^2 + 5$ . Find the equations of the tangent and normal to $C$ at the point where $x = 3$ .

Step 1 — Find the $y$ -coordinate.

y = 3^3 - 4(3)^2 + 5 = 27 - 36 + 5 = -4

So the point is $P = (3, -4)$ .

Step 2 — Differentiate.

\frac{dy}{dx} = 3x^2 - 8x

Step 3 — Find the gradient of the tangent at $x = 3$ .

m = 3(3)^2 - 8(3) = 27 - 24 = 3

Step 4 — Equation of the tangent.

y - (-4) = 3(x - 3) \implies y = 3x - 13

Step 5 — Gradient of the normal.

m_{\text{normal}} = -\frac{1}{3}

Step 6 — Equation of the normal.

y + 4 = -\frac{1}{3}(x - 3) \implies y = -\frac{x}{3} - 3

Example 2 — Increasing and Decreasing Intervals

Find the values of $x$ for which $f(x) = 2x^3 - 3x^2 - 12x + 1$ is a decreasing function.

Step 1 — Differentiate.

f'(x) = 6x^2 - 6x - 12

Step 2 — Set $f'(x) < 0$ (decreasing condition).

6x^2 - 6x - 12 < 0

x^2 - x - 2 < 0

(x-2)(x+1) < 0

Step 3 — Solve the inequality. The roots are $x = -1$ and $x = 2$ . Since the parabola opens upwards, the product is negative between the roots:

-1 < x < 2

Conclusion: $f$ is decreasing for $-1 < x < 2$ .

Example 3 — Connected Rates of Change

A spherical balloon is being inflated so that its radius $r$ cm increases at a constant rate of $0.5$ cm s $^{-1}$ . Find the rate at which the volume $V$ cm³ is increasing when $r = 6$ .

Step 1 — Write the known rate.

\frac{dr}{dt} = 0.5 \text{ cm s}^{-1}

Step 2 — Write the formula for volume.

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

Step 3 — Differentiate with respect to $r$ .

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

Step 4 — Apply the chain rule.

\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4\pi r^2 \times 0.5 = 2\pi r^2

Step 5 — Substitute $r = 6$ .

\frac{dV}{dt} = 2\pi (6)^2 = 72\pi \approx 226 \text{ cm}^3\text{s}^{-1}

Common Mistakes & Examiner Pitfalls

Using the tangent gradient for the normal. Always take the negative reciprocal: $m_{\text{normal}} = -\dfrac{1}{m}$ . Forgetting the negative sign or the reciprocal are both common errors.
Not evaluating $y$ before writing the line equation. You need both coordinates of the point. Substituting only into $\dfrac{dy}{dx}$ gives the gradient, not the $y$ -coordinate — both are required.
Weak inequality vs. strict inequality. When asked for "increasing" or "decreasing", the condition is strict: $\dfrac{dy}{dx} > 0$ or $\dfrac{dy}{dx} < 0$ . A stationary point is neither increasing nor decreasing.
Chain rule inversion errors. In connected rates problems, check carefully whether you need $\dfrac{dA}{dr}$ or $\dfrac{dr}{dA}$ . Setting up $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$ ensures the $dr$ 's cancel correctly.
Not simplifying the line equation. Examiners often award the final mark only for a fully simplified equation ( $y = mx + c$ form or equivalent). Leaving the answer as $y + 4 = 3(x - 3)$ may lose a mark.
Forgetting units in rates of change. In applied problems, quote the units of your answer (e.g. cm³ s⁻¹).

Practice Questions

Q1. The curve $C$ has equation $y = \dfrac{3}{x^2} + 2x$ . Find the equation of the tangent to $C$ at the point $(1, 5)$ .

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Differentiate: $y = 3x^{-2} + 2x \Rightarrow \dfrac{dy}{dx} = -6x^{-3} + 2$

Gradient at $x = 1$ : $m = -6(1)^{-3} + 2 = -6 + 2 = -4$

Tangent equation: $y - 5 = -4(x - 1) \Rightarrow y = -4x + 9$

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Q2. Find the values of $x$ for which $g(x) = x^3 - \dfrac{3}{2}x^2 - 18x + 4$ is increasing.

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Differentiate: $g'(x) = 3x^2 - 3x - 18$

Condition: $3x^2 - 3x - 18 > 0 \Rightarrow x^2 - x - 6 > 0 \Rightarrow (x-3)(x+2) > 0$

Solution: $x < -2$ or $x > 3$

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Q3. The curve $C$ has equation $y = (2x - 1)^4$ . Find the equation of the normal to $C$ at the point where $x = 1$ .

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Find $y$ at $x=1$ : $y = (2(1)-1)^4 = 1^4 = 1$ . Point: $(1, 1)$ .

Differentiate using the chain rule: $\dfrac{dy}{dx} = 4(2x-1)^3 \cdot 2 = 8(2x-1)^3$

Gradient of tangent at $x=1$ : $m = 8(1)^3 = 8$

Gradient of normal: $-\dfrac{1}{8}$

Normal equation: $y - 1 = -\dfrac{1}{8}(x - 1) \Rightarrow y = -\dfrac{x}{8} + \dfrac{9}{8}$

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Q4. The area $A$ cm² of a circle is increasing at a rate of $12$ cm² s⁻¹. Find the rate of increase of the radius when $r = 3$ cm.

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$A = \pi r^2 \Rightarrow \dfrac{dA}{dr} = 2\pi r$

Chain rule: $\dfrac{dA}{dt} = \dfrac{dA}{dr} \cdot \dfrac{dr}{dt} \Rightarrow 12 = 2\pi(3) \cdot \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \dfrac{12}{6\pi} = \dfrac{2}{\pi} \approx 0.637$ cm s⁻¹

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Q5. The curve $C$ has equation $y = x^2 - 5x + 4$ . The tangent at point $P$ on $C$ is parallel to the line $y = 3x - 2$ . Find the coordinates of $P$ and the equation of the normal at $P$ .

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Tangent gradient equals $3$ : $\dfrac{dy}{dx} = 2x - 5 = 3 \Rightarrow x = 4$

$y$ -coordinate: $y = 16 - 20 + 4 = 0$ . So $P = (4, 0)$ .

Normal gradient: $-\dfrac{1}{3}$

Normal equation: $y - 0 = -\dfrac{1}{3}(x - 4) \Rightarrow y = -\dfrac{x}{3} + \dfrac{4}{3}$

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Connections

Prerequisites you should be confident with:

Differentiating Powers of $x$ — the power rule is used in every example above to find $\dfrac{dy}{dx}$ .
The Chain Rule — essential for differentiating composite functions such as $(2x-1)^4$ and for building the connected rates of change formula.

Likely next subtopics to study:

Stationary Points and the Second Derivative — extends the idea of $\dfrac{dy}{dx} = 0$ into classifying maxima, minima and points of inflection.
Further Integration — the reverse process of differentiation, needed for areas and definite integrals.
Optimisation Problems — applies tangents and stationary points to maximise/minimise real-world quantities, a major exam question type in Paper 1.

Tangents, Normals and Rates of Change (Pure Mathematics 1 – 9709)

Introduction

Core Concept

Gradient at a Point

Tangents

Normals

Increasing and Decreasing Functions

Rates of Change

Key Formulae & Definitions

Worked Examples

Example 1 — Tangent and Normal to a Curve

Example 2 — Increasing and Decreasing Intervals

Example 3 — Connected Rates of Change

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

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