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How do you use the derivative to find tangents, classify stationary points, and solve optimisation and connected-rate problems?

Tangents and normals, increasing and decreasing functions, stationary points and their nature using the second derivative, points of inflection, optimisation, and connected rates of change.

A focused answer to the OCR A-Level Mathematics A applications of differentiation content, covering tangents and normals, increasing and decreasing functions, stationary points and their classification by the second derivative, points of inflection, optimisation problems, and connected rates of change.

Generated by Claude Opus 4.811 min answer

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

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

OCR wants you to find the equations of tangents and normals, decide where a function is increasing or decreasing from the sign of the derivative, locate stationary points and classify each as a maximum, minimum or point of inflection (using the second derivative or a sign test), find points of inflection, solve optimisation problems, and link the rates of two changing quantities with the chain rule.

The answer

Tangents and normals

At a point on a curve the derivative gives the gradient of the tangent. The normal is perpendicular, so its gradient is the negative reciprocal.

Increasing and decreasing functions

A function is increasing where dydx>0\dfrac{dy}{dx} > 0 and decreasing where dydx<0\dfrac{dy}{dx} < 0. Stating an interval of increase usually means solving an inequality in the derivative.

Stationary points and their nature

A stationary point occurs where dydx=0\dfrac{dy}{dx} = 0. Classify it with the second derivative.

Points of inflection

A point of inflection is where the curve changes the way it bends, that is where d2ydx2\dfrac{d^2y}{dx^2} changes sign. A stationary point of inflection has dydx=0\dfrac{dy}{dx} = 0 and d2ydx2=0\dfrac{d^2y}{dx^2} = 0 with the same sign of the first derivative either side.

Examples in context

Optimisation

In optimisation you write the quantity to be maximised or minimised as a function of one variable (using a constraint to eliminate the other), differentiate, set the derivative to zero, solve, and confirm the nature of the stationary point. Always finish by answering the question that was asked (the optimal value, not just the variable).

Connected rates of change

When two quantities both depend on time, the chain rule links their rates: dVdt=dVdr⋅drdt\dfrac{dV}{dt} = \dfrac{dV}{dr}\cdot\dfrac{dr}{dt}. Identify the rate you know, the rate you want, and the relationship between the quantities.

Try this

Q1. Find the equation of the tangent to y=x2−3xy = x^2 - 3x at x=2x = 2. [3 marks]

  • Cue. At x=2x = 2, y=−2y = -2 and dydx=2x−3=1\dfrac{dy}{dx} = 2x - 3 = 1, so y+2=1(x−2)y + 2 = 1(x - 2), i.e. y=x−4y = x - 4.

Q2. Find the xx-coordinate of the minimum of y=x2−8x+5y = x^2 - 8x + 5. [2 marks]

  • Cue. dydx=2x−8=0\dfrac{dy}{dx} = 2x - 8 = 0 gives x=4x = 4 (and d2ydx2=2>0\frac{d^2y}{dx^2} = 2 > 0 confirms a minimum).

Exam-style practice questions

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

OCR 20187 marksA closed circular cylinder has total surface area 300π300\pi cm2^2. Show that the volume VV is given by V=150πr−πr3V = 150\pi r - \pi r^3, and hence find the radius that maximises the volume, justifying that it is a maximum.
Show worked answer →

Surface area 2πr2+2πrh=300π2\pi r^2 + 2\pi rh = 300\pi, so h=300π−2πr22πr=150−r2rh = \dfrac{300\pi - 2\pi r^2}{2\pi r} = \dfrac{150 - r^2}{r} (M1).

Volume V=πr2h=πr2⋅150−r2r=πr(150−r2)=150πr−πr3V = \pi r^2 h = \pi r^2 \cdot \dfrac{150 - r^2}{r} = \pi r(150 - r^2) = 150\pi r - \pi r^3 (A1, the printed result).

Differentiate: dVdr=150π−3πr2\dfrac{dV}{dr} = 150\pi - 3\pi r^2 (M1). Set to zero: 150π=3πr2150\pi = 3\pi r^2, so r2=50r^2 = 50 and r=50=52r = \sqrt{50} = 5\sqrt{2} cm (M1, A1).

Second derivative d2Vdr2=−6πr<0\dfrac{d^2V}{dr^2} = -6\pi r < 0 for r>0r > 0, confirming a maximum (M1, A1).

Markers reward eliminating hh, the printed expression, differentiating, solving for rr, and the second-derivative justification.

OCR 20215 marksThe radius of a circular oil slick increases at 0.50.5 m s−1^{-1}. Find the rate at which the area is increasing at the instant the radius is 2020 m.
Show worked answer →

Area A=πr2A = \pi r^2, so dAdr=2πr\dfrac{dA}{dr} = 2\pi r (M1).

The rates are connected by the chain rule (M1): dAdt=dAdr⋅drdt=2πr×0.5=πr\dfrac{dA}{dt} = \dfrac{dA}{dr}\cdot\dfrac{dr}{dt} = 2\pi r \times 0.5 = \pi r (A1).

At r=20r = 20: dAdt=π(20)=20π≈62.8\dfrac{dA}{dt} = \pi(20) = 20\pi \approx 62.8 m2^2 s−1^{-1} (M1, A1).

Markers reward the area derivative, the chain-rule link between the rates, and evaluating at r=20r = 20.

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