AS & A Level · Practice questions · The gradient of a curve

Gradient at a point.

Six original Cambridge-style questions on the gradient of a curve: chords as \(h \to 0\), \(f'(x)\) and \(\dfrac{d^{2}y}{dx^{2}}\), and finding where a curve has a given gradient.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Differentiate, then substitute or solve.

Chord: gradient \(= \dfrac{f(a+h) - f(a)}{h}\); simplify, cancel \(h\), then let \(h \to 0\).
Power rule: \(x^{n} \to nx^{n-1}\); constants vanish; work term by term.
Gradient at a point: substitute \(x\) into \(\dfrac{dy}{dx}\).
Point with a given gradient: solve \(\dfrac{dy}{dx} = m\), then use the curve for \(y\).
Second derivative: differentiate \(\dfrac{dy}{dx}\) again.

Recall

[3 marks]

The points with \(x\)-coordinates \(2\) and \(2 + h\) lie on the curve \(y = x^{2} + x\).

Show that the gradient of the chord joining them is \(5 + h\), and state the gradient of the curve at \(x = 2\).

At \(x = 2\): \(y = 6\). At \(x = 2 + h\): \(y = 4 + 4h + h^{2} + 2 + h = 6 + 5h + h^{2}\). ✓
Gradient \(= \dfrac{5h + h^{2}}{h} = 5 + h\). ✓
As \(h \to 0\), the gradient of the curve at \(x = 2\) is \(5\). ✓

Calculation

[2 marks]

Find \(\dfrac{dy}{dx}\) and \(\dfrac{d^{2}y}{dx^{2}}\) for \(y = 3x^{4} - 2x^{3} + x - 7\).

\(\dfrac{dy}{dx} = 12x^{3} - 6x^{2} + 1\) ✓
\(\dfrac{d^{2}y}{dx^{2}} = 36x^{2} - 12x\) ✓

Calculation

[3 marks]

\(f(x) = x^{3} - 4x^{2} + 2\). Find \(f'(3)\) and \(f''(3)\).

\(f'(x) = 3x^{2} - 8x\), so \(f'(3) = 27 - 24 = 3\). ✓
\(f''(x) = 6x - 8\), so \(f''(3) = 18 - 8 = 10\). ✓

Analysis

[3 marks]

The curve \(y = x^{3} + kx^{2} - 4\), where \(k\) is a constant, has gradient \(4\) at the point where \(x = 2\). Find \(k\).

\(\dfrac{dy}{dx} = 3x^{2} + 2kx\). ✓
At \(x = 2\): \(12 + 4k = 4\). ✓
\(k = -2\). ✓

Analysis

[4 marks]

Find the coordinates of the points on the curve \(y = 2x^{3} - 3x^{2} - 12x + 1\) at which the gradient is zero.

\(\dfrac{dy}{dx} = 6x^{2} - 6x - 12 = 6(x^{2} - x - 2) = 6(x - 2)(x + 1)\). ✓
Zero when \(x = 2\) or \(x = -1\). ✓
\(x = 2\): \(y = 16 - 12 - 24 + 1 = -19\). \(\ x = -1\): \(y = -2 - 3 + 12 + 1 = 8\). ✓
Points \((2, -19)\) and \((-1, 8)\). ✓

Analysis

[5 marks]

A particle moves along a straight line. Its displacement after \(t\) seconds is \(s = t^{3} - 6t^{2} + 15t\) metres. Its velocity is \(\dfrac{ds}{dt}\) and its acceleration is \(\dfrac{d^{2}s}{dt^{2}}\).

(a) Find the velocity when \(t = 1\).

(b) Find the velocity at the instant when the acceleration is zero.

(a) \(\dfrac{ds}{dt} = 3t^{2} - 12t + 15\); at \(t = 1\): \(3 - 12 + 15 = 6\ \text{m s}^{-1}\). ✓
(b) \(\dfrac{d^{2}s}{dt^{2}} = 6t - 12 = 0\) when \(t = 2\); velocity \(= 12 - 24 + 15 = 3\ \text{m s}^{-1}\). ✓
(c) At rest needs \(3t^{2} - 12t + 15 = 0\), i.e. \(t^{2} - 4t + 5 = 0\). ✓
Discriminant \(= 16 - 20 = -4 < 0\), so there are no real roots: the velocity is never zero. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.