AS & A Level · Practice questions · Tangents, normals and rates of change

Lines, slopes and rates.

Six original Cambridge-style questions applying differentiation: tangents and normals, increasing and decreasing functions, and connected rates of change.

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

Point, gradient, line; or know, want, link.

Point: find both coordinates from the curve.
Gradient: substitute \(x\) into \(\dfrac{dy}{dx}\); normal uses \(-\dfrac{1}{m}\).
Line: \(y - y_{1} = m(x - x_{1})\); give it in the form asked.
Increasing/decreasing: factorise \(f'(x)\), sketch it, read off where it is positive or negative.
Connected rates: know one rate, want another, link with a formula, differentiate, then multiply or divide.

Recall

[3 marks]

Find the equation of the tangent to the curve \(y = \dfrac{8}{x^{2}}\) at the point where \(x = 2\).

Point: \(y = \dfrac{8}{4} = 2\), so \((2, 2)\). ✓
\(\dfrac{dy}{dx} = -16x^{-3} = -\dfrac{16}{8} = -2\) at \(x = 2\). ✓
\(y - 2 = -2(x - 2)\), so \(y = -2x + 6\). ✓

Calculation

[4 marks]

Find the equation of the normal to the curve \(y = x^{2} - 5x + 6\) at the point where \(x = 4\). Give your answer in the form \(ax + by = c\).

Point: \(y = 16 - 20 + 6 = 2\), so \((4, 2)\). ✓
\(\dfrac{dy}{dx} = 2x - 5 = 3\); normal gradient \(-\tfrac{1}{3}\). ✓
\(y - 2 = -\tfrac{1}{3}(x - 4)\), so \(3y - 6 = -x + 4\). ✓
\(x + 3y = 10\) ✓

Analysis

[3 marks]

\(f(x) = 2x^{3} + 3x^{2} - 12x + 5\). Find the set of values of \(x\) for which \(f\) is increasing.

\(f'(x) = 6x^{2} + 6x - 12 = 6(x + 2)(x - 1)\). ✓
Positive quadratic, so \(f'(x) > 0\) outside the roots. ✓
\(x < -2\) or \(x > 1\) ✓

Proof

[3 marks]

Show that \(f(x) = x^{3} + 3x^{2} + 5x - 2\) is an increasing function.

\(f'(x) = 3x^{2} + 6x + 5\). ✓
\(= 3(x + 1)^{2} + 2\). ✓
Since \(3(x + 1)^{2} \ge 0\), \(f'(x) \ge 2 > 0\) for all \(x\), so \(f\) is increasing. ✓

Calculation

[4 marks]

The edges of a metal cube expand at \(0.2\ \text{cm s}^{-1}\) when it is heated. Find, at the instant when the edge length is \(5\) cm, the rate of increase of (a) the volume, (b) the total surface area.

(a) \(V = x^{3}\), \(\dfrac{dV}{dx} = 3x^{2} = 75\). \(\dfrac{dV}{dt} = 75 \times 0.2 = 15\ \text{cm}^{3}\,\text{s}^{-1}\). ✓
(b) \(S = 6x^{2}\), \(\dfrac{dS}{dx} = 12x = 60\). \(\dfrac{dS}{dt} = 60 \times 0.2 = 12\ \text{cm}^{2}\,\text{s}^{-1}\). ✓

Analysis

[5 marks]

Water is poured into an inverted conical tank at a constant rate of \(30\ \text{cm}^{3}\,\text{s}^{-1}\). When the depth of water is \(h\) cm, the volume of water is \(V = \tfrac{1}{3}\pi h^{3}\ \text{cm}^{3}\).

(a) Find the rate at which the depth is increasing when \(h = 6\).

(b) Find the depth of water at the instant when the depth is increasing at \(0.15\ \text{cm s}^{-1}\).

(a) \(\dfrac{dV}{dh} = \pi h^{2} = 36\pi\) when \(h = 6\). ✓
\(\dfrac{dh}{dt} = \dfrac{30}{36\pi} = \dfrac{5}{6\pi} = 0.265\ \text{cm s}^{-1}\) (3 s.f.). ✓
(b) \(\dfrac{dh}{dt} = \dfrac{30}{\pi h^{2}} = 0.15\), so \(\pi h^{2} = 200\). ✓
\(h = \sqrt{\dfrac{200}{\pi}} = 7.98\) cm (3 s.f.). ✓

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.