Home · A-Level Further Maths 9231 · Further differentiation · Second derivatives: implicit and parametric · Practice
AS & A Level · Practice questions · Second derivatives: implicit and parametric

Differentiate again, divide again.

Six original Cambridge-style questions on second derivatives of parametric and implicit relations: evaluating at a point, successive differentiation, classifying stationary points and showing concavity.

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 the derivative, then substitute.

01
Recall
[3 marks]

A curve has parametric equations \(x = 3t^{2}\), \(y = 2t^{3}\), for \(t \gt 0\). Find \(\dfrac{dy}{dx}\) and \(\dfrac{d^{2}y}{dx^{2}}\) in terms of \(t\).

  • \(\dfrac{dx}{dt} = 6t\), \(\dfrac{dy}{dt} = 6t^{2}\), so \(\dfrac{dy}{dx} = \dfrac{6t^{2}}{6t} = t\). ✓
  • \(\dfrac{d}{dt}(t) = 1\), so \(\dfrac{d^{2}y}{dx^{2}} = \dfrac{1}{6t}\). ✓
02
Calculation
[4 marks]

Given \(y = \tanh x\), show that \(\dfrac{dy}{dx} = 1 - y^{2}\). Hence show that \(\dfrac{d^{2}y}{dx^{2}} = -2y(1 - y^{2})\), and find the exact value of \(\dfrac{d^{2}y}{dx^{2}}\) when \(x = \ln 2\).

  • \(\dfrac{dy}{dx} = \operatorname{sech}^{2} x = 1 - \tanh^{2} x = 1 - y^{2}\). ✓
  • Differentiate with respect to \(x\): \(\dfrac{d^{2}y}{dx^{2}} = -2y\,\dfrac{dy}{dx} = -2y(1 - y^{2})\). ✓
  • \(\tanh(\ln 2) = \dfrac{2 - \frac{1}{2}}{2 + \frac{1}{2}} = \dfrac{3}{5}\). ✓
  • \(\dfrac{d^{2}y}{dx^{2}} = -2 \times \tfrac{3}{5} \times \left(1 - \tfrac{9}{25}\right) = -\tfrac{6}{5} \times \tfrac{16}{25} = -\tfrac{96}{125}\). ✓
03
Calculation
[5 marks]

The curve \(x^{2}y + y^{3} = 10\) passes through the point \((3, 1)\). Find the values of \(\dfrac{dy}{dx}\) and \(\dfrac{d^{2}y}{dx^{2}}\) at this point.

  • Once: \(2xy + x^{2}y' + 3y^{2}y' = 0\). At \((3, 1)\): \(6 + 9y' + 3y' = 0\), so \(y' = -\tfrac{1}{2}\). ✓
  • Again, using the product rule on \(2xy\), \(x^{2}y'\) and \(3y^{2}y'\): ✓
  • \[\begin{aligned} &2y + 2xy' + 2xy' + x^{2}y'' \\ &\quad + 6y(y')^{2} + 3y^{2}y'' = 0. \end{aligned}\] ✓
  • Substitute \(x = 3\), \(y = 1\), \(y' = -\tfrac{1}{2}\): \(2 - 3 - 3 + 9y'' + \tfrac{3}{2} + 3y'' = 0\). ✓
  • \(12y'' = \tfrac{5}{2}\), so \(y'' = \tfrac{5}{24}\). ✓
04
Analysis
[5 marks]

A cycloid has parametric equations \(x = t - \sin t\), \(y = 1 - \cos t\), for \(0 \lt t \lt 2\pi\).

(a) Show that \(\dfrac{d^{2}y}{dx^{2}} = -\dfrac{1}{(1 - \cos t)^{2}}\).

(b) Deduce that the arch is concave for \(0 \lt t \lt 2\pi\), and find the coordinates and nature of its stationary point.

  • (a) \(\dfrac{dx}{dt} = 1 - \cos t\), \(\dfrac{dy}{dt} = \sin t\), so \(\dfrac{dy}{dx} = \dfrac{\sin t}{1 - \cos t}\). ✓
  • Quotient rule: ✓
  • \[\begin{aligned} \frac{d}{dt}\!\left(\frac{dy}{dx}\right) &= \frac{\cos t(1 - \cos t) - \sin^{2} t}{(1 - \cos t)^{2}} \\ &= \frac{\cos t - 1}{(1 - \cos t)^{2}} = -\frac{1}{1 - \cos t}. \end{aligned}\] ✓
  • Divide by \(\dfrac{dx}{dt} = 1 - \cos t\): \(\dfrac{d^{2}y}{dx^{2}} = -\dfrac{1}{(1 - \cos t)^{2}}\). ✓
  • (b) For \(0 \lt t \lt 2\pi\), \(\cos t \lt 1\), so \((1 - \cos t)^{2} \gt 0\) and \(\dfrac{d^{2}y}{dx^{2}} \lt 0\) throughout: the arch is concave. ✓
  • \(\dfrac{dy}{dt} = \sin t = 0\) at \(t = \pi\) (with \(\dfrac{dx}{dt} = 2 \ne 0\)), the point \((\pi, 2)\). There \(\dfrac{d^{2}y}{dx^{2}} = -\tfrac{1}{4} \lt 0\): a maximum. ✓
05
Analysis
[5 marks]

Find the coordinates of the stationary points of the curve \(2x^{2} + 2xy + y^{2} = 4\), and determine their nature.

  • Once: \(4x + 2y + 2xy' + 2y\,y' = 0\). Setting \(y' = 0\): \(4x + 2y = 0\), so \(y = -2x\). ✓
  • Substitute: \(2x^{2} - 4x^{2} + 4x^{2} = 4\), so \(x^{2} = 2\), \(x = \pm\sqrt{2}\). Points \(\big(\sqrt{2}, -2\sqrt{2}\big)\) and \(\big(-\sqrt{2}, 2\sqrt{2}\big)\). ✓
  • Again: \(4 + 2y' + 2y' + 2xy'' + 2(y')^{2} + 2y\,y'' = 0\). With \(y' = 0\): \(4 + 2(x + y)y'' = 0\), so \(y'' = -\dfrac{2}{x + y}\). ✓
  • At \(\big(\sqrt{2}, -2\sqrt{2}\big)\): \(x + y = -\sqrt{2}\), so \(y'' = \dfrac{2}{\sqrt{2}} = \sqrt{2} \gt 0\): a minimum. ✓
  • At \(\big(-\sqrt{2}, 2\sqrt{2}\big)\): \(x + y = \sqrt{2}\), so \(y'' = -\sqrt{2} \lt 0\): a maximum. ✓
06
Proof
[6 marks]

The part of the hyperbola \(x^{2} - y^{2} = 1\) with \(x \gt 1\) and \(y \gt 0\) is given parametrically by \(x = \cosh t\), \(y = \sinh t\), \(t \gt 0\).

(a) Using the parametric equations, show that \(\dfrac{d^{2}y}{dx^{2}} = -\operatorname{cosech}^{3} t\).

(b) By differentiating \(x^{2} - y^{2} = 1\) implicitly twice, show that \(\dfrac{d^{2}y}{dx^{2}} = -\dfrac{1}{y^{3}}\), and check that this agrees with (a).

(c) Deduce that this part of the curve is concave.

  • (a) \(\dfrac{dx}{dt} = \sinh t\), \(\dfrac{dy}{dt} = \cosh t\), so \(\dfrac{dy}{dx} = \coth t\). Then \(\dfrac{d}{dt}(\coth t) = -\operatorname{cosech}^{2} t\). ✓
  • Divide by \(\sinh t\): \(\dfrac{d^{2}y}{dx^{2}} = \dfrac{-\operatorname{cosech}^{2} t}{\sinh t} = -\operatorname{cosech}^{3} t\). ✓
  • (b) Once: \(2x - 2y\,y' = 0\), so \(y' = \dfrac{x}{y}\). Again: \(1 - (y')^{2} - y\,y'' = 0\), so \(y\,y'' = 1 - \dfrac{x^{2}}{y^{2}}\). ✓
  • \[y'' = \frac{y^{2} - x^{2}}{y^{3}} = \frac{-1}{y^{3}},\ \text{using } x^{2} - y^{2} = 1.\] ✓
  • With \(y = \sinh t\): \(-\dfrac{1}{\sinh^{3} t} = -\operatorname{cosech}^{3} t\), the same as (a). ✓
  • (c) For \(t \gt 0\), \(y = \sinh t \gt 0\), so \(\dfrac{d^{2}y}{dx^{2}} = -\dfrac{1}{y^{3}} \lt 0\) for every point: this part of the curve is concave. ✓

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.