Edexcel A-Level Mathematics A2 P4.5 Differentiation Questions

Practise differentiating implicit and parametric curves, forming tangent or normal equations, and applying connected rates in geometric models.

Exam points

use dy/dx from parametric or implicit curves to form tangent and normal equations
apply the chain rule to linked rates in volumes, areas, radii and lengths
work with diagrams or given formulae to express one variable before differentiating

Covered syllabus points

P4.5.1 - Differentiation of simple functions P4.5.2 - Formation of simple differential

[Maximum number: 7]

Figure 3

Figure 3 shows a sketch of the curve C with parametric equations

x=6 t-3 \sin 2 t \quad y=2 \cos t \quad 0 \leqslant t \leqslant \frac{\pi}{2}

The curve meets the y-axis at 2 and the x-axis at k, where k is a constant.

(a)

Use parametric differentiation to show that

\frac{\mathrm{d} y}{\mathrm{~d} x}=\lambda \operatorname{cosec} t

where $\lambda$ is a constant to be found.

The point P with parameter $t=\frac{\pi}{4}$ lies on C.
The tangent to C at the point P cuts the y-axis at the point N.

[ 4 ]

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-2 \sin t}{6-6 \cos 2 t}$

M1 A1

$=\frac{-2 \sin t}{6-6\left(1-2 \sin ^{2} t\right)}=\frac{-2 \sin t}{12 \sin ^{2} t}=-\frac{1}{6} \operatorname{cosec} t$

M1 A1*

(4)

(b)

Find the exact y coordinate of N, giving your answer in simplest form.

The region bounded by the curve, the x-axis and the y-axis is rotated through $2 \pi$ radians about the x-axis to form a solid of revolution.

[ 3 ]

$P=\left(\frac{3\pi}{2}-3,\sqrt2\right)$
For example,
$y-\sqrt2=-\frac{\sqrt2}{6}\left(0-\left(\frac{3\pi}{2}-3\right)\right)$
or
$\sqrt2=-\frac{\sqrt2}{6}\left(\frac{3\pi}{2}-3\right)+c$ $y=\frac{\pi\sqrt2}{4}+\frac{\sqrt2}{2}$

[Maximum number: 3]

Figure 1

A regular icosahedron of side length $x \mathrm{~cm}$ , shown in Figure 1, is expanding uniformly. The icosahedron consists of 20 congruent equilateral triangular faces of side length $x \mathrm{~cm}$ .

Find the rate of change of the volume of the icosahedron when x=2, giving your answer to 2 significant figures.

$\frac{dA}{dt}=0.025.$ $\frac{dV}{dt} =\frac{dV}{dA}\frac{dA}{dt} =\frac{(3+\sqrt5)x}{8\sqrt3}\cdot0.025.$
When x=2,
$\frac{dV}{dt} =\frac{2(3+\sqrt5)}{8\sqrt3}\cdot0.025 =0.019\ldots\ \mathrm{cm^3s^{-1}}.$

[Maximum number: 7]

Figure 4

Figure 4 shows a sketch of the curve C with parametric equations

x=\sec t \quad y=\sqrt{3} \tan \left(t+\frac{\pi}{3}\right) \quad \frac{\pi}{6}<t<\frac{\pi}{2}

(a)

Find $\frac{\mathrm{d} y}{\mathrm{~d} x}$ in terms of t

[ 3 ]

$\frac{\mathrm{d} x}{\mathrm{~d} t}=\sec t \tan t$ and $\frac{\mathrm{d} y}{\mathrm{~d} t}=\sqrt{3} \sec ^{2}\left(t+\frac{\pi}{3}\right)$ $\left(\frac{\mathrm{d} y}{\mathrm{~d} x}=\right) \frac{\mathrm{d} y}{\mathrm{~d} t} \div \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{\sqrt{3} \sec ^{2}\left(t+\frac{\pi}{3}\right)}{\sec t \tan t}$ oe
(3)

(a) Alt

$x=\sec t \quad y=\sqrt{3} \frac{\tan t+\sqrt{3}}{1-\sqrt{3} \tan t}$ $\frac{\mathrm{d} x}{\mathrm{~d} t}=\sec t \tan t$ and $\frac{\mathrm{d} y}{\mathrm{~d} t}=\sqrt{3} \frac{(1-\sqrt{3} \tan t) \sec ^{2} t-(\tan t+\sqrt{3})\left(-\sqrt{3} \sec ^{2} t\right)}{(1-\sqrt{3} \tan t)^{2}}($ oe )
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} t} \div \frac{\mathrm{d} x}{\mathrm{~d} t}=\sqrt{3} \frac{(1-\sqrt{3} \tan t) \sec ^{2} t-(\tan t+\sqrt{3})\left(-\sqrt{3} \sec ^{2} t\right)}{\sec t \tan t(1-\sqrt{3} \tan t)^{2}}$ oe

(b)

Find an equation for the tangent to C at the point where $t=\frac{\pi}{3}$

Give your answer in the form y=m x+c, where m and c are constants.

[ 4 ]

$t=\frac{\pi}{3}\Rightarrow x=2,\quad y=-3$ $\left.\frac{dy}{dx}\right|_{t=\pi/3} =\frac{\sqrt3\sec^2(2\pi/3)}{\sec(\pi/3)\tan(\pi/3)} =2$ y+3=2(x-2)y=2x-7

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question 3

Question (a)

Question (b)