A-Level Maths (Edexcel 9MA0) · Pure Year 2 · Chapter 9: Differentiation · Lesson 5

Parametric differentiation

113 marks

Hints online, answers free with an account:
mathswariz.uk/worksheets/a-level-parametric-differentiation/

Date

Do Now

Quick questions to start.

A1
Needed today
Differentiate $e^{2 t} cos t$ with respect to $t$ .
[2 marks]
Answer
A2
Needed today
Find an equation of the line with gradient $- \frac{1}{3}$ that passes through $(2, 5)$ . Give it in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.
[2 marks]
Answer
A3
From chapter 8
A curve has parametric equations $x = 2 cos t$ , $y = 3 sin t$ . Find its Cartesian equation.
[2 marks]
Answer
A4
Common slip
A curve has $x = t^{2}$ and $y = t^{3}$ . Write down $\frac{d y}{d t}$ and $\frac{d x}{d t}$ , and hence find $\frac{d y}{d x}$ in terms of $t$ .
[2 marks]
Answer

Example, then your turn

A-Level

Your teacher works through each example with you. Then try the one beside it.

Example 1

A curve has parametric equations $x = t^{3} - 2 t$ , $y = t^{2} + 4$ . Find the gradient of the curve at the point where $t = 2$ .

Hint Differentiate

x

and

y

with respect to

t

. Then

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

: the

y

derivative goes on top.

B1
A curve has parametric equations $x = 2 t^{2} + t$ , $y = t^{3} - 3 t$ . Find $\frac{d y}{d x}$ in terms of $t$ , and find the gradient of the curve at the point where $t = 2$ .
[4 marks]
Answer

Example 2

A curve has parametric equations $x = 4 cos t$ , $y = 3 sin t$ . Find an equation of the normal to the curve at the point where $t = \frac{π}{3}$ . Give it in the form $a x + b y + c = 0$ .

Hint Find the point first. The normal has gradient

- \frac{1}{m}

, where

m

is the gradient of the tangent.

B2
A curve has parametric equations $x = 3 cos t$ , $y = 5 sin t$ . Find an equation of the tangent to the curve at the point where $t = \frac{π}{6}$ . Give it in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are exact.
[5 marks]
Answer

Example 3

A curve has parametric equations $x = t^{2} + 2 t$ , $y = t^{3} - 12 t$ . Find the coordinates of the points where the tangent to the curve is horizontal.

Hint A horizontal tangent needs

\frac{d y}{d x} = 0

, so

\frac{d y}{d t} = 0

while

\frac{d x}{d t}

is not zero.

B3
A curve has parametric equations $x = t^{3} - 3 t$ , $y = t^{2} - 6 t$ . Find the coordinates of the points where the tangent to the curve is vertical.
[5 marks]
Answer

Example 4

A curve has parametric equations $x = ln (t + 2)$ , $y = \frac{1}{t + 1}$ , $t > - 1$ . Find an equation of the tangent to the curve at the point where $x = ln 3$ .

Hint You are given

x

, not

t

. Solve

ln (t + 2) = ln 3

for

t

first.

B4
A curve has parametric equations $x = ln (2 t - 1)$ , $y = \frac{t}{t + 1}$ , $t > \frac{1}{2}$ . Find an equation of the tangent to the curve at the point where $x = 0$ .
[5 marks]
Answer

Example 5

A curve has parametric equations $x = t^{2}$ , $y = 2 t$ . The normal to the curve at the point $P$ , where $t = 2$ , meets the curve again at $Q$ . Find the coordinates of $Q$ .

Hint Find the normal at

P

. Then put

x = t^{2}

and

y = 2 t

into its equation: you get an equation in

t

, and you already know one root,

t = 2

B5
A curve has parametric equations $x = 2 t^{2}$ , $y = 4 t$ . The normal to the curve at the point where $t = 1$ meets the curve again at $Q$ . Find the coordinates of $Q$ .
[6 marks]
Answer

Practice

A-Level

C1
A curve has parametric equations $x = t^{2} - 1$ , $y = t^{3} + t$ . Mia finds the tangent at the point where $t = 1$ . She writes: " $\frac{d y}{d x} = \frac{2 t}{3 t ^{2} + 1} = \frac{1}{2}$ ." Explain her mistake, and find the correct equation of the tangent.
[5 marks]
Answer

C2
A curve has parametric equations $x = 2 t + 1$ , $y = t^{2} - 3 t$ . Leo is asked for the gradient at the point where $x = 5$ . He puts $t = 5$ into $\frac{d y}{d x}$ and gets $\frac{7}{2}$ . Explain his mistake and find the correct gradient.
[4 marks]
Answer
C3
A curve has parametric equations $x = t^{3} - 12 t$ , $y = t^{2} - 6 t$ . To find the stationary point, Zara solves $3 t^{2} - 12 = 0$ . Explain her mistake, and find the coordinates of the stationary point.
[4 marks]
Answer
C4
A curve has parametric equations $x = sin t$ , $y = sin 2 t$ , $0 \leq t \leq \frac{π}{2}$ . Find the gradient of the curve at the point where $t = \frac{π}{3}$ , and find the coordinates of the point where the tangent is horizontal.
[5 marks]
Answer

Exam-style questions

Edexcel exam style

D1
A curve $C$ has parametric equations $x = t^{2} + t$ , $y = t^{3} - 3 t$ .
(a)
Find $\frac{d y}{d x}$ in terms of $t$ .
[2 marks]
Answer
(b)
Find the coordinates of the points on $C$ where the tangent is horizontal.
[3 marks]
Answer
(c)
Find an equation of the normal to $C$ at the point where $t = 2$ . Give it in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.
[4 marks]
Answer
Total for D1: 9 marks

D2
A curve has parametric equations $x = 4 cos t$ , $y = sin 2 t$ , $0 \leq t < 2 π$ . Sam writes: " $\frac{d y}{d x} = \frac{2 cos 2 t}{- 4 sin t}$ . The tangent is horizontal when $sin t = 0$ , so the horizontal tangents are at $(4, 0)$ and $(- 4, 0)$ ."
Sam is wrong.
(a)
Explain Sam's mistake, and describe the tangent at $(4, 0)$ .
[2 marks]
(b)
Find the coordinates of all the points on the curve where the tangent is horizontal.
[5 marks]
Answer
(c)
The curve passes through the origin twice. Find the gradient of the curve each time it passes through the origin.
[3 marks]
Answer
Total for D2: 10 marks
D3
A curve has parametric equations $x = t + \frac{1}{t}$ , $y = t - \frac{1}{t}$ , $t > 0$ . Find an equation of the tangent to the curve that is parallel to the line $y = 3 x$ . Give your answer in the form $y = 3 x + k$ , where $k$ is an exact constant.
[7 marks]
Answer
D4
A curve has parametric equations $x = 3 t^{2}$ , $y = 6 t$ . The point $P$ on the curve has parameter $t = p$ .
(a)
Show that the normal to the curve at $P$ has equation $y + p x = 6 p + 3 p^{3}$ .
[4 marks]
(b)
Given that $p = 2$ , the normal at $P$ meets the curve again at $Q$ . Find the coordinates of $Q$ .
[3 marks]
Answer
(c)
Hence find the area of triangle $O P Q$ , where $O$ is the origin.
[3 marks]
Answer
Total for D4: 10 marks
D5
The diagram shows the curve with parametric equations $x = t^{2} - 4$ , $y = t^{3} - 9 t$ . The curve meets the $y$ -axis at $A$ and $B$ .
(a)
Find $\frac{d y}{d x}$ in terms of $t$ .
[2 marks]
Answer
(b)
Find the coordinates of $A$ and $B$ .
[2 marks]
Answer
(c)
Show that the tangents to the curve at $A$ and at $B$ meet on the $x$ -axis, and find the point where they meet.
[5 marks]
Answer
Total for D5: 9 marks
D6
A curve has parametric equations $x = e^{2 t}$ , $y = e^{t} - t$ .
(a)
Find $\frac{d y}{d x}$ in terms of $t$ .
[2 marks]
Answer
(b)
Find the coordinates of the stationary point on the curve.
[2 marks]
Answer
(c)
Find an equation of the tangent to the curve at the point where $t = ln 2$ . Give it in the form $y = m x + c$ , where $m$ and $c$ are exact.
[4 marks]
Answer
Total for D6: 8 marks

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
The diagram shows the curve with parametric equations $x = 3 cos t$ , $y = sin 2 t$ , $0 \leq t < 2 π$ . The curve passes through the origin twice. The two tangents to the curve at the origin and the tangent to the curve at the point where $t = 0$ form a triangle. Find the area of this triangle.
[9 marks]
Hint 1 · What to try
The curve is at the origin when $x = 0$ and $y = 0$ . Find the two values of $t$ .
Hint 2 · The first line
$\frac{d y}{d x} = \frac{2 cos 2 t}{- 3 sin t}$ . Find its value at each of those values of $t$ .
Hint 3 · The full method
At $t = 0$ , $\frac{d x}{d t} = 0$ , so that tangent is vertical. Where does it meet the other two?
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can find dy/dx as (dy/dt) ÷ (dx/dt) when x and y are given in terms of t.
I can find the value of t at a point before working out a gradient there.
I can find equations of tangents and normals to a parametric curve.
I can find where the tangent is horizontal (dy/dt = 0) or vertical (dx/dt = 0).

Answers and mark scheme

The worked answers and the full mark scheme are free with a Mathswariz account. Teachers and students can both sign up.

Get the answers