Parametric Differentiation

SPEC 7.4b • DIFFERENTIATION

Finding gradients when curves are defined by parametric equations

What Are Parametric Equations?

Definition

Parametric equations express

x

and

y

separately in terms of a third variable called the parameter, usually

t

or

θ

.

Instead of writing

y

directly in terms of

x

, we write:

x = f (t)

— the x-coordinate as a function of

t

y = g (t)

— the y-coordinate as a function of

t

Example

The parametric equations

x = t^{2}

,

y = 2 t

define a parabola.

When

t = 1

: point

(1, 2)

. When

t = 2

: point

(4, 4)

. When

t = - 1

: point

(1, - 2)

.

Key Concept

Each value of

t

gives a unique point

(x, y)

on the curve. The parameter

t

links the coordinates together.

The Parametric Differentiation Formula

Key Formula

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

Or equivalently:

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

This formula comes from the chain rule:

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

Method

Differentiate

x

and

y

separately with respect to

t

, then divide

\frac{d y}{d t}

by

\frac{d x}{d t}

.

Important

Make sure you divide the right way round! It's

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

, not

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

.

Method Steps

1

Differentiate $y$ with respect to $t$ to get

\frac{d y}{d t}

2

Differentiate $x$ with respect to $t$ to get

\frac{d x}{d t}

3

Calculate

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

4

Substitute the parameter value if finding the gradient at a specific point

Watch Out

\frac{d x}{d t}

must not equal zero when dividing — this would make the gradient undefined (vertical tangent).

Worked Example01

A curve has parametric equations $x = t^{3}$ , $y = t^{2} - 4 t$ . Find $\frac{d y}{d x}$ in terms of $t$ .

1

Find

\frac{d x}{d t}

:

x = t^{3} ⟹ \frac{d x}{d t} = 3 t^{2}

2

Find

\frac{d y}{d t}

:

y = t^{2} - 4 t ⟹ \frac{d y}{d t} = 2 t - 4

3

Apply the formula:

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{2 t - 4}{3 t ^{2}}

Answer

\frac{d y}{d x} = \frac{2 t - 4}{3 t ^{2}}

or

\frac{2 ( t - 2 )}{3 t ^{2}}

Worked Example02

A curve has parametric equations $x = 3 cos θ$ , $y = 5 sin θ$ . Find $\frac{d y}{d x}$ in terms of $θ$ .

1

Find

\frac{d x}{d θ}

:

x = 3 cos θ ⟹ \frac{d x}{d θ} = - 3 sin θ

2

Find

\frac{d y}{d θ}

:

y = 5 sin θ ⟹ \frac{d y}{d θ} = 5 cos θ

3

Apply the formula:

\frac{d y}{d x} = \frac{5 cos θ}{- 3 sin θ} = - \frac{5 cos θ}{3 sin θ}

4

Simplify:

\frac{d y}{d x} = - \frac{5}{3} cot θ

or

- \frac{5}{3 tan θ}

Answer

\frac{d y}{d x} = - \frac{5 cot θ}{3}

or

- \frac{5}{3 tan θ}

Horizontal and Vertical Tangents

Horizontal Tangent

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

This occurs when:

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

(provided

\frac{d x}{d t} \neq = 0

)

Vertical Tangent

\frac{d y}{d x}

is undefined

This occurs when:

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

(provided

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

)

Method

Find parameter values where the numerator (

\frac{d y}{d t}

) or denominator (

\frac{d x}{d t}

) equals zero.

Special Case

If both

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

and

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

at the same value of

t

, further analysis is needed (use L'Hopital's rule or limits).

Worked Example03

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

1

Find derivatives:

\frac{d x}{d t} = 2 t - 2

,

\frac{d y}{d t} = 3 t^{2} - 3

2

For horizontal tangent:

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

3 t^{2} - 3 = 0 ⟹ 3 (t^{2} - 1) = 0 ⟹ t = \pm 1

3

Check

\frac{d x}{d t} \neq = 0

: At

t = 1

:

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

(excluded!)

At

t = - 1

:

\frac{d x}{d t} = - 4 \neq = 0

✓

4

Find coordinates at

t = - 1

:

x = (- 1)^{2} - 2 (- 1) = 1 + 2 = 3

y = (- 1)^{3} - 3 (- 1) = - 1 + 3 = 2

AnswerPoint

(3, 2)

Equations of Tangents and Normals

Tangent

Gradient:

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

Use:

y - y_{1} = m (x - x_{1})

Normal

Gradient:

m_{N} = - \frac{d x / d t}{d y / d t}

(Negative reciprocal of tangent gradient)

1

Find the point

(x_{1}, y_{1})

by substituting the parameter value into

x

and

y

2

Find the gradient at that parameter value (tangent:

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

, normal: negative reciprocal)

3

Write the equation using

y - y_{1} = m (x - x_{1})

Tip

The parameter gives you both the gradient AND the point — make sure you find both!

Worked Example04

A curve has parametric equations $x = t^{2}$ , $y = 2 t$ . Find the equation of the tangent at the point where $t = 3$ .

1

Find the point at

t = 3

:

x = 3^{2} = 9

,

y = 2 (3) = 6

Point:

(9, 6)

2

Find derivatives:

\frac{d x}{d t} = 2 t

,

\frac{d y}{d t} = 2

3

Find gradient:

\frac{d y}{d x} = \frac{2}{2 t} = \frac{1}{t}

At

t = 3

:

\frac{d y}{d x} = \frac{1}{3}

4

Tangent equation:

y - 6 = \frac{1}{3} (x - 9)

y - 6 = \frac{x}{3} - 3

y = \frac{x}{3} + 3

or

x - 3 y + 9 = 0

Answer

y = \frac{1}{3} x + 3

or

x - 3 y + 9 = 0

Worked Example05

A curve has parametric equations $x = 4 sin t$ , $y = 3 cos t$ . Find the equation of the normal at $t = \frac{π}{6}$ .

1

Find the point at

t = \frac{π}{6}

:

x = 4 sin \frac{π}{6} = 4 \times \frac{1}{2} = 2

y = 3 cos \frac{π}{6} = 3 \times \frac{3}{2} = \frac{3 3}{2}

2

Find derivatives:

\frac{d x}{d t} = 4 cos t

,

\frac{d y}{d t} = - 3 sin t

3

Tangent gradient:

\frac{d y}{d x} = \frac{- 3 sin t}{4 cos t} = - \frac{3}{4} tan t

At

t = \frac{π}{6}

:

\frac{d y}{d x} = - \frac{3}{4} \times \frac{1}{3} = - \frac{3}{4}

4

Normal gradient

= - \frac{1}{( - 3 /4 )} = \frac{4}{3} = \frac{4 3}{3}

5

Normal:

y - \frac{3 3}{2} = \frac{4 3}{3} (x - 2)

Answer

y - \frac{3 3}{2} = \frac{4 3}{3} (x - 2)

Practice

Question 1: A curve has parametric equations

x = t^{2} + 2 t

,

y = t^{3} - t

. Find the value(s) of

t

at points where the tangent is parallel to the x-axis.

1

Find derivatives:

\frac{d x}{d t} = 2 t + 2

,

\frac{d y}{d t} = 3 t^{2} - 1

2

Tangent parallel to x-axis means

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

, so

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

:

3 t^{2} - 1 = 0

t^{2} = \frac{1}{3}

t = \pm \frac{1}{3} = \pm \frac{3}{3}

3

Check

\frac{d x}{d t} \neq = 0

at these values (it doesn't equal zero) ✓

Answer

t = \pm \frac{3}{3}

(or

\pm \frac{1}{3}

)

Practice

Question 2: A curve has parametric equations

x = 2 t

,

y = \frac{8}{t}

. Find the equation of the tangent at the point where

t = 2

.

1

Find the point at

t = 2

:

x = 2 (2) = 4

,

y = \frac{8}{2} = 4

Point:

(4, 4)

2

Find derivatives:

\frac{d x}{d t} = 2

,

\frac{d y}{d t} = - \frac{8}{t ^{2}}

3

Find gradient at

t = 2

:

\frac{d y}{d x} = \frac{- 8/ t ^{2}}{2} = \frac{- 4}{t ^{2}}

At

t = 2

:

\frac{d y}{d x} = \frac{- 4}{4} = - 1

4

Tangent equation:

y - 4 = - 1 (x - 4)

y = - x + 8

or

x + y = 8

Answer

y = - x + 8

or

x + y = 8

Summary

Key Formulas

ConceptFormula

First Derivative

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

Horizontal Tangent

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

(with

\frac{d x}{d t} \neq = 0

)

Vertical Tangent

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

(with

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

)

Normal Gradient

- \frac{d x / d t}{d y / d t} = - \frac{1}{d y / d x}

Key Exam Skills

Find gradient at specific parameter values

Find equations of tangents and normals

Locate horizontal/vertical tangent points

Always check

\frac{d x}{d t} \neq = 0

for horizontal tangents