PARAMETRICEQUATIONS

An equation which is given as a relationship between x and y onlyis a Cartesian equation.

Sometimes it is more convenient to express x and y in terms of athird variable, often t.

e.g. x = t2 – 3t 1 ty = t +

These are called parametric equations.

A point on the curve can be found by choosing a value of t.

For example, if we let t = 2, the point found is: ( –2, 52

)

Example 1: Find the cartesian equation of the curves with parametric equations: a) x = t + 2 ; y = t2 – 5 b) x = 1 + sin t ; y = cos t – 4

a) Since x = t + 2 t = x – 2

Substitute in y = t2 – 5 y = ( x – 2 )2 – 5

y = x2 – 4x – 1

b) Using the identity: sin2 t + cos2 t = 1

( x – 1 )2 + ( y + 4 )2 = 1

If possible, make t the subject of one of the equations.

If the parametric equations involve trigonometric functions, look for an appropriate identity.

1t2

t2

–x =

1t2

t2

+y =

. . . (1)

. . . (2)

(1) + (2): x + y = t

Multiply (2) by 2t2 : 2t2y = t3 + 2

2(x + y)2y = ( x + y )3 + 2

( x + y )2 ( 2y – ( x + y ) ) = 2

Example 2: Find the cartesian equation of the curves with parametric equations: 1

t2

t2

–x =1t2

t2

+y =;

( x + y )2 ( y – x ) = 2

Substitute in (3):

. . . (3)

To find the gradient of a curve with parametric equations

the chain rule is used.

i.e. d yd x

=dyd t

×d td x

Example 1: Find d yd x

for the curve with parametric equations:

x = t2 + t ; y = t3 – 2t

d xd t

= 2t + 13t2 – 2

dyd t

= 3t2 – 2

× 12t + 1

=3t2 – 2 2t + 1

d yd x

=

Example 3: Find the gradient of the curve with parametric equations x = t2 ; y = 5 + 6t at the point ( 4, – 7 )

d yd x

=dyd t

×d td x

Using:

6 × 12 t

=3 t

At the point ( 4, – 7 ), t = – 2

d yd x

=

32

–

Note, care is needed here.From the x co-ordinatet2 = 4, it would be easyto deduce t = 2.

d xd t

= 2t d yd t

= 6

d yd x

=

Example 4: Find the equation of the tangent to the curve with parametric equations x = 4t ; y = t3 + 1 at the point where t = 2.

dyd t

=d xd t

=

d yd x

=

3t2 4

3t2 14

× At t = 2, d yd x

= 3(4) ×14

= 3

y = (2)3 + 1 = 9

x = 4(2) = 8

The equation of the tangent is: y – 9 = 3 ( x – 8 )

y = 3x – 15

d yd x

=dyd t

×d td x

Using:

Example 5: Find the equation of the tangent to the curve with parametric equations x = t2 ; y = t3 at the point P where t = 2.The tangent intersects the curve again at the point Q as shown. Find the coordinates of Q.

x

y

P

Q

( 1, – 1 )

d yd x = ×

12t

3t2

At t = 2, d yd x = 3

P is ( 4, 8 )

The equation of the tangent is:

y – 8 = 3 ( x – 4 ) y = 3x – 4 . . . (1)

For intersection at Q: sub. original equation in (1): t3 = 3t2 – 4

t3 – 3t2 + 4 = 0 ( t – 2 )( t2 – t – 2 ) = 0 Hence at Q, t = – 1,

( t – 2 )( t – 2 )( t + 1 ) = 0 so Q is

Summary of key points:

This PowerPoint produced by R.Collins ; Updated Nov. 2012

• If possible, make t the subject of one of the equations.

• If the parametric equations involve trigonometric functions, look for an appropriate identity.

i.e. d yd x

=dyd t

×d td x

To find the gradient of a curve with parametric equations,the chain rule is used:

To find the cartesian equation of a curve given in parametric form: