PARAMETRIC EQUATIONS DIFFERENTIATION

WJEC PAST PAPER PROBLEM

(OLD P3) JUNE 2003

PAST PAPER P3 JUNE 2003

A curve has parametric equations

2cos

3sin

x t

y t

(2sin ) (3cos ) 6 0p y p x

Show that the tangent to the curve at the point P, whose parameter is p, has equation:

2cos

2sin

x t

dxt

dt

3sin

3cos

y t

dyt

dt

dy dy dt

dx dt dx

1

2sin

dt

dx t

First find the gradient of the tangent at the point where the parameter is p

13cos

2sin

dyt

dx t

13cos

2sin

3cos

2sin

dyp

dx p

p

p

where the parameter is p we simply replace t with p

This is the gradient of the TANGENT at the required point with parameter p

The equation of the tangent is found using the standard equation of a straight line:

1 1( ) ( )y y m x x 3cos

2sin

pm

p

1

1

2cos

3sin

x p

y p

Where t=p

3cos( 3sin ) ( 2cos )

2sin

py p x p

p

The equation of the tangent is

2 2

2 2

3cos( 3sin ) ( 2cos )

2sin

2sin ( 3sin ) 3cos ( 2cos )

( 2sin ) 6sin (3cos ) 6cos

0 2(sin ) (3cos ) 6(sin cos )

0 2(sin ) (3cos ) 6

py p x p

p

p y p p x p

p y p p x p

p y p x p p

p y p x

So the equation of the TANGENT can be simplified to:

THE QUESTION CONTINUES TO SAY:

The tangent meets the x axis at A.

Find the least value of the length OA, where O is the origin.

When a line crosses the x axis we have the y coordinate as zero.

USE y=0 in the equation of the tangent that we have just found.

2(sin ) (3cos ) 6 0p y p x When y=0

2(sin )0 (3cos ) 6 0

(3cos ) 6 0

6

3cos

2

cos

p p x

p x

xp

xp

This will be a least value when cos p=1(the most that cos p can be)

Because the most denominator gives the least fraction.

Of course the x coordinate IS the distance OA

CONCLUDE BY SAYING:

The least value of the distance OA is 2