Higher Engineering Mathematics

DIFFERENTIATION OF PARAMETRIC EQUATIONS 315

G

differentiation (from Chapter 27):

dy

dx

=

dy

dθ

×

dθ

dx

It may be shown that this can be written as:

dy

dx

=

dy

dθ

dx

dθ

(1)

For the second differential,

d^2 y

dx^2

=

d

dx

(

dy

dx

)

=

d

dθ

(

dy

dx

)

·

dθ

dx

or

d^2 y

dx^2

=

d

dθ

(

dy

dx

)

dx

dθ

(2)

Problem 1. Givenx= 5 θ−1 and

y= 2 θ(θ−1), determine

dy

dx

in terms ofθ

x= 5 θ−1, hence

dy

dθ

= 5

y= 2 θ(θ−1)= 2 θ^2 − 2 θ,

hence

dy

dθ

= 4 θ− 2 = 2 ( 2 θ− 1 )

From equation (1),

dy

dx

=

dy

dθ

dx

dθ

=

2 ( 2 θ− 1 )

5

or

2

5

( 2 θ− 1 )

Problem 2. The parametric equations of a

function are given byy=3 cos 2t,x=2 sint.

Determine expressions for (a)

dy

dx

(b)

d^2 y

dx^2

(a)y=3 cos 2t, hence

dy

dt

=−6 sin 2t

x=2 sint, hence

dx

dt

=2 cost

From equation (1),

dy

dx

=

dy

dt

dx

dt

=

−6 sin 2t

2 cost

=

−6(2 sintcost)

2 cost

from double angles, Chapter 18

i.e.

dy

dx

=−6 sint

(b) From equation (2),

d^2 y

dx^2

=

d

dt

(

dy

dx

)

dx

dt

=

d

dt

(−6 sint)

2 cost

=

−6 cost

2 cost

i.e.

d^2 y

dx^2

=− 3

Problem 3. The equation of a tangent drawn to

a curve at point(x 1 ,y 1 )is given by:

y−y 1 =

dy 1

dx 1

(x−x 1 )

Determine the equation of the tangent drawn to

the parabolax= 2 t^2 , y= 4 tat the pointt.

At pointt,x 1 = 2 t^2 , hence

dx 1

dt

= 4 t

and y 1 = 4 t, hence

dy 1

dt

= 4

From equation (1),

dy

dx

=

dy

dt

dx

dt

=

4

4 t

=

1

t

Hence, the equation of the tangent is:

y− 4 t=

1

t

(

x− 2 t^2

)

Problem 4. The parametric equations of a

cycloid arex=4(θ−sinθ),y=4(1−cosθ).

Determine (a)

dy

dx

(b)

d^2 y

dx^2