Higher Engineering Mathematics, Sixth Edition

318 Higher Engineering Mathematics

29.4 Further worked problems on

differentiation of parametric

equations

Problem 5. The equation of the normal drawn to

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

y−y 1 =−

1

dy 1

dx 1

(x−x 1 )

Determine the equation of the normal drawn to the

astroidx=2cos^3 θ,y=2sin^3 θat the pointθ=

π

4

x=2cos^3 θ,hence

dx

dθ

=−6cos^2 θsinθ

y=2sin^3 θ,hence

dy

dθ

=6sin^2 θcosθ

From equation (1),

dy

dx

=

dy

dθ

dx

dθ

=

6sin^2 θcosθ

−6cos^2 θsinθ

=−

sinθ

cosθ

=−tanθ

Whenθ=

π

4

,

dy

dx

=−tan

π

4

=− 1

x 1 =2cos^3

π

4

= 0 .7071 andy 1 =2sin^3

π

4

= 0. 7071

Hence,the equation of the normal is:

y− 0. 7071 =−

1

− 1

(x− 0. 7071 )

i.e. y− 0. 7071 =x− 0. 7071

i.e. y=x

Problem 6. The parametric equations for a

hyperbola arex=2secθ,y=4tanθ.Evaluate

(a)

dy

dx

(b)

d^2 y

dx^2

, correct to 4 significant figures,

whenθ=1radian.

(a) x=2secθ, hence

dx

dθ

=2secθtanθ

y=4tanθ, hence

dy

dθ

=4sec^2 θ

From equation (1),

dy

dx

=

dy

dθ

dx

dθ

=

4sec^2 θ

2secθtanθ

=

2secθ

tanθ

=

2

(

1

cosθ

)

(

sinθ

cosθ

) =

2

sinθ

or 2cosecθ

Whenθ=1rad,

dy

dx

=

2

sin1

=2.377, correct to 4

significant figures.

(b) From equation (2),

d^2 y

dx^2

=

d

dθ

(

dy

dx

)

dx

dθ

=

d

dθ

(2cosecθ)

2secθtanθ

=

−2cosecθcotθ

2secθtanθ

=

−

(

1

sinθ

)(

cosθ

sinθ

)

(

1

cosθ

)(

sinθ

cosθ

)

=−

(

cosθ

sin^2 θ

)(

cos^2 θ

sinθ

)

=−

cos^3 θ

sin^3 θ

=−cot^3 θ

When θ=1rad,

d^2 y

dx^2

=−cot^31 =−

1

(tan1)^3

=−0.2647, correct to 4 significant figures.

Problem 7. When determining the surface

tension of a liquid, the radius of curvature,ρ,of

part of the surface is given by:

ρ=

√

√

√

√

[

1 +

(

dy

dx

) 2 ]^3

d^2 y

dx^2

Find the radius of curvature of the part of the

surface having the parametric equationsx= 3 t^2 ,

y= 6 tat the pointt=2.