Higher Engineering Mathematics

DIFFERENTIATION OF PARAMETRIC EQUATIONS 317

G

From equation (1),

dy

dx

=

dy

dθ

dx

dθ

=

6 sin^2 θcosθ

−6 cos^2 θsinθ

=−

sinθ

cosθ

=−tanθ

Whenθ=

π

4

,

dy

dx

=−tan

π

4

=− 1

x 1 =2 cos^3

π

4

= 0 .7071 andy 1 =2 sin^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=2 secθ,y=4 tanθ. Evaluate

(a)

dy

dx

(b)

d^2 y

dx^2

, correct to 4 significant figures,

whenθ=1 radian.

(a)x=2 secθ, hence

dx

dθ

=2 secθtanθ

y=4 tanθ, hence

dy

dθ

=4 sec^2 θ

From equation (1),

dy

dx

=

dy

dθ

dx

dθ

=

4 sec^2 θ

2 secθtanθ

=

2 secθ

tanθ

=

2

(

1

cosθ

)

(

sinθ

cosθ

) =

2

sinθ

or 2 cosecθ

Whenθ=1 rad,

dy

dx

=

2

sin 1

=2.377, correct to

4 significant figures.

(b) From equation (2),

d^2 y

dx^2

=

d

dθ

(

dy

dx

)

dx

dθ

=

d

dθ

(2 cosecθ)

2 secθtanθ

=

−2 cosecθcotθ

2 secθtanθ

=

−

(

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=1 rad,

d^2 y

dx^2

=−cot^31 =−

1

(tan 1)^3

=−0.2647, correct to 4 significant figures.

Problem 7. When determining the surface ten-

sion 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.

x= 3 t^2 , hence

dx

dt

= 6 t

y= 6 t, hence

dy

dt

= 6

From equation (1),

dy

dx

=

dy

dt

dx

dt

=

6

6 t

=

1

t

From equation (2),

d^2 y

dx^2

=

d

dt

(

dy

dx

)

dx

dt

=

d

dt

(

1

t

)

6 t

=

−

1

t^2

6 t

=−

1

6 t^3