Higher Engineering Mathematics, Sixth Edition

Differentiation of parametric equations 319

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

Hence, radius of curvature, ρ=

√√

√

√

[

1 +

(

dy

dx

) 2 ] 3

d^2 y

dx^2

=

√

√

√√

[

1 +

(

1

t

) 2 ]^3

−

1

6 t^3

When t=2, ρ=

√

√

√

√

[

1 +

(

1

2

) 2 ]^3

−

1

6 ( 2 )^3

=

√

( 1. 25 )^3

−

1

48

=− 48

√

( 1. 25 )^3 =−67.08

Now try the following exercise

Exercise 127 Further problemson

differentiation of parametric equations

1. A cycloid has parametric equations
   x= 2 (θ−sinθ),y= 2 ( 1 −cosθ). Evaluate, at
   θ= 0 .62 rad, correct to 4 significant figures,

(a)

dy

dx

(b)

d^2 y

dx^2

.

[(a) 3.122 (b)−14.43]

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 )

Use this in Problems 2 and 3.

2. Determine the equation of the normal drawn
   to the parabolax=

1

4

t^2 ,y=

1

2

tatt=2.

[y=− 2 x+3]

3. Find the equation of the normal drawn to
   the cycloidx= 2 (θ−sinθ),y= 2 ( 1 −cosθ)
   atθ=

π

2

rad. [y=−x+π]

4. Determine the value of

d^2 y

dx^2

, correct to 4 sig-

nificant figures, atθ=

π

6

rad for the cardioid

x= 5 ( 2 θ−cos2θ),y= 5 (2sinθ−sin2θ).

[0.02975]

5. The radius of curvature,ρ,ofpartofasur-
   face when determining the surface tension of
   a liquid is given by:

ρ=

[

1 +

(

dy

dx

) 2 ] 3 / 2

d^2 y

dx^2

Find the radius of curvature (correct to 4 sig-

nificant figures) of the part of the surface

having parametric equations

(a)x= 3 t,y=

3

t

at the pointt=

1

2

(b)x=4cos^3 t,y=4sin^3 tatt=

π

6

rad.

[(a) 13.14 (b) 5.196]