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
- 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.
- Determine the equation of the normal drawn
to the parabolax=
1
4
t^2 ,y=
1
2
tatt=2.
[y=− 2 x+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]
- 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]
- 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]