Differentiation of parametric equations 319
x = 3t 2 , hence
dx
dt
= 6t
y = 6t , hence
d y
dt
= 6
From equation (1),
d y
dx
=
d y
dt
dx
dt
=
6
6t
=
1
t
From equation (2),
d
2 y
dx 2 =
d
dt
d y
dx
dx
dt
=
d
dt
1
t
6t
=
−
1
t 2
6t
= −
1
6t 3
Hence, radius of curvature, ρ =
1 +
d y
dx
2
3
d 2 y
dx 2
=
1 +
1
t
2
3
−
1
6t 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 problems on
differentiation of parametric equations
1. A cycloid has parametric equations
x = 2(θ − sin θ), y = 2(1 −cos θ). Evaluate, at
θ = 0.62 rad, correct to 4 significant figures,
(a)
d y
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
d y 1
dx 1
(x − x 1 )
Use this in Problems 2 and 3.
2. Determine the equation of the normal drawn
to the parabola x =
1
4
t 2 , y =
1
2
t at t = 2.
[y =−2x + 3]
3. Find the equation of the normal drawn to
the cycloid x = 2(θ − sin θ), y = 2(1 − cos θ)
at θ =
π
2
rad.
[y =−x + π]
4. Determine the value of
d 2 y
dx 2 , correct to 4 significant figures, at θ =
π
6
rad for the cardioid
x = 5(2θ − cos 2θ), y = 5(2 sin θ − sin 2θ).
[0.02975]
5. The radius of curvature, ρ, of part of a surface when determining the surface tension of
a liquid is given by:
ρ =
1 +
d y
dx
2
3/2
d 2 y
dx 2
Find the radius of curvature (correct to 4 significant figures) of the part of the surface
having parametric equations
(a) x = 3t , y =
3
t
at the point t =
1
2
(b) x = 4 cos 3 t, y = 4 sin 3 t at t =
π
6
rad.
[(a) 13.14 (b) 5.196]