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
d y 1
dx 1
(x − x 1 )
Determine the equation of the normal drawn to the
astroid x = 2 cos 3 θ, y = 2 sin 3 θ at the point θ =
π
4
x = 2 cos
3
θ, hence
dx
dθ
=−6 cos
2
θ sin θ
y = 2 sin
3
θ, hence
d y
dθ
= 6 sin
2
θ cos θ
From equation (1),
d y
dx
=
d y
dθ
dx
dθ
=
6 sin
2
θ cos θ
−6 cos 2 θ sin θ
=−
sin θ
cos θ
= −tanθ
When θ =
π
4
,
d y
dx
=−tan
π
4
=−1
x 1 = 2 cos 3 π
4
= 0.7071 and y 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 are x = 2 sec θ, y = 4 tanθ. Evaluate
(a)
d y
dx
(b)
d 2 y
dx 2 , correct to 4 significant figures,
when θ = 1 radian.
(a) x = 2 secθ, hence
dx
dθ
= 2 sec θ tan θ
y = 4 tan θ, hence
d y
dθ
= 4 sec 2 θ
From equation (1),
d y
dx
=
d y
dθ
dx
dθ
=
4 sec 2 θ
2 secθ tan θ
=
2 secθ
tan θ
=
2
1
cos θ
sin θ
cos θ
=
2
sin θ
or 2 cosec θ
When θ = 1 rad,
d y
dx
=
2
sin 1
= 2.377, correct to 4
significant figures.
(b) From equation (2),
d 2 y
dx 2 =
d
dθ
d y
dx
dx
dθ
=
d
dθ
(2 cosec θ)
2 secθ tan θ
=
−2 cosecθ cot θ
2 secθ tan θ
=
−
1
sin θ
cos θ
sin θ
1
cos θ
sin θ
cos θ
= −
cos θ
sin 2 θ
cos 2 θ
sin θ
= −
cos 3 θ
sin 3 θ
= −cot
3
θ
When θ = 1 rad,
d 2 y
dx 2 =−cot 3 1 =−
1
(tan 1)
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 +
d y
dx
2
3
d
2 y
dx 2
Find the radius of curvature of the part of the
surface having the parametric equations x = 3t 2 ,
y = 6t at the point t = 2.
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
d y 1
dx 1
(x − x 1 )
Determine the equation of the normal drawn to the
astroid x = 2 cos 3 θ, y = 2 sin 3 θ at the point θ =
π
4
x = 2 cos
3
θ, hence
dx
dθ
=−6 cos
2
θ sin θ
y = 2 sin
3
θ, hence
d y
dθ
= 6 sin
2
θ cos θ
From equation (1),
d y
dx
=
d y
dθ
dx
dθ
=
6 sin
2
θ cos θ
−6 cos 2 θ sin θ
=−
sin θ
cos θ
= −tanθ
When θ =
π
4
,
d y
dx
=−tan
π
4
=−1
x 1 = 2 cos 3 π
4
= 0.7071 and y 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 are x = 2 sec θ, y = 4 tanθ. Evaluate
(a)
d y
dx
(b)
d 2 y
dx 2 , correct to 4 significant figures,
when θ = 1 radian.
(a) x = 2 secθ, hence
dx
dθ
= 2 sec θ tan θ
y = 4 tan θ, hence
d y
dθ
= 4 sec 2 θ
From equation (1),
d y
dx
=
d y
dθ
dx
dθ
=
4 sec 2 θ
2 secθ tan θ
=
2 secθ
tan θ
=
2
1
cos θ
sin θ
cos θ
=
2
sin θ
or 2 cosec θ
When θ = 1 rad,
d y
dx
=
2
sin 1
= 2.377, correct to 4
significant figures.
(b) From equation (2),
d 2 y
dx 2 =
d
dθ
d y
dx
dx
dθ
=
d
dθ
(2 cosec θ)
2 secθ tan θ
=
−2 cosecθ cot θ
2 secθ tan θ
=
−
1
sin θ
cos θ
sin θ
1
cos θ
sin θ
cos θ
= −
cos θ
sin 2 θ
cos 2 θ
sin θ
= −
cos 3 θ
sin 3 θ
= −cot
3
θ
When θ = 1 rad,
d 2 y
dx 2 =−cot 3 1 =−
1
(tan 1)
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 +
d y
dx
2
3
d
2 y
dx 2
Find the radius of curvature of the part of the
surface having the parametric equations x = 3t 2 ,
y = 6t at the point t = 2.