316 Higher Engineering Mathematics
(a) Ellipse
(b) Parabola
(c) Hyperbola
(d) Rectangular hyperbola
(e) Cardioid
(f) Astroid
(g) Cycloid
Figure 29.1
It may be shown that this can be written as:
dy
dx
=
dy
dθ
dx
dθ
(1)
For the second differential,
d 2 y
dx 2 =
d
dx
d y
dx
=
d
dθ
d y
dx
·
dθ
dx
or
d 2 y
dx 2 =
d
dθ
dy
dx
dx
dθ
(2)
Problem 1. Given x = 5θ − 1 and
y = 2θ (θ − 1), determine
d y
dx
in terms of θ.
x = 5θ − 1, hence
d y
dθ
= 5
y = 2θ(θ − 1) = 2θ 2 − 2θ,
hence
d y
dθ
= 4θ − 2 =2 (2θ − 1)
From equation (1),
d y
dx
=
d y
dθ
dx
dθ
=
2(2θ − 1)
5
or
2
5
(2θ − 1)
Problem 2. The parametric equations of a
function are given by y = 3 cos2t , x = 2 sint .
Determine expressions for (a)
d y
dx
(b)
d
2 y
dx 2 .
(a) y = 3 cos 2t , hence
d y
dt
=−6 sin2t
x = 2 sin t , hence
dx
dt
= 2 cos t
From equation (1),
d y
dx
=
d y
dt
dx
dt
=
−6 sin2t
2 cos t
=
−6(2 sin t cos t )
2 cos t
from double angles, Chapter 17
i.e.
dy
dx
=−6 sin t
(b) From equation (2),
d 2 y
dx 2 =
d
dt
d y
dx
dx
dt
=
d
dt
(−6 sint)
2 cost
=
−6 cost
2 cost
i.e.
d 2 y
dx 2 =−3
Problem 3. The equation of a tangent drawn to a
curve at point (x 1 , y 1 ) is given by:
y − y 1 =
d y 1
dx 1
(x − x 1 )
(a) Ellipse
(b) Parabola
(c) Hyperbola
(d) Rectangular hyperbola
(e) Cardioid
(f) Astroid
(g) Cycloid
Figure 29.1
It may be shown that this can be written as:
dy
dx
=
dy
dθ
dx
dθ
(1)
For the second differential,
d 2 y
dx 2 =
d
dx
d y
dx
=
d
dθ
d y
dx
·
dθ
dx
or
d 2 y
dx 2 =
d
dθ
dy
dx
dx
dθ
(2)
Problem 1. Given x = 5θ − 1 and
y = 2θ (θ − 1), determine
d y
dx
in terms of θ.
x = 5θ − 1, hence
d y
dθ
= 5
y = 2θ(θ − 1) = 2θ 2 − 2θ,
hence
d y
dθ
= 4θ − 2 =2 (2θ − 1)
From equation (1),
d y
dx
=
d y
dθ
dx
dθ
=
2(2θ − 1)
5
or
2
5
(2θ − 1)
Problem 2. The parametric equations of a
function are given by y = 3 cos2t , x = 2 sint .
Determine expressions for (a)
d y
dx
(b)
d
2 y
dx 2 .
(a) y = 3 cos 2t , hence
d y
dt
=−6 sin2t
x = 2 sin t , hence
dx
dt
= 2 cos t
From equation (1),
d y
dx
=
d y
dt
dx
dt
=
−6 sin2t
2 cos t
=
−6(2 sin t cos t )
2 cos t
from double angles, Chapter 17
i.e.
dy
dx
=−6 sin t
(b) From equation (2),
d 2 y
dx 2 =
d
dt
d y
dx
dx
dt
=
d
dt
(−6 sint)
2 cost
=
−6 cost
2 cost
i.e.
d 2 y
dx 2 =−3
Problem 3. The equation of a tangent drawn to a
curve at point (x 1 , y 1 ) is given by:
y − y 1 =
d y 1
dx 1
(x − x 1 )
