Differentiation of parametric equations 317
Determine the equation of the tangent drawn to the
parabolax= 2 t^2 ,y= 4 tat the pointt.
At pointt,x 1 = 2 t^2 , hence
dx 1
dt
= 4 t
and y 1 = 4 t, hence
dy 1
dt
= 4
From equation (1),
dy
dx
=
dy
dt
dx
dt
=
4
4 t
=
1
t
Hence, the equation of the tangent is:
y− 4 t=
1
t
(
x− 2 t^2
)
Problem 4. The parametric equations of a cycloid
arex= 4 (θ−sinθ),y= 4 ( 1 −cosθ). Determine
(a)
dy
dx
(b)
d^2 y
dx^2
(a) x= 4 (θ−sinθ),
hence
dx
dθ
= 4 −4cosθ= 4 ( 1 −cosθ)
y= 4 ( 1 −cosθ), hence
dy
dθ
=4sinθ
From equation (1),
dy
dx
=
dy
dθ
dx
dθ
=
4sinθ
4 ( 1 −cosθ)
=
sinθ
( 1 −cosθ)
(b) From equation (2),
d^2 y
dx^2
=
d
dθ
(
dy
dx
)
dx
dθ
=
d
dθ
(
sinθ
1 −cosθ
)
4 ( 1 −cosθ)
=
( 1 −cosθ)(cosθ)−(sinθ)(sinθ)
( 1 −cosθ)^2
4 ( 1 −cosθ)
=
cosθ−cos^2 θ−sin^2 θ
4 ( 1 −cosθ)^3
=
cosθ−
(
cos^2 θ+sin^2 θ
)
4 ( 1 −cosθ)^3
=
cosθ− 1
4 ( 1 −cosθ)^3
=
−( 1 −cosθ)
4 ( 1 −cosθ)^3
=
− 1
4 ( 1 −cosθ)^2
Now try the following exercise
Exercise 126 Further problemson
differentiation of parametric equations
- Givenx= 3 t−1andy=t(t− 1 ), determine
dy
dx
in terms oft.
[
1
3
( 2 t− 1 )
]
- A parabola has parametric equations:x=t^2 ,
y= 2 t.Evaluate
dy
dx
whent= 0 .5. [2]
- The parametric equations for an ellipse
arex=4cosθ,y=sinθ. Determine (a)
dy
dx
(b)
d^2 y
dx^2
.
[
(a)−
1
4
cotθ(b)−
1
16
cosec^3 θ
]
- Evaluate
dy
dx
at θ=
π
6
radians for the
hyperbola whose parametric equations are
x=3secθ,y=6tanθ.[4]
- The parametric equations for a rectangular
hyperbola are x= 2 t, y=
2
t
.Evaluate
dy
dx
whent= 0 .40. [−6.25]
The equation of a tangent drawn to a curve at
point(x 1 ,y 1 )is given by:
y−y 1 =
dy 1
dx 1
(x−x 1 )
Use this in Problems 6 and 7.
- Determine the equation of the tangent drawn
to the ellipsex=3cosθ,y=2sinθatθ=
π
6
.
[y=− 1. 155 x+4]
- Determine the equation of the tangent drawn
to the rectangular hyperbolax= 5 t,y=
5
t
at
t=2. [
y=−
1
4
x+ 5
]