Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

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

= 4 −4cosθ= 4 ( 1 −cosθ)

y= 4 ( 1 −cosθ), hence

dy

=4sinθ
From equation (1),

dy
dx

=

dy

dx

=

4sinθ
4 ( 1 −cosθ)

=

sinθ
( 1 −cosθ)

(b) From equation (2),


d^2 y
dx^2

=

d

(
dy
dx

)

dx

=

d

(
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


  1. Givenx= 3 t−1andy=t(t− 1 ), determine
    dy
    dx


in terms oft.

[
1
3

( 2 t− 1 )

]


  1. A parabola has parametric equations:x=t^2 ,
    y= 2 t.Evaluate


dy
dx

whent= 0 .5. [2]


  1. 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

]


  1. Evaluate


dy
dx

at θ=

π
6

radians for the
hyperbola whose parametric equations are
x=3secθ,y=6tanθ.[4]


  1. 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.


  1. Determine the equation of the tangent drawn
    to the ellipsex=3cosθ,y=2sinθatθ=


π
6

.
[y=− 1. 155 x+4]


  1. Determine the equation of the tangent drawn
    to the rectangular hyperbolax= 5 t,y=


5
t

at
t=2. [

y=−

1
4

x+ 5

]
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