differentiation

318 DIFFERENTIAL CALCULUS
Hence, radius of curvature, ρ =
When t = 2, ρ =
[
√
1 +
=
[
( ]
√ 1 2 3
1 +
2)
Now try the following exercise
− 1
6 (2) 3 =
( ) ]
dy
2 3
dx
d 2 y
dx 2
[
( ) ]
√ 1 2 3
1 +
t
− 1
6t 3
√
(1.25)
3
− 1
48
=−48 √ (1.25) 3 =−67.08
Exercise 129 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) dy
dx (b) d2 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
dy 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 t2 , y = 1 t at t = 2.
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 rad. [y =−x + π]
2
4. Determine the value of d2 y
, correct to 4 significant
figures, at θ = π rad for the cardioid
dx2 6
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:
[ ( ) ]
dy
2 3/2
1 +
dx
ρ =
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]