The volume of a cone with radius r and height h is given by the formula,
V =

1

3

πr2h
According to the question,
h =

1

6

r ⇒ r = 6h
Substituting in the formula,
∴ V =

1

3

π(6h)2h = 12 πh3
The rate of change of the volume with respect to time is

dV

dt

= 12 π

d

dh

(h3) ×

dh

dt

[By chain rule]
= 12 π (3h)2×

dh

dt

36 π h2×

dh

dt

Given that

dV

dt

= 12 π cm3/s
Substituting the values

dV

dt

= 12 and h=4 in equation (1), we have,
12 = 36 π (4)2×

dh

dt

⇒

dh

dt

=

12

36π×16

⇒

dh

dt

=

1

48π

Hence, the height of the sand cone is increasing at the rate of

1

48π

cm/s
OR
Let P(x, y) be any point on the given curve x2+y2 – 2x – 3 = 0.
Tangent to the curve at the point (x, y) is given by

dy

dx

Differentiating the equation of the curve w .r. t. x we get
2x + 2y

dy

dx

- 2 = 0
⇒

dy

dx

=

2−2x

2y

=

1−x

y

Let P(x1,y1) be the point on the given curve at which the tangents are parallel to the x axis
∴

dy

dx

|(x1,y1) = 0
⇒

1−x1

y1

= 0
⇒ 1 - x1 = 0
⇒ x1 = 1
To get the value of y1 just substitute x1 = 1 in the equation x2+y2 – 2x – 3 = 0, we get
(1)2+y12 - 2 × 1 - 3 = 0
⇒ y12 - 4 = 0
⇒ y12 = 4
⇒ y1 = ± 2
So, the points on the given curve at which the tangents are parallel to the x-axis are (1, 2) and (1, -2).