The volume of a cone with radius r and height h is given by the formula,
V = $\frac{1}{3}\mathit{\pi }{\mathit{r}}^2\mathit{h}$
According to the question,
h = $\frac{1}{6}$ r ⇒ r = 6h
Substituting in the formula,
∴ V = $\frac{1}{3}\mathit{\pi }{\left(6\mathit{h}\right)}^2\mathit{h}$ = 12 $\mathit{\pi }{\mathit{h}}^3$
The rate of change of the volume with respect to time is
$\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{t}}$ = 12 π $\frac{\mathit{d}}{\mathit{d}\mathit{h}}\left({\mathit{h}}^3\right)$ × $\frac{\mathit{d}\mathit{h}}{\mathit{d}\mathit{t}}$ [By chain rule]
= 12 π ${\left(3\mathit{h}\right)}^2\times \frac{\mathit{d}\mathit{h}}{\mathit{d}\mathit{t}}$
36 π ${\mathit{h}}^2\times \frac{\mathit{d}\mathit{h}}{\mathit{d}\mathit{t}}$
Given that $\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{t}}$ = 12 π $\mathit{c}{\mathit{m}}^3$/s
Substituting the values $\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{t}}$ = 12 and h=4 in equation (1), we have,
12 = 36 π ${\left(4\right)}^2\times \frac{\mathit{d}\mathit{h}}{\mathit{d}\mathit{t}}$
⇒ $\frac{\mathit{d}\mathit{h}}{\mathit{d}\mathit{t}}$ = $\frac{12}{36\mathit{\pi }\times 16}$
⇒ $\frac{\mathit{d}\mathit{h}}{\mathit{d}\mathit{t}}$ = $\frac{1}{48\mathit{\pi }}$
Hence, the height of the sand cone is increasing at the rate of $\frac{1}{48\mathit{\pi }}$ cm/s
OR
Let P(x, y) be any point on the given curve ${\mathit{x}}^2+{\mathit{y}}^2$ – 2x – 3 = 0.
Tangent to the curve at the point (x, y) is given by $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$
Differentiating the equation of the curve w .r. t. x we get
2x + 2y $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ - 2 = 0
⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = $\frac{2-2\mathit{x}}{2\mathit{y}}$ = $\frac{1-\mathit{x}}{\mathit{y}}$
Let P(${\mathit{x}}_1,{\mathit{y}}_1$) be the point on the given curve at which the tangents are parallel to the x axis
∴ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}{|}_{\left({\mathit{x}}_1,{\mathit{y}}_1\right)}$ = 0
⇒ $\frac{1-{\mathit{x}}_1}{{\mathit{y}}_1}$ = 0
⇒ 1 - ${\mathit{x}}_1$ = 0
⇒ ${\mathit{x}}_1$ = 1
To get the value of ${\mathit{y}}_1$ just substitute ${\mathit{x}}_1$ = 1 in the equation ${\mathit{x}}^2+{\mathit{y}}^2$ – 2x – 3 = 0, we get
${\left(1\right)}^2+{\mathit{y}}_1^2$ - 2 × 1 - 3 = 0
⇒ ${\mathit{y}}_1^2$ - 4 = 0
⇒ ${\mathit{y}}_1^2$ = 4
⇒ ${\mathit{y}}_1$ = ± 2
So, the points on the given curve at which the tangents are parallel to the x-axis are (1, 2) and (1, -2).