The volume of a cone with radius r and height h is given by the formula, V = $\frac{1}{3}\pi r^2 h$ According to the question, h = $\frac{1}{6}$ r ⇒ r = 6h Substituting in the formula, ∴ V = $\frac{1}{3}\pi (6h)^2 h$ = 12 $\pi h^3$ The rate of change of the volume with respect to time is $\frac{dV}{dt}$ = 12 π $\frac{d}{dh}(h^3)$ × $\frac{dh}{dt}$ [By chain rule] = 12 π $(3h)^2 \times \frac{dh}{dt}$ 36 π $h^2 \times \frac{dh}{dt}$ Given that $\frac{dV}{dt}$ = 12 π $\mathrm{cm}^3$/s Substituting the values $\frac{dV}{dt}$ = 12 and h=4 in equation (1), we have, 12 = 36 π $(4)^2 \times \frac{dh}{dt}$ ⇒ $\frac{dh}{dt}$ = $\frac{12}{36\pi \times 16}$ ⇒ $\frac{dh}{dt}$ = $\frac{1}{48\pi}$ Hence, the height of the sand cone is increasing at the rate of $\frac{1}{48\pi}$ cm/s OR Let P(x, y) be any point on the given curve $x^2 + y^2$ – 2x – 3 = 0. Tangent to the curve at the point (x, y) is given by $\frac{dy}{dx}$ Differentiating the equation of the curve w .r. t. x we get 2x + 2y $\frac{dy}{dx}$ - 2 = 0 ⇒ $\frac{dy}{dx}$ = $\frac{2-2x}{2y}$ = $\frac{1-x}{y}$ Let P($x_1, y_1$) be the point on the given curve at which the tangents are parallel to the x axis ∴ $\left.\frac{dy}{dx}\right|_{(x_1,y_1)}$ = 0 ⇒ $\frac{1-x_1}{y_1}$ = 0 ⇒ 1 - $x_1$ = 0 ⇒ $x_1$ = 1 To get the value of $y_1$ just substitute $x_1$ = 1 in the equation $x^2 + y^2$ – 2x – 3 = 0, we get $(1)^2 + y_1^2$ - 2 × 1 - 3 = 0 ⇒ $y_1^2$ - 4 = 0 ⇒ $y_1^2$ = 4 ⇒ $y_1$ = ± 2 So, the points on the given curve at which the tangents are parallel to the x-axis are (1, 2) and (1, -2).