**Toolbox:**

- If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
- $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
- The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
- It $y=f(x)$ and both depend on 't' then $\large\frac{dy}{dt}$$=f'(x).\large\frac{dx}{dt}$

The gravel falls from the conveyor belt to fall in the shape of a cone with base diameter x and height x.

Therefore $radius =\large\frac{x}{2}$

The volume of gravel at the instant 't' is

$V=\large\frac{1}{3} $$\pi r^2 h=\large\frac{1}{3}$$\pi \bigg(\large\frac{x}{2}\bigg)^2$$.x$

$\quad=\large\frac{1}{12}$$\pi x^3$----(i)

Differentiating (i) with respect to t we have

$\large\frac{dv}{dt}=\frac{1}{2} $$\times 3 \pi x^2 \large\frac{dx}{dt}$

$\qquad=\large\frac{\pi x^2}{4} \frac{dx}{dt}=>\frac{dx}{dt}=\frac{4}{\pi x^2} \frac{dv}{dt}$

When $\large\frac{dv}{dt}$$=30 ft^3/min$ and $x=10 ft$

$\large\frac{dx}{dt}=\frac{4}{\pi \times 100}$$ \times 30=\large\frac{6}{5 \pi}$

The rate of change in height is $\large\frac{dx}{dt}=\frac{6}{5 \pi} $$ft/min$