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Gravel is being dumped from a conveyor belt at a rate of $30\;ft^{3}/min$ and its coarsened such that it forms a pile in the shape of the cone whose base diameter and hight are always equal. How fast is the hight of the pile increasing when the pile is $10$ ft hight?

1 Answer

  • 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$


answered Jul 23, 2013 by meena.p
edited Jul 23, 2013 by meena.p

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