Answer to Question #104831 in Calculus for Yosan Fesahazion

Question

The height of an inverted cone as it is filled with sand is always four times the radius. The radius is increasing at the rate of 3 cm/s. Calculate the rate at which the volume is increasing at the instant the volume is 800 cm^3

Answer 1

Let us assume that:

The radius of inverted cone be $r$ cm. and height be $h$ cm.

Since:

The height of the cone is always four times the radius.

So:

We can write:

$h = 4r$

We are given:

The increasing rate of radius is:

$\frac{dr}{dt} = 3$ cm/s.

And:

The volume of the cone is:

$V = 800 \,\, cm^3$

Now:

We know that, the volume of a cone is:

$V = \frac{1}{3} \pi r^2 h \\ \,\,\,\,\,\,= \frac{4}{3} \pi r^3 \hspace{1 cm} \left[ \because h = 4r \right]$

Again, we can also write:

$800 = \frac{4}{3} \pi r^3 \hspace{1 cm} \left[ \because V = 800 \right] \\ \Rightarrow r^3 = \frac{600}{\pi}$

$\Rightarrow r = \left( \frac{600}{\pi} \right)^{\frac{1}{3}}$

Now:

Derivative of $V$ with respect to time $t$ is:

$\frac{dV}{dt} = \frac{4}{3} \pi \frac{d}{dt} (r^3)$

$\,\,\,\,\,\,\,\,\,= \frac{4}{3} \pi \left( 3r^2 \frac{dr}{dt} \right)$

$\,\,\,\,\,\,\,\,\,= 4 \pi \left(\frac{600}{\pi} \right)^{\frac{2}{3}} (3)$ $\hspace{1cm} \left[ \because \frac{dr}{dt} = 3 \, and \, r = \left( \frac{600}{\pi} \right)^{\frac{1}{3}} \right]$

$\,\,\,\,\,\,\,\,\,= 12 (\pi)^{\frac{1}{3}} (600)^{\frac{2}{3}}$

$\,\,\,\,\,\,\,\,\,\approx 1250$

Hence:

The volume of the cone is increasing approximately at the rate of $1250$ cubic centimeters per second.