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.