Chapter 4, Problem 24P

To determine

The water level raising at the instant the cone is completely submerged.

Explanation of Solution

Given information:

A cone of radius $r$ and height $h$ is lowered point first at a rate of $1cm/s$ in a tall cylinder of radius $R$ cm which is partially filled with water.

Calculations:

Let $a$ be the depth that the cone is submerged.

We are given that $\frac{da}{dt}=1$

The volume of water that is displaced is given by the formula $V=\frac{1}{3}\pi b^{2}a$ ,

where $b$ is obtained from the relation

  $\begin{array}{l}\frac{h}{r}=\frac{a}{b}\\ or,\\ b=\frac{ar}{h}\end{array}$

Thus $V=\frac{1}{3}\pi {\left(\frac{ar}{h}\right)}^{2}a$

Now differentiating the volume we get,

  $\frac{dV}{dt}=\frac{\pi r^2{a}^2}{h^2}.\frac{da}{dt}$

When the cone is completely submerged, $a=h$ and we have $\frac{dV}{dt}=\pi r$ .

Let $l$ denote the water level in the cylinder.

Then the volume of water in the cylinder is given by $V=\pi R^{2}l$ ,

and we have $\frac{dV}{dt}=\pi R\frac{dl}{dt}$ .

Putting in the value of $\frac{dV}{dt}$ , and solving for $\frac{dl}{dt}$ , we get,

  $\frac{dl}{dt}=\frac{r^2}{R^2}$ .

Thus we can say that the water level is rising at a rate of $\frac{r^2}{R^2}$ cm/s at the instant the cone is completely submerged.