Chapter 6, Problem 5P

To determine

To show: The depth of the water decreases at a constant rate without regarding the shape of the bowl.

Answer to Problem 5P

The depth of the water decreases at a constant rate.

Explanation of Solution

Given information:

Water in an open bowl evaporates at a rate proportional to the area of the surface of the water.

Calculation:

The rate of decrease of the volume is proportional to the area of the surface.

$\frac{dV}{dt}=-kA\left(x\right)$

Here, k is the positive constant and $A\left(x\right)$ is the area of the surface at the water depth of x.

Consider that the rate of change of depth with respect to time as $\frac{dx}{dt}$.

Apply Chain rule as shown below.

$\frac{dV}{dt}=\frac{dV}{dx}\frac{dx}{dt}$

Substitute $-kA\left(x\right)$ for $\frac{dV}{dt}$ in the above Equation.

$\frac{dV}{dx}\frac{dx}{dt}=-kA\left(x\right)$ (1)

Expression for the volume of water up to a depth x as shown below.

$V\left(x\right)={\displaystyle {\int }_0^{x}A\left(s\right)ds}$

Here, $A\left(s\right)$ is the area of the cross section at the water depth of s.

Differentiate both sides of the Equation with respect to x.

$\frac{dV}{dx}=A\left(x\right)$

Substitute $A\left(x\right)$ for $\frac{dV}{dx}$ in Equation (1).

$\begin{array}{l}A\left(x\right)\frac{dx}{dt}=-kA\left(x\right)\\ \frac{dx}{dt}=-k\\ \frac{dx}{dt}=\text{constant}\end{array}$

Therefore, the depth of the water decreases at a constant rate without regarding the shape of the bowl. Hence, it is proved.