Water Evaporation and Change in Depth

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Water in an open bowl evaporates at a rate proportional to the area of the surface of the water (this means that the rate of decrease of the volume is proportional to the area of the surface). Show that the depth of the water decreases at a constant rate, regardless of the shape of the bowl.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by chisigma, magneto, MarkFL, and Pranav. You can find Pranav's solution below.

[sp]Let at any time $t$, the area of water open to the atmosphere be $A(h)$ where $h$ is the height of water level from the bottom of container.

Clearly, volume of the remaining water is given by $\displaystyle V(h)=\int_0^h A(y)\,dy\,\,\, (*)$. As per the question:

$$\frac{dV}{dt}=-kA(h)$$

where $k$ is some positive constant.

But from (*), we have:
$$\frac{dV}{dt}=A(h)\frac{dh}{dt}$$
Hence,
$$A(h)\frac{dh}{dt}=-kA(h)$$
$$\Rightarrow \frac{dh}{dt}=-k$$
which shows that the rate of change of height is independent of shape of bowl and decreases at constant rate.[/sp]