Welcome to Calculus. I'm Professor Ghrist, and we're about to

begin lecture 32, Bonus Material. Let us revisit Torricelli's Law that we

first saw in chapter 3. Recall, Torricelli's Law tells you about

the rate of change of the height h of fluid in a leaking tank where the cross

sectional area at the top is A. The version of Torricelli's Law that we

stated was that dh, dt is proportional to the square root of h and inversely

proportional to A. However, this is not the original version

of Torricelli's law. Torricelli's law originally stated

something about the volumetric rate of change, the flow rate going out from the

tank. And this more primal version states that

dV dt, the rate of change of volume, is proportional to the square root of h,

with a negative constant of proportionality since the fluid is

leaking out. Now the reasons for this version of

Torricelli's law come from physical arguments.

One way to get this is to assume that you have a single particle fluid at the top

of the tank, and you release it under the action of gravity, how fast will it exit

the tank. Well since distance traveled is a

quadratic function of time but velocity is a linear function of time then the

flow rate, that which depends on the velocity of the exiting fluid, is

proportional to the square root of h. If we take that law as a given, how can

we derive the version that tells you something about the rate of change of the

height of the fluid? Well we know what the volume v is.

It is of course, the integral of the volume element, and that volume element

implicates the cross sectional area. We can write that as the integral of a of

y d y, where y is the distance from the bottom and the limits are y goes from

zero to h. Both of these functions v and h are

functions of time. What happens if we differentiate with

respect to time? Well, I think you can see where we're

going with this. We're going to use the fundamental

theorem of integral calculus to show that this equal a of h substituting in the

limit times dhdt, invoking the chain rule.

This is dvdt and we know, from physical arguments, that that is equal to negative

kappa times the square root of h. If we solve that equation for dhdt, then

we see the inverse dependence on a. Here's another example with a very

similar solution. Consider what happens if you have a

beverage in a glass, and you assume that, that liquid is evaporating.

At what rate is it evaporating? Well, you might reasonably assume that

the rate is proportional to the exposed area a the top.

Then what is happening to the height of this fluid, how quickly or how slowly Is

it changing? Of course this is going to depend on the

shape of your glass. Well again, we need to have some sort of

relationship between the volume and the height, and the exposed area at the top.

And so, just like before, we will write down the volume as the integral of the

volume element. The volume element dV is going to be A of

y dy, or again, y is distance from the bottom.

Then in this case, writing the volume as the integral element, we obtain the

integral of a of y d y is y goes from zero to h, where h is the height of the

fluid as before, all of these are functions of time.

And so if we differentiate with respect to t and invoke the fundamental theorem

of integral calculus. We obtain d v d t is what?

A at h times d h d t. Now what do we know about this?

The evaporation of the beverage is telling you about the rate of change of

volume. So, if that rate is proportional to the

exposed area, then that means that what we've computed is equal to some constant,

C, times A of H, a cross sectional area across the top.

What do you notice on both sides of the equation, we have?

The cross sectional area at the top A of H.

Therefore these terms cancel and we are left with the result that the d h d t is

equal to a constant. That is under these assumptions about

evaporation, the rate of change of the height is independent, the shape, and the

glass. That is a little bit remarkable, and it's

just one example of how being familiar and indeed dexterous with the computation

of volume elements and the fundamental theorem of integral calculus can lead us

to some wonderfully useful, and sometimes surprising results.