Torricelli's Law: Calculus problem hidden in the last blog post

Buzzles
1

I have despaired of anyone noticing it in the hovertext of the image so I’m putting it here:

Torricelli’s Law says that if there’s a hole with radius r at the bottom of a vat then fluid flows out at a rate of

1/2\tau r^2\sqrt{2gh}

where g is 9.81 \text{m}/\text{s}^2 and h is the depth of the hole. If the vat is a cylinder with height H and radius R, how long does it take for the vat to drain?

2

Drum roll… :drum:

Here’s my answer:

First, the volume of the partially-drained cylinder as a function of h is just the area of the base times the height:

V(h) = 1/2\tau R^2 h.

(By the way, \tau is 2\pi if you didn’t get the memo.)

The volume of an infinitesimally thin disk at height h is 1/2\tau R^2 dh.

If you divide that volume by the rate of change of the volume at that height, you get the amount of time it takes for that particular disk’s worth to drain. (Volume divided by volume-over-time = time.) And volume-over-time at height h is what we were given by Torricelli’s Law.

So we just need to sum up the amounts of time for all the disks as h goes from H to 0. (Or 0 to H, doesn’t matter.) Easy calculus problem! (Or extremely mind-bendingly hard problem without calculus. Calculus sure was a good idea. Thanks Newton and Leibniz!)

\int_0^H\frac{1/2\tau R^2 dh}{1/2\tau r^2\sqrt{2gh}} = \frac{R^2}{r^2\sqrt{2g}}\int_0^H\frac{dh}{\sqrt{h}} = \left. \frac{R^2}{r^2\sqrt{2g}} \cdot 2\sqrt{h} \;\right\rvert_0^H = \sqrt{\frac{2H}{g}}\left(\frac{R}{r}\right)^2

QED

(If you’re seeing this in email, the good stuff is in spoiler tags so I presume you’ll have to click through. This parenthetical will self-destruct.)

PS: Hmm, the spoiler tags are messing up the math! The fraction bars are missing. Maybe I’ll just stop with the spoiler tags… PPS: I’m now compromising and putting the main LaTeX part outside the spoiler tags.