# Puddle Puzzle Solution

The following is a solution to Puzzle #15 – Puddle Puzzle. Take a look if you haven’t already.

We base our answer on the assumption that the liquid is incompressible. By this we mean the volume of the liquid is conserved, such that its cross-sectional area does not change.

Let’s look at the cross-section of the liquid using the diagram given in the problem. It can be split into two smaller triangles by drawing a line perpendicular to the ground through the cube’s edge, as pictured above. This allows us to split the width of the triangle into two segments of lengths $a$ and $b$. From the diagram, we can see that

$\displaystyle \tan\theta=\frac{a}{s}\to a=s\tan\theta$

$\displaystyle \tan\theta=\frac{s}{b}\to b=s\cot\theta$

Using the fact that the area of a triangle is half its base length times its height, the cross-sectional area of the liquid is

$A=\frac{1}{2}\cdot s\cdot\left(s\tan\theta+s\cot\theta\right)$

$A=\frac{1}{2}s^2\left(\tan\theta+\cot\theta\right)$

The sum of the trigonometric terms actually simplifies very nicely if we write them in terms of fractions. Using the definitions of the tangent and cotangent functions we have

$\displaystyle \tan\theta+\cot\theta=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}$

Putting the two fractions over a common denominator gives

$\displaystyle \tan\theta+\cot\theta=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$

$\displaystyle \tan\theta+\cot\theta=\frac{1}{\sin\theta\cos\theta}$

Now we use a double angle formula, which tells us

$\sin 2\theta=2\sin\theta\cos\theta$

Hence

$\displaystyle \tan\theta+\cot\theta=\frac{2}{\sin 2\theta}$

Substituting this simpler expression into that for $A$ gives

$\displaystyle A=\frac{s^2}{\sin 2\theta}$

The cross-sectional area of the liquid can be determined using the initial condition $s\left(\theta=\frac{\pi}{4}\right)=s_0$:

$\displaystyle A=\frac{s_0^2}{1}$

This gives us

$s(\theta)=s_0\sqrt{\sin 2\theta}$

Neat! I like the answer to this problem because it contains a function you don’t often encounter; the square root of a trig function. Its form tells us what we already knew: as the cube tips over, the liquid becomes more spread out horizontally, so its vertical dimension has to decrease. In the limit that the cube tips over completely, the liquid has infinite horizontal dimensions, so its ‘thickness’ becomes infinitesimal in order to conserve its volume.

Here’s an animation showing the cross-section of the liquid as the cube tips over.

Of course, the animation looks slightly unnatural since the liquid cannot instantaneously adapt to the motion of the cube – there will be some sloshing about before the liquid settles into its ground state. But in the original problem, I stated the cube tipped ‘slowly’, which is the physics-exam shorthand for “infinitesimally slowly”.

The cube is assumed to tip so slowly that the system is always in its ground state. That is, one could imagine suddenly stopping the cube in its motion, and at that moment we will always find the liquid in equilibrium. The system is said to evolve adiabatically.