## Wednesday, March 28, 2012

### Caustics and implicit differentiation

A couple of days ago I give my Calculus class an exam about differentiation and I was looking for some interesting problems involving implicit differentiation. I had in mind something that had real life applications, maybe like finding the rate of change between two variables in a chemical reaction, or a physical phenomenon, something where the variables were related by an equation such that one cannot explicitly solve one in terms of the other (which is the spirit of implicit differentiation).

After browsing for a while, I couldn't find any nice looking equation to put in my exam, but I came across a family of interesting curves that appear mainly in optics. They are called caustics and basically they are the result of reflection and refraction of light rays through the boundary of an object.

Some usual places where we can see these type of curves are coffee cups (a mathematician's best friend), wine glasses, ponds, fountains, etc. Some of the most famous curves that arises as caustics are cardioids which in general will satisfy an equation like

$(x^2+y^2-x)^2=x^2+y^2$.

This looked like a nice equation on which one could do some implicit differentiation business, but just finding a rate of change between $x$ and $y$ would not have been fun at all, even calculating the equation of the tangent line is nothing more but a regular calculus problem, so I thought about combining it with an optimization problem.

If we want to find the widest part of the cardioid along the vertical direction, we have to look for the maximum and minimum values of $y$. This can be calculated by finding $\frac{dy}{dx}=0$, which using implicit differentiation gives

$\frac{dy}{dx}=\frac{(x^2+y^2-x)(2x-1)-x}{y(1-2(x^2+y^2-x))}=0$

and therefore we have $(x^2+y^2-x)(2x-1)-x=0$. From here we obtain that $y^2=\frac{x}{2x-1}-x^2+x$ and by substituting back in the equation of the caustic we find that the extrema happen at $x=3/4$ and $y=\pm \frac{3\sqrt{3}}{4}$.

Therefore we have that the widest part happens at $x=3/4$ and has a total width of $\frac{3\sqrt{3}}{2}$. Similarly, for $\frac{dx}{dy}=0$ we have that $y(1-2(x^2+y^2-x))=0$ from where we have that $y=0$  with $x=0, 2$, and $y^2=1/2-x^2+x$, and then putting that into the original equation gives $x=-1/4$ and $y=\pm \frac{\sqrt{3}}{4}$.

After doing this, a natural question would be to calculate the diameter of the caustic. This seems to be a harder question if one tries to it analytically, writing down the equations and solving the optimization problem. Instead, a more geometrical approach can solve the problem easily.

Recalling the geometric nature of a cardiod, it is obtained as the locus of a fixed point of a circle that rotates around another fixed circle. By studying this, it is not difficult to convince oneself that the diameter is achieved in either $x=3/4, y=\frac{3\sqrt{3}}{4}$ to $x=3/4, y=-\frac{3\sqrt{3}}{4}$, or at $x=2, y=0$ and some other point. The first pair of points give a distance of $d=\frac{3\sqrt{3}}{2}$. When considering the second case, is not difficult to see that the maximum distance occurs when the second point is $x=-1/4, y=\frac{\sqrt{3}}{4}$, where we have a distance of $d=\frac{\sqrt{21}}{2}$ which is smaller that the previous one. Therefore we have that the diameter of the caustic is $\frac{3\sqrt{3}}{4}$ which intuitively make sense since there would be the place where rays of light would be reflected closer to the cup.

#### 1 comment:

1. Very knowledgeable and impressive blog.Partial differentiation and implicit differentiation are two basic different techniques with a slight difference of keeping other variable constant or not.Its really worthy to read this blog.