A Cool looking Curve
Random number: 93
20th November 2016
Remember the good old days?
Back when I doodled random paraphernalia on the sides of my Chinese Homework?
There's one classic that I'll never forget. I'm about 72% sure that it was Sunny that taught me how to draw this during tutoring in grade 3/4 at the South Pacific Education Center in Parramatta, but I'm really not certain, it's just a feeling I get.
Drawing this beauty is easy enough that an 8-year old me could understand it. You start with two perpendicular lines, mark some equally spaced intervals on them, and connect them so that 1 goes to 1, 2 goes to 2, 3 goes to 3, and so on. It was easy! It was fun! It made a nice looking curve! If you increase the number of lines you draw, it takes much longer, but it is soo worth it:
I'm not quite sure why it took another 10 years before I would even question what curve that this process creates. I'm sure I must have drawn the curve multiple times throughout those 10 years. Regardless, the question has been bestowed to me on this cold lonely night, so I shall attack it!
Suppose that I'm drawing the curve on the \(xy\) plane in the first quadrant square \(0\leq x \leq 1\) and \(0 \leq y \leq 1\). Now suppose that I draw the line connecting \((0,1-i)\) and \((i,0)\) for some \(i \in \mathbb{R} : 0 \leq i \leq 1\). The equation of the line I draw is:
$$y = (1 - \frac{1}{i})x + (1 - i)$$
To find the curve in question however, we can fix \(x\) and vary over \(i\) to find the maximum \(y\) for a given \(x\). So, let's differentiate \(y\) w.r.t \(i\) and set this to 0.
$$\frac{\partial y}{\partial i} = -1 + \frac{x}{i^2} = 0$$
Leads to the solution that \(i = \sqrt{x}\). Substituting this into our original line equation, we end up with the equation:
$$y = 1 + x - 2\sqrt{x}$$
The red line is that function! Seems like it's right!