Serious Doodles

Calculus Level 2

Most of us have probably discovered some neat shapes simply by doodling, such as this curve you get when connecting the dots from one line to a perpendicular one in inverse dot order. (Source: Vihart)

But what shape is this curve anyway? It's not something built with discrete points like most graphs, rather a pattern formed by parts of many straight lines.

Say the lines of dots are perpendicular with a length of 1. What equation do we get as $a$ is incremented continuously from $0$ to $1$ , in terms of $x$ , that models this curve?

e^{\frac{1}{x+1}}-e+1

1-\sqrt{1-(1-x)^2}

\frac{1}{x+\frac{\sqrt{5}-1}{2}}-\frac{\sqrt{5}-1}{2}

(1-\sqrt{x})^2

1 solution

Marley McWilliams
Aug 31, 2019

Let's start by converting that parametric equation to a function of $a$ and $x$ . $(a(1-t),(1-a)t) \to y=\frac{(a-x)(1-a)}{a}$

The slope is formed by the highest connecting line at every given point $x$ . What value of $a$ maximises the height of the line at a given $x$ ? $\frac{dy}{da}=\frac{x-a^2}{a^2}$ Setting to zero, $x=a^2$ . Since $a$ is between $0$ and $1$ and is positive, $a=\sqrt{x}$ . Plugging in: $y=\frac{(\sqrt{x}-x)(1-\sqrt{x})}{\sqrt{x}}=(1-\sqrt{x})^2$

Serious Doodles

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