# Mathematical Toy Part 2

Warning: the following post contains some intense mathematics, caution is advised.

In the previous post, I investigated a simple toy that generated an ellipse. However, how did I know it was an ellipse? I proved it, and here is how.

First, we need to model the two pivot points of the toy. They are both fixed on the x and y-axis. The toy also moves in a circular motion. The pivot on the y-axis is starts at 1 and the pivot on the x-axis is at 0. Hence, we will model the motion of the y pivot by the cosine function and the motion of the x pivot by the sine function. Taken together, this information gives us two points:

$(0, \cos(t))$

$(\sin(t), 0)$

I am using t to denote how the position of the pivots change in time. Due to boundary complications, I will only model one half of a complete revolution. However, since the resulting shape is symmetrical, this is model sufficient to determine the curve the toy traces. Thus, 0 ≤ t ≤ π

The handle is attached to a rigid piece of wood. The rigid piece of wood can be modeled by a line segment. Both of the pivot points lie on this line segment. To determine the equation of the line segment, we must use the good old slope formula:

$m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} = \dfrac{\cos(t) - 0}{0-\sin(t)} = \dfrac{\cos(t)}{-\sin(t)}$

Then we can use the slope-point formula:

$y - y_1 = m(x - x_1)$

$y - \cos(t) = \dfrac{\cos(t)}{-\sin(t)}(x-0)$

$y = \dfrac{\cos(t)}{-\sin(t)}x + \cos(t)$

Alright! Now we are getting somewhere! The next step is to model the handle. For simplicity, we will model the handle as a point 1 unit away from the x-axis pivot. To determine this point, we can use the distance formula:

$d = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}$

$1 = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}$

$1^2 = (x_2 - \sin(t))^{2} + (y_2 - 0)^{2}$

$1 = (x_2)^2 - 2x_2\sin(t) + \sin^2(t) + (y_2)^{2}$

We can use the previous formula for the equation of the line to determine y2 in terms of x2:

$y = \dfrac{\cos(t)}{-\sin(t)}x + \cos(t)$

Thus:

$1 = (x_2)^2 - 2x_2\sin(t) + \sin^2(t) + \left( \dfrac{\cos(t)}{-\sin(t)}x_2 + \cos(t) \right)^2$

$1 = (x_2)^2 - 2x_2\sin(t) + \sin^2(t) + \dfrac{\cos^2(t)}{\sin^2(t)}(x_2)^2 - 2\dfrac{\cos^2(t)}{\sin(t)}x_2 + \cos^2(t)$

Rearranging we find that:

$1 = \sin^2(t) + \cos^2(t) + (x_2)^2 + \dfrac{\cos^2(t)}{\sin^2(t)}(x_2)^2 - 2x_2\sin(t) - 2\dfrac{\cos^2(t)}{\sin(t)}x_2$

$1 = \sin^2(t) + \cos^2(t) + (x_2)^2 \left(1 + \dfrac{\cos^2(t)}{\sin^2(t)} \right) - 2x_2 \left( \sin(t) + \dfrac{\cos^2(t)}{\sin(t)} \right)$

Using well-known trigonometry identities, the above reduces to:

$1 = 1 + (x_2)^2 \left(\dfrac{1}{\sin^2(t)} \right) - 2x_2 \left( \dfrac{1}{\sin(t)} \right)$

$0 = (x_2)^2 \left( \dfrac{1}{\sin^2(t)} \right) - 2x_2 \left( \dfrac{1}{\sin(t)} \right)$

$0 = x_2 \left( \dfrac{x_2}{\sin^2(t)} - 2\dfrac{1}{\sin(t)} \right)$

Ignoring the trivial solution, we find that x2 is equal to:

$x_2 = 2\sin(t)$

And solving for y2 yields:

$y_2 = \dfrac{\cos(t)}{-\sin(t)}2\sin(t) + \cos(t) = -\cos(t)$

Hence, the point in parametric form is:

$(2\sin(t), - \cos(t))$

We would like to identify what curve these parametric equations will trace. This is equivalent to solving the follow 2 equations in terms of x and y only:

$x = 2\sin(t)$

$y = -\cos(t)$

By squaring both sides of both equations, we get the following:

$x^2 + 4y^2 = 4\sin^2(t) + 4\cos^2(t) = 4$

$x^2 + 4y^2 = 4$

$\dfrac{x^2}{4} + \dfrac{y^2}{1} = 1$

Which is the equation of an ellipse. Hence, the handle of the toy traces a path equivalent to an ellipse. Below is an animation I constructed using the parametric equations.

https://www.desmos.com/calculator/sz3roeuszm

A copious amount of work? Maybe. But what a beautiful proof.