Convert $x(t)$ and $y(t)$ into Cartesian form

Question

Consider the parametric equations given by \begin{align*} x(t)&=\sin{t}-t,\\ y(t) & = 1-\cos{t}.\end{align*}

I want to write these parametric equations in Cartesian form.

In order to eliminate the sine and cosine terms I think I probably need to consider some combination of $x(t),y(t), x(t)^2$ and $y(t)^2$ but I can't see exactly how to do this.

This link may help you https://en.wikipedia.org/wiki/Cycloid — yahoo, Sep 27 '17 at 14:50
What do you get when you compute those four quantities? Are there any familiar trigonometric identities that you can use can use to combine them in a way that results in a simpler expression? — Travis Willse, Sep 27 '17 at 14:52

score 1 · Accepted Answer · answered Sep 27 '17 at 14:52

1

From the second equality, you'll get $t=\arccos(1-y)$. So,\begin{align}x&=\sin\bigl(\arccos(1-y)\bigr)-\arccos(1-y)\\&=\sqrt{1-(1-y)^2}-\arccos(1-y)\\&=\sqrt{2y-y^2}-\arccos(1-y).\end{align}

answered Sep 27 '17 at 14:52

José Carlos Santos

427,504

score 1 · Answer 2 · answered Sep 27 '17 at 14:53

1

$$t=\arccos(1-y)$$ $$x=\sin(\arccos(1-y))-\arccos(1-y)$$ $$=\sqrt{1-(1-y)^2}-\arccos(1-y)$$

answered Sep 27 '17 at 14:53

velut luna

9,961

score 0 · Answer 3 · answered Sep 27 '17 at 14:54

0

$$ y(t) = 1-\cos{t}\implies t= \arccos(1-y) \implies x= \arccos(\sin(1-y)) -\arccos(1-y)$$ that is

$$x= \sqrt{1-(1-y)^2} -\arccos(1-y)$$

answered Sep 27 '17 at 14:54

Guy Fsone

23,903

Convert $x(t)$ and $y(t)$ into Cartesian form

3 Answers3