# How do I find the 2 slopes at which this parametric function crosses itself?

I have a parametric function. If you graph it, you'll find that it looks like a figure 8.

x(t) = 2sin(2t)
y(t) = 8sin(t)


How do I find the slopes of the function when it intersects with itself again (graph this to see what I mean)? Should I convert it to a non-parametric function first?

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There’s no need to abandon parametric form:

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{8\cos t}{4\cos 2t}=\frac{2\cos t}{\cos 2t}\;.$$

Now just find (in terms of $t$) where it crosses itself.

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Thanks, how do I find where it crosses? Setting it equal to itself feels odd... – Minden Petrofsky Nov 17 '12 at 6:01
@Minden: You need to find distinct values $t_1$ and $t_2$ of $t$ such that $2\sin 2t_1=2\sin 2t_2$ and $8\sin t_1=8\sin t_2$. Clearly this happens when $\sin 2t_1=\sin 2t_2$ and $\sin t_1=\sin t_2$. Since the sine is periodic with period $2\pi$, you need only consider values of $t$ in $[0,2\pi)$; for which pairs of values is it true that $\sin t_1=\sin t_2$ and $\sin 2t_1=\sin 2t_2$? You may find it helpful to write $\sin 2t$ as $2\sin t\cos t$; then you can see that you want pairs such that $\sin t_1=\sin t_2$ and $\cos t_1=\cos t_2$. – Brian M. Scott Nov 17 '12 at 6:09

Let $x_i=2\sin 2t_i$ and $y_i=8\sin t_i,$

to find the points where the curve crosses it self, we need $x_i=x_j$ and $y_i=y_j$ for some $i\ne j$

So, $\sin t_i=\sin t_j--->(1)$ and $\sin 2t_i=\sin 2t_j--->(2)$

$(1)$ gives, $t_i=n\pi+(-1)^nt_j$

$(2)$ gives, $2t_i=m\pi+(-1)^m2t_j$ where $m,n$ are integers.

If $n=2q, t_i=2q\pi+t_j-->(3)$

If $n=2q+1, t_i+t_j=(2q+1)\pi--->(4)$

If $m=2r, 2t_i=2r\pi+2t_j\implies t_i=r\pi+t_j--->(5)$

If $m=2r+1, 2t_i=(2r+1)\pi-2t_j\implies t_i+t_j=(2r+1)\frac{\pi}2--->(6)$

We can only combine, $(3)$ and $(5),$ so that $t_i=2s\pi+t_j$

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