# Thread: Intersection of parametric curve in 2d plane, and winplot q's

1. ## Intersection of parametric curve in 2d plane, and winplot q's

The attachment is a set of equations that define a parametric curve in the xy plane. My instructions are to find the lowest point on the curve. I can see how my solution my solution is correct, but when I have winplot give me the derivative curve, which is an inverse parabola with vertex (-3,1). It's intercepts are nowhere in the -1/2 neighborhood on either axis. So what am i seeing in the derivative curve? What does its vertex represent?

So while dinking about with this thing, i became curious how i would find the point of intersection on the curve (just the original curve) . I know how to find the intersection of two parametric curves, but not sure I've learned the
process of finding it with one. Can it be done?

Thanx!

2. Hello, boardguy67!

I can help with one of your questions . . .

The set of equations defines a parametric curve in the xy-plane.

. . $\begin{array}{ccc}x &=& t^3 - 3t \\ y &=& t^2 + t + 1 \end{array}$

My instructions are to find the lowest point on the curve.

It is at the very bottom of the loop, where there is a horizontal tangent.

So while dinking about with this thing,
i became curious how i would find the point of intersection on the curve.
Let $s$ be a value of the parameter.
. . We have a point: . $P(s^3-3s,\;s^2+s+1)$

Let $t$ be another value of the parameter $(s \neq t)$.
. . We have another point: . $Q(t^3-3t,\;t^2+t+1)$

When will $P = Q$ ?

We have: . $\begin{array}{ccccc}s^3-3s &=& t^3-3t & [1] \\ s^2+s+1 &=& t^2+t+1 & [2] \end{array}$

From [1]: . $s^3-t^3 \:=\: 3s - 3t \quad\Rightarrow\quad (s-t)(s^2+st + t^2) \:=\: 3(s-t)$

. . Since $s\neq t$, we have: . $s^2 + st + t^2 \:=\:3$ .[1]

From [2]: . $s^2-t^2 + s - t \:=\:0 \quad\Rightarrow\quad (s-t)(s+t) + (s-t) \:=\:0 \quad\Rightarrow\quad (s-t)(s+t+1) \:=\:0$

. . Since $s \neq t$, we have: . $s + t + 1 \:=\:0 \quad\Rightarrow\quad t \:=\:\text{-}s-1$ .[2]

Substitute into [1]: . $s^2 + s(\text{-}s-1) + (\text{-}s-1)^2 \:=\:3 \quad\Rightarrow\quad s^2 - s^2 - s + s^2 + 2s + 1 \:=\:3$

. . . $s^2 + s - 2 \:=\:0 \quad\Rightarrow\quad (s-1)(s+2) \:=\:0 \quad\Rightarrow\quad s \:=\:1,\:-2$

Substitute into [2]: . $t \:=\:-2,\:1$

If $s = 1\!:\;\begin{Bmatrix}x &=& 1^3 - 3(1) &=& \text{-}2 \\ y &=& 1^2 + 1 + 1 &=& 3 \end{Bmatrix}$

The curve intersects itself at $(\text{-}2,\:3)$.

3. That is awesome. Great detail Tank you very much. From the sounds of it, it's not terribly likely i can return the favor in mathematical question, but if you ever need a skateboard, i'll give you a smokin discount.

Be well.