Finding the self intersection point of two parametric equations.

Question

Find the values of t where the graph of the parametric equations crosses itself: $x = t^3-4t^2+t+7$ and $y = t^2-t$.

I am asking this problem on behalf of me and my Calculus 3 class. We understand the idea of $t$ being two different times, so making them say $s$ and $t$, then setting $x(s) = x(t)$ and $y(s) = y(t)$.

So we get:

$$x(t) = t^3-4t^2+t+7 = s^3-4s^2+s+7 = x(s)$$

$$y(t) = t^2-t = s^2-s = y(s)$$

However, past that we are completely lost. Our textbook goes over this type of problem for all of one paragraph, summarizing the process heavily. We asked our professor to go over this type of problem again but it still makes next to no sense. We also looked on YouTube and here on StackExchange for examples but they are either not the same type of problem, or the answers given are heavily summarized with no clear process. We are looking for a concrete process or set of ideas to follow in order to solve a problem like this.

Any help to this is greatly appreciated!

Answer 1

By ploting the path we can get a pretty big hint of where it should cross itself. By looking at the graph it becomes easy to see that the path crosses itself with something like x=1 and y=2. But we wanna do this algebraically, not graphically. So, lets try another method. We can easily see that the path crosses itself if 2 or more points with different values of t, have the same vallues of both x and y (s and t as you were calling them). We know that s and t must be different, but x and y must be iqual, lets try once again to visualize this. We know that x²-x = y²-y, so we get the following plot: The places where the functions intersect are the places where x²-x=y²-y, so, lets try to find the equation to those intersections: y²-y+(x-x²)=0, using something as simple as the bhaskara's formula, we get that y=0.5±0.5√(4x²-4x+1). This can be simplified to y=0.5±|x-0.5|. If you do the same for the equation x³-4x²+x+7=y³-4y²+y+7, we should end up with something like this: We only want the points where s≠t, so, we're only interested in the points where (s,t) = (-1,2) and (2,-1). This gives us our solution, with t=-1 and t=2 the coordinates (x,y) of your path are the same, so, the path intersects itself.

Answer 2

First attack $y,$ the weakest link. Substitute $s=t+u$ in $y$ to obtain $0=2tu+u^2-u.$ Cancel $u$ to obtain $u=1-2t.$ Therefore, $s=1-t.$ If $t=s,$ then $t={1\over 2},$ a case we must reject. Put $s =1-t$ in $x$ to obtain $$f(t)=2t^3-3t^2-3t+2=0.$$ The rejected case tells us that $2t-1$ must divide $f(t).$ In fact, $$f(t)=(2t-1)(t^2-t-2)=(2t-1)(t+1)(t-2).$$ Thus, the curve crosses itself at $(x,y)=(1,2)$ when $t$ is $-1$ and $2.$

Answer 3

After we know that $(1,2)$ is a double point, it's even clearer if we notice that \begin{eqnarray} x-1 &=& (t-3)(t-2)(t+1) \\ y-2 &=& (t-2)(t+1) \end{eqnarray} a parametrization of the cubic $-35 + 12 x - x^2 + 33 y - 5 x y - 10 y^2 + y^3=0$ with a singular point $(1,2)$.