# At what point does this parametric curve cross itself?

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• Dec 5th 2013, 09:50 AM
s3a
At what point does this parametric curve cross itself?
Problem:The parametric curve r = (–t^2 – 4t – 4, –3cos(πt), t^3 – 28t) crosses itself at one and only one point P.a) What are the coordinates of this point P?b) Letting θ be the acute angle between the two tangent lines to the curve at the crossing point, what is the value of cos(θ)?Answers:a) P(–16,–3,–48)b) cos(θ) = 0.934487734928968I'm trying to follow this link ( Pauls Online Notes : Calculus II - Tangents with Parametric Equations ), but I'm having trouble adapting that to this particular problem.Could someone please tell me how to go about solving this problem?
• Dec 5th 2013, 11:28 AM
Plato
Re: At what point does this parametric curve cross itself?
Quote:

Originally Posted by s3a
Problem:The parametric curve r = (–t^2 – 4t – 4, –3cos(πt), t^3 – 28t) crosses itself at one and only one point P.a) What are the coordinates of this point P?b)

I can tell you that the most difficult part of this problem is finding the point of self intersection.
Find $t_1\ne t_2$ such that $r(t_1)=r(t_2)$. I frankly see no clear solution to that.

Finding the angle is easy: $\theta = \arccos \left( {\frac{{ {r'({t_1}) \cdot r'({t_2})} }}{{\left\| {r'({t_1})} \right\|\,\left\| {r'({t_2})} \right\|}}} \right)$
• Dec 5th 2013, 03:29 PM
BobP
Re: At what point does this parametric curve cross itself?
If the point of intersection has parameters $t_{1}$ and $t_{2},$ then

$-t_{1}^{2}-4t_{1}-4 = -t_{2}^{2}-4t_{2}-4,$

and

$t_{1}^{3}-28t_{1}=t_{2}^{3}-28t_{2}.$

Solve simultaneously to get the values $-6$ and $2.$
• Dec 5th 2013, 07:24 PM
s3a
Re: At what point does this parametric curve cross itself?
Thanks for the replies, guys!

BobP, I'm a little too sleepy to do this manually now, so I used software and got t_1 = -2, t_1 = 2, t_1 = -6, as well as t_2 = -2, t_2 = 2, t_2 = -6. How do we know to reject t_2 = -2?
• Dec 6th 2013, 01:40 AM
BobP
Re: At what point does this parametric curve cross itself?
I don't know why your software came up with this -2. I did this on paper and found just the two values, -6 and +2.
• Dec 6th 2013, 04:48 AM
BobP
Re: At what point does this parametric curve cross itself?
OK., it's because it's accepting $t_{1}=t_{2}$ as a possible solution.
• Dec 6th 2013, 08:15 AM
s3a
Re: At what point does this parametric curve cross itself?
Hello, again.

That answer doesn't sit well with me, because, in my work (which is attached in this post), I get t_2 = –2 by examining t_1 = –t_2 – 4 and not t_1 = t_2.

If I examine t_1 = t_2, then I get the set of all solutions (which makes sense to me).

Having said that, t_2 = –2 IS wrong (because it doesn't give the correct answer when I plug it into r), but can you help me see why? Am I supposed to plug in all points, and reject what disagrees with the majority, or is there a better way than that?

Could you also tell me why finding t_2 ≠ t_2 such that r(t_1) = r(t_2) yields the final answer I'm looking for, in the first place?
• Dec 6th 2013, 10:11 AM
BobP
Re: At what point does this parametric curve cross itself?
You've arrived at the correct result haven't you ?

I agree with your first equation $t_{1}=-t_{2}-4,$ and if you substitute the first of the three possible values you have at the bottom of the page $t_{2}=-2,$ you find that $t_{1}=-2$ also. So, you reject that since the two values cannot be the same. That leaves you with the other two.

My algebra is different from yours.

That first equation can be written as

$t_{2}^{2}-t_{1}^{2}+4t_{2}-4t_{1}=0,$ so

$(t_{2}-t_{1})(t_{2}+t_{1})+4({t_{2}-t_{1})=0,$

$(t_{2}-t_{1})(t_{2}+t_{1}+4)=0$

and rejecting $t_{2}=t_{1},$

we have $t_{2}+t_{1}+4=0.$

The same technique can be used for the second equation.
It needs the identity

$a^{3}-b^{3}=(a-b)(a^{2}+ ab +b^{2}).$

Then substitute from the first into the second to arrive at a quadratic.
• Dec 6th 2013, 12:04 PM
s3a
Re: At what point does this parametric curve cross itself?
Yes, I have arrived at the correct answer. :)

However, I have arrived at the correct answer by just assuming what Plato said, without understanding why that is the case.

To be specific, I made a typo in my previous post when I asked “Could you also tell me why finding t_2 ≠ t_2 such that r(t_1) = r(t_2) yields the final answer I'm looking for, in the first place?”.; I meant to ask “Could you also tell me why finding t_1 ≠ t_2 such that r(t_1) = r(t_2) yields the final answer I'm looking for, in the first place?”

Basically, I now know to reject t_2 = –2, because I don't want t_1 to equal to t_2, but what is the specific, formal reasoning for not being allowed to have t_1 equal to t_2, and why does finding t_1 and t_2 such that t_1 ≠ t_2 and r(t_1) = r(t_2) allow us to find the point P?
• Dec 6th 2013, 12:13 PM
Plato
Re: At what point does this parametric curve cross itself?
Quote:

Originally Posted by s3a
I meant to ask “Could you also tell me why finding t_1 ≠ t_2 such that r(t_1) = r(t_2) yields the final answer I'm looking for, in the first place?”

For a curve to self-intersect means the actual graph must "loop".
If one thinks of $t$ as a time variable, then the plot is in the exact same place at different times..
• Dec 6th 2013, 02:05 PM
s3a
Re: At what point does this parametric curve cross itself?
Oh, yes, I now get part a! :D

For part b, however, I get dr(t)/dt = (-2t – 4, 3πsin(πt), 3t^2 – 28)

dr(-6)/dt = (8, 0, 80) = r'(-6)

dr(2)/dt = (-8,0,-4) = r'(2)

cos(θ) = [r'(-6) ⋅ r'(2)]/[||r'(-6)|| ||r'(2)||] = -384/[sqrt(80) * sqrt(6464)] = -0.53399299138798174728
≠ 0.934487734928968
(which is the correct final answer).

What am I doing wrong?
• Dec 6th 2013, 02:37 PM
BobP
Re: At what point does this parametric curve cross itself?
• Dec 6th 2013, 03:21 PM
s3a
Re: At what point does this parametric curve cross itself?
I think I found the mistake you speak of.:
r'(2) = (-8,0,-16) and not (-8,0,-4).

I get the rest. :D

Thank you both very much!
• Dec 6th 2013, 03:47 PM
Plato
Re: At what point does this parametric curve cross itself?
Quote:

Originally Posted by s3a
I think I found the mistake you speak of.:
r'(2) = (-8,0,-16) and not (-8,0,-4).
I get the rest.

To get the acute angle we need to use absolute value $|R'(t_1)\cdot R'(t_2)|$