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?
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?
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?
You've arrived at the correct result haven't you ?
I agree with your first equation and if you substitute the first of the three possible values you have at the bottom of the page you find that 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
so
and rejecting
we have
The same technique can be used for the second equation.
It needs the identity
Then substitute from the first into the second to arrive at a quadratic.
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?
Oh, yes, I now get part a!
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?
Actually, I found the angle that was roughly 160 degrees, and I converted it to the angle that was roughly 20 degrees, which yields the same answer, but I prefer my way, because it makes me understand what is going on geometrically a little better, but thank you for emphasizing that the answer must be positive. (When drawing the tangents, there are two pairs of angles, where each pair consists of two equivalent angles, and I chose (one of) the acute one(s).)