Question

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) = 0.934487734928968

Could someone please tell me how to go about solving this problem?

Answer 1

For A: set –t^2 – 4t – 4=0 and find a solution for T, then plug T into all three components to get your point.

Answer 2

@FutureMathProfessor, thanks for the answer. Did you chose the first component, because it was the easiest to compute, and had the fewest number of solutions?

–t^2 – 4t – 4 = 0
t^2 + 4t + 4 = 0
(t+2)(t+2) = 0
(t+2)^2 = 0
t = -2

Component 1:
-(-2)^2 – 4(-2) – 4 = 0 ≠ -16 (which is just simply wrong)

Component 2:
-3cos(-2π) = -3*1 = -3 = -3 (which is what we want)

Component 3:
(-2)^3 – 28(-2) = -8 + 56 = 48 ≠ -48 (which is the negative of what we want)

What about part b?

Lastly, why would you say that works? Could you also explain the background theory for this problem instead of just how to compute it, please?

Answer 3

Ha! it's tough.!! even going backward from the answer, I can't get. Please, anyone have any idea. I'm curious

Answer 4

i have an idea ... not a good idea tho

Answer 5

if it meets at one point, then there will be 2 tangents at that point

Answer 6

since the second part is asking for tangents ... im thinking that this is how they expect you to flow from a to b

Answer 7

what is the derivative of r? and can we define a single point with 2 tangents? maybe from that t^3 element

Answer 8

http://www.wolframalpha.com/input/?i=-t%5E2-4t-4%3D+-u%5E2-4u-4%2C+cos%28pi*t%29+%3D+cos%28pi*u%29%2C+t%5E3-28t+%3D+u%5E3+-+28u

Answer 9

@ganeshie8, I just, simply, don't understand any of that. :P

@amistre64, I have a feeling that I must do something like that. I found this ( http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx ), but I'm still having trouble with this problem, and I would rather do a before b. :)

How would I know how to graph this without cheating/using software?

Also, for one input, there are two outputs, which violates the vertical line test, such that this is not a function. Do we just ignore that one x coordinate of 0, or do we define a specific y coordinate for that x coordinate of 0?

Lastly, I get dr/dt = (–2t – 4, 3πsin(πt), 3t^2 – 28).

Answer 10

@TheRealMeeeee

Answer 11

we could possible reduce this to 2d workings.

the projection of the curve onto the xy plane, will either cross or not cross
the projection of the curve onto the zx plane, will either cross or not cross
the projection of the curve onto the zy plane, will either cross or not cross



the issue being that the projection may cross at more than one point in a particular plane, but at least then you have a sense of where it MIGHT be crossing and can verify it with another plane projection of with the original curve