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Determine if and where the curve intersects with itself

  1. Dec 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Given the parametric equations:


    [itex]x= t^{5}- 4t^{3}[/itex] and [itex]y= t^{2}[/itex]


    A)Determine whether or not the curve described by the parametric equations crosses itself at some point.

    B)If so, find the (x,y) coordinate point where the curve crosses itself.



    Where would I begin? How would I even begin to thing about this?
     
  2. jcsd
  3. Dec 30, 2013 #2

    Ray Vickson

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    If the curve (x(t).y(t)) crosses itself, there must be values ##t_1 < t_2## such that ##x(t_1) = x(t_2)## and ##y(t_1) = y(t_2)##.
     
  4. Dec 30, 2013 #3
    That makes sense. However how can the t values actually be determined in order to find the point at which the curve intersects? Or, should the (x,y) values somehow be found first, at which time the t values can be determined? Chicken or the egg?
     
  5. Dec 30, 2013 #4
    The way I did it was to write ##t_2=t_1+z##. Use Ray Vickson's suggestion to create two equations. Solve the system for ##t_1##. Then calculate x and y.
     
  6. Dec 30, 2013 #5
    scurty,

    sorry, but it makes very little sense to me at this point.

    Solving [itex]t_{2}= t_{1} + z[/itex] for [itex]t_{1}[/itex]

    where does the z come from?

    assuming I know where z comes from or what z is..

    [itex]t_{1} = t_{2} - z[/itex]


    Now that I solved for [itex]t_{2}[/itex] how does this allow someone to solve a system? And what system was being referred to?

    What values for t_{1} and t_{2} were you able to find? Does setting up the system allow you to find these t values?

    Again, my apologies that I have very little understanding of the information you guys are generously communicating here. I feel bad asking you to dumb it down a bit.
     
  7. Dec 30, 2013 #6

    Dick

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    There's probably a lot of ways to think about problems like this and about the only thing they have in common is the need to think about it first. I would start with the condition ##y(t_1)=y(t_2)##, since it's the simplest. What is the relation between ##t_1## and ##t_2## that could make that work? What do you think? Then apply that to try to solve ##x(t_1)=x(t_2)##.
     
  8. Dec 30, 2013 #7
    I might have been too unclear, sorry about that.

    Ray Vickson suggested starting with ##t_1 < t_2## (since you are looking for a point at two different times). Well, that means there is some real number ##z## such that ##t_2 = t_1+z##.

    As Dick said, starting with the y equation is easier. What I meant by solving for ##t_1## was plugging in ##t_2 = t_1+z## into the y equations: ##t_1^2=(t_1+z)^2##. Likewise your second equation will be ##t_1^5-4t_1^3 = (t_1+z)^5-4(t_1+z)^3##.

    Solving this system for ##t_1## would involve solving for ##z## in one equation and subbing it into the other equation.
     
    Last edited: Dec 30, 2013
  9. Dec 30, 2013 #8

    Dick

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    That's not so much unclear as being way too explicit as to stating the actual answer you got. We try not to do that here, don't we? Takes all the fun away from the OP discovering it.
     
    Last edited: Dec 30, 2013
  10. Dec 30, 2013 #9
    You're right, of course. It did feel like a bit too much information to be given, sorry.
     
  11. Dec 30, 2013 #10

    Dick

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    That's ok, everyone feels great about finding the answer and wants to announce it. But don't. I redacted the solution from my quote, could you take it out of your post? The cat is probably out of the bag, but let's just clean up a little.
     
  12. Dec 30, 2013 #11
    My apologies as well for inquiring about the solution. This question is derived from an example problem from "Paul's Online Math Notes" - Example 1 of the parametric equations calc 2 section.

    The question is to find the tangent lines at a point. The parametric equations and the point (0,4) are given. In the process of finding the tangent lines at (0,4) we find the t values through calculus.

    However, I wanted a better understanding of the topic, so I wanted to figure out how the point (0,4) was provided. How did they choose to give (0,4) as the point of intersection. Perhaps in the real world, the point of intersection of two parametric equations won't be so conveniently provided. They couldn't have just pulled that point out of a hat. That is the basis of my question.

    Too bad there are the occasional student who only seeks the immediate answer for the purpose of finishing their web based homework the day before the due date. Heck, I know there are times when I wish I could get through the webassign homework at the last minute. But my intention here is to ponder this topic for a real understanding. Having the solution will not deter my effort to continue thinking about the concept here. Appreciate all you guys do.
     
  13. Dec 30, 2013 #12

    Dick

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    Thanks. I'm a little unsure how the two are related, since I don't have "Paul's Online Math Notes", but that's ok. Don't apologize for posting a perfectly well stated problem. You didn't have to be told the answer is (0,4), you could derive it. If you understand how to solve the problem then then all is good. If not, keep asking questions.
     
  14. Dec 31, 2013 #13

    Ray Vickson

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    Don't agonize over it; just start to write it out. Equality of the y values says ##t_1^2 = t_2^2##. What does that tell you? Now look at the x-equations.
     
  15. Dec 31, 2013 #14
    First Attempt:

    [itex]\sqrt{t_{1}^{2}} = \sqrt{t_{1}+z}^{2}[/itex]

    [itex]t_{1}=t_{1} +z[/itex]

    z = 0 (doesn't seem to make sense because that would mean [itex]t_{1} = t_{2}[/itex], which it shouldn't.

    And subbing z = 0 into the other equation results in 0 = 0.

    Which is a true statement but does not result in a value for [itex]t_{1}[/itex].


    Second Attempt:

    [itex]t_{1}^{2} = (t_{1} + z)^{2}[/itex]

    [itex]t_{1}^{2} = (t_{1} + z)(t_{1} +z)[/itex]

    [itex]t_{1}^{2} = (t_{1}^{2} + 2(t_{1})z +z^{2})[/itex]

    [itex] 0 = 2(t_{1})z +z^{2})[/itex]

    [itex] \frac{-2(t_{1})z}{z} = \frac{z^{2}}{z}[/itex]

    [itex] -2t_{1}=z[/itex]

    Now subbing into the other equation:

    [itex]t_{1}^{5}-4t_{1}^{3} = (t_{1} - 2t_{1})^{5} - 4(t_{1}-2t_{1})^{3}[/itex]

    [itex]t_{1}^{5}-4t_{1}^{3} = -t_{1}^{5} + 4t_{1}^{3}[/itex]

    [itex]2t_{1}^{5}=4t_{1}^{3} + 4t_{1}^{3}[/itex]

    [itex] \frac{t_{1}^{5}}{t_{1}^{3}}= \frac{4t_{1}^{3}}{t_{1}^{3}} [/itex]


    [itex]t_{1}^{2}=4[/itex]

    [itex]t_{1} = +2, -2[/itex]

    This looks like its on the right track, but one time can't produce two values.

    Now I have to look at how to find [itex]t_{2}[/itex]

    perhaps [itex]t_{2}=t_{1} + z[/itex]

    [itex]t_{2} = t_{1} + (-2t_{1})[/itex]

    [itex]t_{2} = t_{1} - 2t_{1})[/itex]

    [itex]t_{2} = t_{1}(1-2) )[/itex] and since we know [itex]t_{1} = 2, -1[/itex]

    [itex]t_{2} = (+/-2)(-1)[/itex] =

    [itex]t_{2} = 2, -2[/itex]

    Which is also confusing because [itex]t_{2}[/itex] represents a single value (a given time for example). How can a single point in time contain two different values? In other words, if the t values represent the horizontal axis on a coordinate plane, how can one t variable carry two different values? [itex]t_{1}[/itex] can only be one point on the horizontal axis and [itex]t_{2}[/itex] can only be one value on the horizontal axis. Why did I get 2 numbers when finding each value of t?

    Anyhow, not that I have

    [itex]t_{1} = 2, -2[/itex]

    [itex]t_{2} = 2, -2[/itex]

    I can plug them into the parametric equations

    x = t^5 - 4t^3
    x = (+/-)2^5 - 4((+/-)2)^3
    x=0

    y =t^2
    y= (+/-)2^2
    y = 4

    So the point of intersection is (0,4)

    Thanks.

    My question is, it would make more sense intuitively if [itex]t_{1} = -2[/itex] and [itex]t_{2} = 2[/itex]

    Since having multiple values for each does not satisfy the original assumption: [itex]t_{2} = t_{1} + z[/itex]

    Overall I think I was on the right track, but the additional +/- values have me questioning the approach I took to solving this system.
     
  16. Dec 31, 2013 #15

    Ray Vickson

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    Your approach is OK but is much longer than it needs to be. Part of the problem is the substitution ##t_2 = t_1 + z##, which just makes everything longer and more complicated. From equality of the y's we get ##t_1^2 = t_2^2##, so ##t_1 = t_2## or ##t_1 = -t_2##. Only the second case counts (why?). Using it in the x-equation we have ##-t_2^5 + 4 t_2^3 = t_2^5 - 4 t_2^3##, so ##2t_2^3 (t_2^2-4) = 0##. The relevant root is ##t_2 = 2##.
     
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