1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Intersection of two curves

  1. Nov 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that these curves do not intersect.

    z=(1/a)(a-y)^2
    y^2+z^2=a^2/4

    Where a is the radius of the circle and other shape.

    2. Relevant equations

    There aren't any.

    3. The attempt at a solution

    I tried setting them equal to each other but got the equation, which doesn't simplify into anything solve able.

    a^2/4-y^2=a^2-4ay+6y^2-4y^3/a+y^4/a^2
     
  2. jcsd
  3. Nov 19, 2013 #2

    Mark44

    Staff: Mentor

    I think that the easiest way to do this is to start with a sketch of the two curves. The 2nd equation is of a circle - what is its radius?

    What sort of figure does the first equation represent?
     
  4. Nov 19, 2013 #3
    The first equation looks similar to z=1/x. The radius of the circle is a/2. I've already sketched out the equations.
     
  5. Nov 19, 2013 #4

    Mark44

    Staff: Mentor

    No, not at all. The 1/a part is just a constant. The variable part, (a - y)2, is what's important.
    That's correct for the circle's radius.
     
  6. Nov 19, 2013 #5
    It is a decreasing, concave curve is what I should have said. How does one show that there is not a solution for both equations?
     
  7. Nov 19, 2013 #6

    Mark44

    Staff: Mentor

    It is concave up, yes, but it's decreasing on part of its domain and increasing on the other part.

    What geometric figure is y = K(a - x)2? (I changed the variables so that you might recognize what this is more easily.)
    Like I said before, a sketch of both curves is a good start. So far, you have only one of them.
     
  8. Nov 19, 2013 #7
    It is a parabola. The book gives us the graph of both functions, but limits it to the first quadrant. I should have soda that too, my bad.
     
  9. Nov 19, 2013 #8

    Mark44

    Staff: Mentor

    You could try this: Write a function that represents the distance between an arbitrary point on the circle and one on the left side of the parabola. Find the minimum value of that function. If the minimum value is greater than zero, the two curves don't intersect.

    Trying to solve the equation that you found seems like a dead end to me.
     
  10. Nov 19, 2013 #9
    Am I wrong in thinking the formula for the distance between the functions is the top curve minus the bottom curve?
     
  11. Nov 19, 2013 #10

    Mark44

    Staff: Mentor

    Yes, you're wrong. That would give you the vertical distance between the two curves, which isn't the same as the distance I described.

    A slightly different approach would be to find the point on the parabola that's closest to the origin. Then the line segment from the origin to the parabola would intersect the circle at some point, and you could easily find the distance between the circle and the parabola.

    I'm sorta making this up as I go along, but I think this is a workable way to go.
     
  12. Nov 19, 2013 #11

    Mark44

    Staff: Mentor

    Offtopic, but do you happen to be in Pensacola?
     
  13. Nov 19, 2013 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Since the second equation is a circle centered at the origin, the distance from a point on the parabola to the origin is going to give you a fourth order equation, not too much different from what you get trying to intersect them. It might pay to be more clever. If pensNAS has got a good picture, think about drawing a tangent line to the circle that has the circle on one side and the parabola on the other. Then if you write down an equation for the tangent line and show it doesn't intersect the parabola, you would be done.
     
  14. Nov 23, 2013 #13
    Thanks for the help! I actually went and talked to my professor, since this is part of a statics problem and not calculus. He was okay with just plotting a few points around where the curves get closest.

    @Mark44, I'm from Pensacola.
     
  15. Nov 23, 2013 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok. For the record, I think it's pretty easy to show that the tangent line to the circle passing through x=a/(2sqrt(2)), y=a/(2sqrt(2)) doesn't intersect the parabola. Since the two curves are on different sides of that line, they can't intersect.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted