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Parabola question

  1. Jul 4, 2008 #1
    This question seems easy enough, but I'm having a hard time getting my mind around it. I want to find the solution to a second order polynomial (parabola) defined by the following criteria...

    1) It passes through points a and b
    2) At point a the slope of the curve is zero (horizontal)
    3) At point b the slope of the curve is infinite (vertical)

    The primary axis of the parabola will obviously need to be rotated by a certain angle to satisfy the slope requirements.

    Thanks in advance for any help you can provide.
     
  2. jcsd
  3. Jul 5, 2008 #2
    The axis has to be rotated by[tex]\pi[/tex]/4 here.Also i think you need to know the
    co-ordinates of A & B. then you can find the eqn of the parabola using derivatives and integration.
     
  4. Jul 5, 2008 #3

    arildno

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    Well, to help you along a bit, you ought to start out with the general formula of CONICS, and see if you can find some solutions here, prior to restricting your solution set to the realm of parabolae.

    Thus, you are interested in determing A,B,C, D, E and F in the following general equation:
    [tex]Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0[/tex]

    Requirement 1) gives you then 2 linear equations in your unknowns.

    For 2) and 3) you'll need to use the implicit differentiation theorem; for 3), restate this as the requirement dx/dy=0
     
  5. Jul 5, 2008 #4
    Here is a picture of what I am trying to represent.

    http://www.swied.com/pics/parabola.png

    This isn't a homework problem. I'm actually trying to design a template for a wood working project. Of all the conic sections I think that the parabola is the most appropriate.
     
    Last edited: Jul 5, 2008
  6. Jul 5, 2008 #5
    Here are the base equations that I am starting from...

    y' = a0 + a1 x' + a2 x'^2

    The rotation would be defined by the following two equations...

    y' = -x sin(theta) + y cos(theta)
    x' = x cos(theta) + y sin(theta)

    Given point a at (0,0),
    and point b at (10, 3).

    I'm tying to solve for a0, a1, a2, and theta.

    So far I have scribbled on several pages trying to solve this problem without success.
     
  7. Jul 5, 2008 #6

    HallsofIvy

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    Your first formula is only true for a parabola having vertical axis. It might be simpler to start from the general conic: [itex]Ax^2+ Bxy+ Cy^2+ Dx+ Ey+ F= 0[/itex]. Knowing that one point on the graph is (0,0) tells us that F= 0. Knowing that (10, 3) is on the graph tells us that 100A+ 30B+ 9C+ 10D+ 3E= 0.

    Differentiating with respect to x, 2Ax+ By+ Bxy'+ 2Cyy'+ D+ Ey'= 0. Knowing that y'= 0 at (0,0) says D= 0. The simplest way to handle "vertical" is to divide the equation by y': 2Ax/y'+ By/y'+ Bx+ 2Cy+ D/y'+ E= 0. At (10,3), all terms with y' in the denominator will become 0 so 3B+ 6C+ E= 0. At this point we have 4 equations for 6 unknowns. Since we could always divide through by any non-zero coefficient, it is sufficient to solve for 5 of the unknowns in terms of the last, then choose whatever non-zero value we want for that. The fifth equation comes from the condition that this must be a parabola: B2- 4AC= 0.
     
  8. Jul 5, 2008 #7

    Redbelly98

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    An ellipse would be much easier to solve. Why is the parabola the most appropriate?
     
  9. Jul 5, 2008 #8
    I already explored using an ellipse. Yes, it is much easier to solve, but it just doesn't look right. The ellipse is too arcing out towards point b. I need to find a curve that is fairly flat and then curves in quickly at the edge towards point a.

    I'm heading out of town in a few minutes, and won't be online again until tomorrow night.

    Thanks,

    Scottt
     
  10. Jul 5, 2008 #9

    Redbelly98

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    Correction, x=10, so 10B + 6C + E = 0.

    I'm working on a solution (which is how I caught the typo) and will report back when it's complete.

    Regards,

    Mark
     
  11. Jul 5, 2008 #10

    arildno

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    Just for the record:
    For the general case, with [itex]a=(x_{a},y_{a}),b=(x_{b},y_{b})[/itex], note that your system of equations will be:
    [tex]Ax_{a}^{2}+Bx_{a}y_{a}+Cy_{a}^{2}+Dx_{a}+Ey_{a}+F=0[/tex]
    [tex]Ax_{b}^{2}+Bx_{b}y_{b}+Cy_{b}^{2}+Dx_{b}+Ey_{b}+F=0[/tex]
    [tex]2Ax_{a}+By_{a}+D=0[/tex]
    [tex]2Cy_{b}+Bx_{b}+E=0[/tex]
    [tex]B^{2}-4AC=0[/tex]

    By multiplying the third with [itex]-x_{a}[/itex], adding to the first, and multiplying the fourth with [itex]-y_{b}[/tex], adding to the second, yields a somewhat simpler system:
    [tex]-Ax_{a}^{2}+Cy_{a}^{2}+Ey_{a}+F=0[/tex]
    [tex]Ax_{b}^{2}-Cy_{b}^{2}+Dx_{b}+F=0[/tex]
    [tex]2Ax_{a}+By_{a}+D=0[/tex]
    [tex]2Cy_{b}+Bx_{b}+E=0[/tex]
    [tex]B^{2}-4AC=0[/tex]
     
  12. Jul 5, 2008 #11

    tiny-tim

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    Welcome to PF!

    Hi swied ! Welcome to PF! :smile:

    Solve it the way Euclid would have …

    A parabola focuses lines from infinity which are parallel to its axis, onto a single point (called, obviously, the focus! :biggrin:) …

    so draw some parallel lines and see what happens … :wink:
     
  13. Jul 5, 2008 #12

    Redbelly98

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    As this is not a homework question, I've gone ahead and solved it using Hall's equations from post #6. Note, the parabola's orientation is different than in the figure given in post #4, but that shouldn't be a problem as this is a woodworking project template.

    The solution I got is

    [tex]
    9 x^2 + 60 x y + 100 y^2 - 1200y = 0
    [/tex]
    (Hooray! No pesky irrational coefficients!!!)

    I've attached a graph of the parabola, along with an ellipse (for comparison) which meets the same slope conditions.

    Swied, to aid you in making a template, I'm also including coordinates for some points along the curve:

    x, y

    0.000, 0.000
    1.000, 0.008
    2.000, 0.033
    3.000, 0.080
    3.500, 0.113
    4.000, 0.152
    4.500, 0.200
    5.000, 0.257
    5.500, 0.325
    6.000, 0.405
    6.500, 0.500
    7.000, 0.614
    7.500, 0.750
    8.000, 0.917
    8.214, 1.000
    8.743, 1.250
    9.142, 1.500
    9.442, 1.750
    9.663, 2.000
    9.821, 2.250
    9.889, 2.400
    9.924, 2.500
    9.952, 2.600
    9.974, 2.700
    9.989, 2.800
    9.997, 2.900
    10.000, 3.000
     

    Attached Files:

  14. Jul 5, 2008 #13

    Redbelly98

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    I've come up with another equation form which has the advantages:
    1. It's as easy to solve as the ellipse equation
    and
    2. It is virtually indistinguishable from the parabola.

    The equation is similar to that of an ellipse:

    [tex]
    |\frac{x-h}{A}|^N + |\frac{y-k}{B}|^N = 1
    [/tex]

    For N=2, you get an ellipse. For the current problem, this "hyper-ellipse" has center at (0,3) with A=10, B=3, so that

    [tex]
    |\frac{x}{10}|^N + |\frac{y-3}{3}|^N = 1
    [/tex]

    Solving for y, we get

    [tex]
    y = 3 - 3 (1-|\frac{x}{10}|^N)^{1/N}
    [/tex]

    N=2.355 gives a pretty good approximation to the parabola solution (see attached graph). One could also put the formula into Excel or other graphing software, and adjust the N parameter until you like what you see.
     

    Attached Files:

  15. Jul 6, 2008 #14
    Thanks to everyone for the great help. You nailed it!

    I think I have enough information to make my template now. I'll post it here when I'm all finished with the design. It is going to be a wooden surfboard. The template is going to be of the boards cross section.

    Redbelly98: Thanks for the suggestion of using a hyper ellipse. This is the first time I had heard of such a function. After reading up on it I discovered that it is commonly used in architectural design and woodworking.

    I'm seriously thinking about changing my design to use the hyper ellipse.

    Here is a nice Java applet that lets you explore the different shapes produced by a super ellipse (note: a hyper ellipse is a subset of the super ellipse).

    http://www.activeart.de/dim-shops/training/SuperEllipse/
     
  16. Jul 7, 2008 #15
    Now you have me curious about hyper ellipses, I looked it up on wolfram research and all it gave was a formula with no derivation (wolfram is super bad about showing how they arrive at a conclusion). In math you always have some function f which is usually through actual data points and the data points are matched to some predetermined function (linear, power, exponential) but to deviate from that to create a function f is extremely difficult, like finding the function to a line integral for some crazy surface. My goal here is to get back into calculus in a much deeper level than when I was in college and to understand the underlying principals of everything, everything in engineering is based on calculus, differential equations and physics. Is a a hyper ellipse just a modification of the parametric equations to a regular ellipse?
     
  17. Jul 7, 2008 #16

    arildno

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    Yes.
     
  18. Jul 7, 2008 #17

    Redbelly98

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