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Minimizing distances between points of curves

  1. Oct 30, 2012 #1
    PROBLEM STATEMENT:

    I'm looking for a somewhat general method to find the expression for the distance (in [itex]\R^2[/itex] mortal, euclidean space) between a point in a certain curve and some point outside the line.

    ATTEMPTS TO SOLVE THE PROBLEM:

    In the case of the distance between the origin and some point (x,y), the length is just [itex]\sqrt{x^2+y^2}[/itex]. This follows from the Pythagorean theorem, and can be used to minimize the distance between the origin and some point on a curve. We also observe that this formula can be used to minimize the distance between som point (a,b) and some point on a curve.
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    In the case of the distance between a line Ax+By+C=0 and some point (x,y), it's magnitude is given by [itex]\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}[/itex]. This can be proven through a combination of the ``length formula'' (Pythagorean Theorem) and a theorem which says that the slope of a line which is perpendicular to a line y=kx+m is equal to [itex]\frac{-1}{k}[/itex]. I found this proof on http://math.ucsd.edu/~wgarner/math4c/derivations/distance/distptline.htm [Broken] I suspect that it can also be proven with some vector algebra reasoning as well. It is not a coincidence that the expression for this distance contains the length of a vector orthogonal to the line in its denominator.

    In the case of minimizing the distance between a point in the circumference of a circle and some curve, things get pretty fuzzy for me. I know that once again, you use the Pythagorean theorem. What's more: since we know that the smallest distance from a line to a point is the one which forms a line perpendicular to the line in question, we know that the smallest distance from a point on the circumference on a circle and a point of some curve is the one which is perpendicular to the TANGENT LINE of the circle.

    The gradient of a function is a vector which is perpendicular to the tangent line of the contour line, and therefore it shows the direction which the line must have to form the smallest distance to some other curve. How do I use this to find the expression of the smallest distance of a point on the circumference of a circle to some other point of a curve?

    Other questions that I would like to find answers to are:
    -what is the expression for the distance between a point of a parabola and a point?
    -what is the expression for the distance between a point of a hyperbola and a point?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 30, 2012 #2

    LCKurtz

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    The distance from a point ##(a,b)## to ##(x,y)## is given by$$
    d=\sqrt{(x-a)^2+(y-b)^2}$$To minimize this quantity as ##(x,y)## varies along some curve it is usually easier to minimize$$
    D=d^2 =(x-a)^2+(y-b)^2$$A general way to do this is, given a curve ##C## in the xy plane, to parameterize ##C##. For example, if ##C## is the circle of radius 2 centered at the origin, you could express ##C## as ##x=2\cos t,\, y=2\sin t##, or if ##C## is the parabola ##y = x^2##, you could parameterize it as ##x = x, y = x^2##. So, in the first example you would be minimizing$$
    D = (2\cos t -a)^2+(2\sin t - b)^2$$Then you can find the critical values of ##t## using the usual calculus techniques.
     
  4. Oct 30, 2012 #3
    Ok. But is it given that any curve in the plane can be parametized? If so, what theorems say this? For instance, what about the hyperbolae?
     
  5. Oct 30, 2012 #4

    LCKurtz

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    Any function you can express as ##y = f(x)## can be parameterized as ##x=x,\, y = f(x)##. Something like the hyperbola ##x^2-y^2=1## could be parameterized by ##x=\cosh t,\, y=\sinh t## for the left half and ##x=-\cosh t,\, y=\sinh t## for the right half. Similar things like that arise in most situations you are likely to encounter.

    In theory, any rectifiable curve can be parameterized in terms of arc length, so the answer is theoretically, yes. But you may have heard the old saying: "In theory, theory and practice are the same. But in practice, they aren't".
     
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