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Curvature and Tangent Line Distance Relationship

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data

    Let T be the tangent line at the point P(x,y) to the graph of the curve ##x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}, a>0##. Show that the radius of curvature at P is three times the distance from the origin to the tangent line T.


    2. Relevant equations
    R=1/K
    ##R=\frac{\left ( 1+\left ( y' \right )^2 \right )^{2/3}}{y''}##

    3. The attempt at a solution
    I'm having quite a lot of trouble figuring out where to begin this problem. I know that I need to find the curvature and radius of curvature of the given curve, but just find that is confusing to me.

    To begin I used Wolfram Alpha to evaluate what the curvature and radius of curvature would be. I get the result that
    ##K=3\sqrt{\frac{1}{x^{\frac{2}{3}}}+\frac{1}{y^{\frac{2}{3}}}}\left ( x\cdot y \right )^{\frac{3}{2}}##
    ##R=\frac{1}{3\sqrt{\frac{1}{x^{\frac{2}{3}}}+\frac{1}{y^{\frac{2}{3}}}}\left ( x\cdot y \right )^{\frac{3}{2}}}##

    I'm not sure how this conclusion is made considering it is unclear to me how to actually differentiate the original equation to find K or R.

    I know that the distance from the origin to the point P could be found by integrating the magnitude of the derivative of th original equation equation, but I am not sure how to integrate this way either.

    I was thinking maybe I could give "a" an arbitrary value like 1 so that I might actually be able to differentiate it, yet I am not sure about it. Is there any differentiation technique that I should be looking specifically at to help me find the derivative of ##x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}, a>0##?

    Or am i going about this problem completely wrong and should think of curvature and the tangent line in a different sense?
     
  2. jcsd
  3. Sep 21, 2015 #2
    Does anyone have a hint? I have been working on this problem in between my breaks at school, and can't understand where to start.
     
  4. Sep 21, 2015 #3

    Mark44

    Staff: Mentor

    Use implicit differentiation to find dy/dx in your equation ##x^{2/3} + y^{2/3} = a^{2/3}##.
    The distance from the origin to P(x, y) is ##\sqrt{x^2 + y^2}##. This is a point on the graph whose equation is given, so I would look for using that equation to get the distance in terms of x alone.
     
  5. Sep 21, 2015 #4
    When differentiating is it okay if I say that a^(2/3) will be a constant?
     
  6. Sep 21, 2015 #5

    Mark44

    Staff: Mentor

    Yes
     
  7. Sep 22, 2015 #6
    For differentiating the first equation I get##-\frac{\sqrt[3]{y}}{\sqrt[3]{x}}##. And when I do the second derivative, and I end up with ##\left ( \frac{1}{3} \right )\sqrt[3]{\frac{y}{x^4}}##.

    Using these values I get

    ##\frac{3\left ( \frac{x^\frac{2}{3}+y^{\frac{3}{2}}}{x^{\frac{2}{3}}} \right )^{\frac{3}{2}}}{\left ( \frac{1}{3} \right )\sqrt[3]{\frac{y}{x^4}}}##

    as the curvature, but I think that looks like it is going in the wrong direction still.

    Considering that when I typed the original equation. I got the result that the curvature of the radius is

    ##3\sqrt{\frac{1}{x^{\frac{2}{3}}}+\frac{1}{y^{\frac{2}{3}}}}(xy)^{2/3}##

    Is there anything that you can suggest I am making a mistake on?
     
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