- #1
TyroneTheDino
- 46
- 1
Homework Statement
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.
Homework Equations
R=1/K
##R=\frac{\left ( 1+\left ( y' \right )^2 \right )^{2/3}}{y''}##
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?