Curvature and Tangent Line Distance Relationship

[TyroneTheDino]

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?

 

[TyroneTheDino]

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.

 

[Mark44]

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.

Use implicit differentiation to find dy/dx in your equation $x^{2/3} + y^{2/3} = a^{2/3}$.

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.

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.

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?

 

[TyroneTheDino]

Use implicit differentiation to find dy/dx in your equation x2/3+y2/3=a2/3x^{2/3} + y^{2/3} = a^{2/3}.

When differentiating is it okay if I say that a^(2/3) will be a constant?

 

[Mark44]

When differentiating is it okay if I say that a^(2/3) will be a constant?

Yes

 

[TyroneTheDino]

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?