Show the radius of curvature at (a,a) on the curve x*3+y*3=2a*3 is a/√2
Answers
Answer:
This curve is somewhat well-known; it is called the folium of Descartes.
There might be a sleeker way to do this, but I find that the usual curvature formula for plane curves works nicely enough: to find the curvature of the curve [math]y = f(x)[/math] at the point [math](x_0, f(x_0))[/math], we have
[math]\displaystyle \kappa(x_0) = \frac{|f’’(x_0)|}{[1 + f’(x_0)^2]^{3/2}}[/math]
Even if our curve is not exactly a one-to-one function, locally, we can say it is, and implicit differentiation will save us here. Let’s pretend that [math]y = f(x)[/math] for some [math]f[/math] around the point [math](3a/2, 3a/2)[/math]. Then, by differentiating both sides of the equation with respect to x, we get
[math]\displaystyle 3x^2 + 3y^2y’ = 3ay + 3axy’[/math]
and isolating y’ yields
[math]\displaystyle y’ = \frac{3x^2 - 3ay}{3ax - 3y^2} = \frac{x^2 - ay}{ax - y^2}[/math]
Then, to get [math]y’’[/math], we just differentiate [math]y’[/math] with respect to [math]x[/math]:
[math]\displaystyle y’’ = \frac{(ax - y^2)(2x - ay’) - (x^2 - ay)(a - 2yy’)}{(ax - y^2)^2}[/math]
(Exercise: show that [math]y’’ = \dfrac{2ax^2y}{(ax - y^2)^3}[/math].)
Substituting [math]x = y = \frac{3a}{2}[/math] gives us [math]y’ = -1[/math], [math]y’’ = -\frac{32}{3a}[/math] and so
[math]\displaystyle \kappa\left(\frac{3a}{2}\right) = \frac{\frac{32}{3a}}{2\sqrt{2}} = \frac{8\sqrt{2}}{3a}[/math]
The radius of curvature is, of course, [math]\displaystyle \frac{1}{\kappa} = \frac{3a}{8\sqrt{2}}[/math].
You can also do it using a parametrization and the usual curvature formulas for parametric space curves, but I’m not too thrilled about all those quotient rules. Then again, all things considered, it might be the same amount of work. For your information, the usual parametrization is
[math]\displaystyle x = \frac{3at}{1 + t^3}[/math]
[math]\displaystyle y = \frac{3at^2}{1 + t^3}[/math]
and the desired point is given by [math]t = 1[/math].