What is the equation of the osculating circle that intersects the parabola
{f(x; x^2) : x E R} at the point (0; 0)? Find an arc length parametrization of this circle.

Question

What is the equation of the osculating circle that intersects the parabola
{f(x; x^2) : x E R} at the point (0; 0)? Find an arc length parametrization of this circle.

amistre64 · Answer

i recall something about a "k" ...

amistre64 · Answer

$$r = \frac{1}{|k(t)|}$$where k is curvature

amistre64 · Answer

$$k=\frac{y''}{(1+y'^2)^{3/2}}$$

anonymous · Answer

But would i need the equation of the circle to find K? like at least know the radius?

amistre64 · Answer

k can be determined from y = x^2, and k is used to determine the radius

the equation for the circle in this problem is centered at (0,r)

amistre64 · Answer

y = x^2
y' = 2x
y'' = 2

k = 2/(1+(2x)^2)^(3/2), at x=0;  k = 2

r = 1/k = 1/2

x^2 + (y-1/2)^2 = 1/4 is the equation of the circle, but not in parametric form

amistre64 · Answer

http://www.wolframalpha.com/input/?i=x%5E2+%2B+%28y-1%2F2%29%5E2+%3D+1%2F4%2C+y+%3D+x%5E2