How to find the center of curvature?

First we need to find the curvature k = l(y'')l / (1 + (y')^2)^(3/2).

With y = x^2 we have y' = 2x and y '' = 2. So k = 2 / (1 + 4*x^2)^(3/2).

At the point (1,1) we then have k = 2 / 5^(3/2) and thus the radius of curvature

rho = (1/2)*5^(3/2).

Next, we need to find the unit normal vector to the curve at (1,1).

Since the curve can be parameterized as x = t, y = t^2 we have position vector

<t, t^2> and velocity vector <1, 2t>. So the unit tangent vector is

T(t) = <1, 2t> / sqrt(1 + 4*t^2), which at t = 1 is <(1/sqrt(5)), (2/sqrt(5))>.

The unit normal vector will then be N = <(-2/sqrt(5)), (1/sqrt(5))>, as this

makes the dot product N o T = 0 and orients N upward, reflecting the fact that

y = x^2 is concave upward.

The position vector of the center of curvature is then

<1,1> + rho*T = <1,1> + (1/2)*5*sqrt(5)*<(-2/sqrt(5)), (1/sqrt(5))> =

<1,1> + <-5, (5/2)> = <-4, (7/2)>.

So the center of curvature is (-4, (7/2)), and the equation of the osculating circle is

(x + 4)^2 + (y - (7/2))^2 = (rho)^2 = 125/4.

Radius of curvature at any point, R(x) = [{(1+y'²)^(3/2)}/(|y''| )}

y'(x) = 2x and y''(x) = 2 or

y'(1) = 2 y''(1) = 2 or

R(1) = [{(1+2²)^(3/2)}/(2 )} = (1/2)*5√5 = 2.5*√5

Slope of normal at (1, 1) = -(1/2)

Equation to normal

(y-1) = (-1/2)*(x-1)

If (h,k) is the center, then we have

(k-1)/(h-1) = -1/2 or 2k - 2 = -h+1 or h= 3-2k ---------------------- (1)

also

(k-1)² + (h-1)² = {R(1)}² = (2.5²)*5 or

h² + k² + 2 -2h -2k = 7.8125, substituting from (1) we ge

9+4k² -12k +2+ k² -2(3-2k) -2k = 31.25 or

5k²-10k - 35.25 = 0 or

k² -2k - 7.05 = 0

Only positive value of k will be considered as we know the center lies in 1st or second quadrant only.