Curvature — Question 4

Question 4 – Solution

Strategy. The center lies one radius of curvature from the point in the principal-normal direction.

Step 1: Curvature For $y=x^2$, $$y'=2x,\qquad y''=2,$$ so $$\kappa(0)=\frac{|2|}{(1+0)^{3/2}}=2, \qquad \rho(0)=\frac 12.$$ At the vertex the curve bends upward, so $\mathbf N=\left\langle 0,1\right\rangle$.

Step 2: Center With $P=(0,0)$, $$C=P+\rho\mathbf N=(0,0)+\frac 12(0,1) =\boxed{\left(0,\frac 12\right)}.$$

Step 3: Circle equation Radius $1/2$ and center $(0,1/2)$ give $$\boxed{x^2+\left(y-\frac 12\right)^2=\frac 14}.$$ The circle contains $(0,0)$. Implicit differentiation gives $$2x+2\left(y-\frac 12\right)y'=0,$$ so at $(0,0)$, $y'=0$, matching the parabola’s horizontal tangent. The common curvature at contact gives second-order agreement.