Find the curvature and radius of curvature.
y = 2x^2 + 3, x = -1
r (t) = < t, 2t^2 + 3, 0 >
r'(t) = < 1, 4t, 0 >
The magnitude of r'(t) = root {16t^2 + 1}.
Is this correct so far?
I suspect that the "K" in the title was supposed to be "kappa", $\displaystyle \kappa$, the curvature, and R the radius of curvature. The first, and easiest, step of one method of finding those would be to find the length of the tangent vector.
If a curve is given by a vector function, $\displaystyle \vec{r}(t)= x(t)\vec{i}+ y(t)\vec{j}+ z(t)\vec{k}$, then the tangent vector at any point is the derivative: $\displaystyle \vec{T}(t)= x'(t)\vec{i}+ y'(t)\vec{j}+ z'(t)\vec{k}$. If, in particular, the parameter, t, is the arc-length, s, then that tangent vector will have length 1 and is the "unit tangent vector". If t is not the arc-length then we still have relation $\displaystyle s= \int |T|dt$.
With arc-length as parameter, so that the tangent vector has unit length, the derivative of the tangent vector is perpendicular to the tangent vector (Since $\displaystyle |\vec{T|}= \sqrt{\vec{T}\cdot \vec{T}}= 1$, $\displaystyle \vec{T}\cdot \vec{T}= 1$, a constant, so $\displaystyle \frac{d\vec{T}\cdot \vec{T}}{ds}= 2\vec{T}\cdot\frac{d\vec{T}}{ds}= 0$) and is called the "normal vector" to the curve at that point. The "curvature" is the length of that normal vector. The "radius of curvature" is 1 over the curvature.
Where t is not the arc-length, we have, by the chain rule, $\displaystyle \frac{d\vec{T}}{ds}= \frac{d\vec{T}}{dt}\frac{dt}{ds}$. And, since $\displaystyle s= \int \left| \frac{d\vec{T}}{dt}\right| dt$, $\displaystyle \frac{ds}{dt}= |\vec{T}|$, $\displaystyle \frac{d\vec{T}}{ds}= \frac{d\vec{T}}{dt}\frac{ds}{dt}$.