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?
If a curve is given by a vector function, , then the tangent vector at any point is the derivative: . 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 .
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 , , a constant, so ) 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, . And, since , , .