Question

Find center of curvature with Normal=((1+0.5t²)⁻¹(-t, 1-0.5t², t)) and r=((t,0.5t², t³/6)) and Kappa=((1+0.5t²)⁻²)

a) (t, 0.5t², t³/6)
b) (-t, 1-0.5t², t)
c) (1-0.5t², t, -t)
d) (t³/6, t, 0.5t²)

Answer 1

The center of curvature is found by adding the reciprocal of the curvature (1/Kappa) times the normal vector to the position vector r. However, none of the provided options match the resulting computation for the given normal vector and curvature.

Step-by-step explanation:

The student is asking to find the center of curvature given a normal vector Normal and position vector r, as well as the curvature Kappa. Since the normal vector points towards the center of curvature, the center of curvature C can be found using the formula:

C = r + (1/Kappa) * Normal

For a parametric curve, the radius of curvature R is the reciprocal of the curvature Kappa, R = 1/Kappa. Given that the normal vector is normalized, the center of curvature is at a distance R = 1/Kappa from the point on the curve along the direction of the normal.

So, by substituting the given values for the normal vector and Kappa, we find that:

C = r(t) + ((1 + 0.5t²)²) * Normal(t)

Which simplifies to:

C = (t, 0.5t², t³/6) + (1, -t, 1-0.5t²).

Therefore, the correct answer is none of the provided options, as they do not match the resulting expression after the computation.