Curvature — Question 7

Question 7 – Solution

Strategy. Since the curve is regular, curvature is zero exactly when $\mathbf r'$ and $\mathbf r''$ are parallel.

Step 1: Derivatives $$\mathbf r'=\left\langle 1,2t,3t^2\right\rangle,\qquad \mathbf r''=\left\langle 0,2,6t\right\rangle.$$ The first component of $\mathbf r'$ is always $1$, so the curve is regular.

Step 2: Cross product $$\begin{align*} \mathbf r'\times\mathbf r'' &=\left\langle 1,2t,3t^2\right\rangle\times\left\langle 0,2,6t\right\rangle\\ &=\left\langle 12t^2-6t^2,-6t,2\right\rangle\\ &=\left\langle 6t^2,-6t,2\right\rangle. \end{align*}$$ Its squared magnitude is $$36t^4+36t^2+4.$$ Every term is nonnegative and the constant term is $4>0$. Therefore the cross product never vanishes.

Step 3: Conclusion $$\boxed{\kappa(t)>0\text{ for every real }t;\text{ there are no zero-curvature points}.}$$ Geometrically, the tangent direction is changing everywhere. In particular, no point of this twisted cubic is locally straight in the curvature sense.