Curvature — Question 4

Question 4 – Solution

Strategy Use the parameter-independent cross-product formula.

Derivatives $$\vec r'=\left\langle 1,2t,3t^2\right\rangle,\qquad \vec r''=\left\langle 0,2,6t\right\rangle,$$ $$\vec r'\times\vec r''=\left\langle 6t^2,-6t,2\right\rangle.$$ At $t=1$, its magnitude is $2\sqrt{19}$, while $\|\vec r'(1)\|=\sqrt{14}$. Therefore $$\boxed{\kappa(1)=\frac{2\sqrt{19}}{(\sqrt{14})^3} =\frac{\sqrt{19}}{7\sqrt{14}}}.$$

Check The cross product is nonzero, so the tangent is genuinely changing direction and the curvature is positive.