Curvature — Question 10

Question 10 – Solution

Strategy Let $\theta=t^3$ and use the turning-angle relation $\kappa=\|d\mathbf T/dt\|/(ds/dt)$.

Tangent and rates Since $\theta'=3t^2>0$, $$\mathbf T=\left\langle -\sin\theta,\cos\theta\right\rangle,\qquad \left\|\frac{d\mathbf T}{dt}\right\|=3t^2.$$ The speed is $ds/dt=\|\vec r'\|=2(3t^2)=6t^2$.

Curvature $$\boxed{\kappa=\frac{\|d\mathbf T/dt\|}{ds/dt} =\frac{3t^2}{6t^2}=\frac 12}.$$

Interpretation Curvature measures direction change per unit distance, not per unit time. Both direction-change rate and speed acquire the same parameter factor, which cancels.