Curvature — Question 5

Question 5 – Solution

Strategy. Apply the cross-product formula twice and simplify all powers before comparing.

Step 1: First parametrization $$\mathbf r'=\left\langle 1,2t,0\right\rangle,\qquad \mathbf r''=\left\langle 0,2,0\right\rangle.$$ Thus $$\boxed{\kappa_r(t)=\frac 2{(1+4t^2)^{3/2}}}.$$

Step 2: Second parametrization $$\mathbf R'=\left\langle 3u^2,6u^5,0\right\rangle,\qquad \mathbf R''=\left\langle 6u,30u^4,0\right\rangle.$$ The cross product has $k$-component $$(3u^2)(30u^4)-(6u^5)(6u)=54u^6.$$ Also $$\|\mathbf R'\|=3u^2\sqrt{1+4u^6}$$ because $u>0$. Hence $$\kappa_R(u)=\frac{54u^6}{27u^6(1+4u^6)^{3/2}} =\boxed{\frac 2{(1+4u^6)^{3/2}}}.$$

Step 3: Compare Substituting $t=u^3$ into $\kappa_r$ gives exactly $\kappa_R$. Curvature depends on the geometric curve, not its regular parametrization. The restriction $u>0$ ensures $\mathbf R'\ne\mathbf 0$ and selects the positive-parameter branch. The speed simplification uses $u^2\ge 0$, which also holds for negative $u$.