Curvature — Question 2

Question 2 – Solution

Strategy. Compute the speed and cross product in symbolic form before simplifying.

Step 1: Derivatives and speed $$\mathbf r'=\left\langle -a\sin t,a\cos t,b\right\rangle,\qquad \mathbf r''=\left\langle -a\cos t,-a\sin t,0\right\rangle,$$ and $$\|\mathbf r'\|=\sqrt{a^2+b^2}.$$

Step 2: Cross product $$\mathbf r'\times\mathbf r'' =\left\langle ab\sin t,-ab\cos t,a^2\right\rangle.$$ Its magnitude is $$\sqrt{a^2b^2(\sin^2t+\cos^2t)+a^4} =a\sqrt{a^2+b^2}.$$ Thus $$\boxed{\kappa(t)=\frac{a\sqrt{a^2+b^2}}{(a^2+b^2)^{3/2}} =\frac{a}{a^2+b^2}}.$$

Step 3: Interpretation No $t$ remains, so curvature is constant. When $b=0$, the helix becomes a circle of radius $a$ and $\kappa=1/a$. As $|b|$ grows, $a/(a^2+b^2)$ decreases: the helix stretches vertically and bends less sharply.