Curvature — Question 8

Question 8 – Solution

Strategy. Use the planar graph formula and maximize the even function by working with $u=t^2$.

Step 1: Curvature $$y'=3t^2,\qquad y''=6t,$$ so $$\boxed{\kappa(t)=\frac{6|t|}{(1+9t^4)^{3/2}}}.$$ At $t=0$, $\mathbf r'(0)=\left\langle 1,0,0\right\rangle\ne\mathbf 0$, so the curve is regular, but $y''(0)=0$ and therefore $\kappa(0)=0$. This is an inflection point.

Step 2: Maximize for $t>0$ There $|t|=t$. Logarithmic differentiation gives $$\frac{\kappa'}{\kappa}=\frac 1t-\frac{54t^3}{1+9t^4}.$$ Setting this equal to zero, $$1+9t^4=54t^4 \quad\Longrightarrow\quad t^4=\frac 1{45}.$$ Thus $t=45^{-1/4}$ on the positive side. By even symmetry, the other maximizer is its negative.

Step 3: Points With $\alpha=45^{-1/4}$, the two curve points are $$\boxed{(\alpha,\alpha^3,0)\quad\text{and}\quad(-\alpha,-\alpha^3,0)}.$$ Curvature tends to zero as $|t|\to\infty$ and is zero at $t=0$, so these critical points are global maxima.