Curvature — Question 2

Question 2

Find the curvature of the parabola $\vec r(t)=\left\langle t,t^2\right\rangle$ for general $t$, and evaluate it at the point $(1,1)$. Where is the curvature largest?

Question 2 – Solution

Strategy Apply the planar curvature formula and then analyze its denominator.

Calculation With $x'=1$, $y'=2t$, $x''=0$, and $y''=2$, $$\kappa(t)=\frac{|x'y''-y'x''|}{(x'^2+y'^2)^{3/2}} =\boxed{\frac{2}{(1+4t^2)^{3/2}}}.$$ At $t=1$, $\boxed{\kappa=2/(5\sqrt 5)}$.

Maximum The numerator is constant and the denominator is minimized at $t=0$. Hence the maximum curvature is $\kappa(0)=2$, at the vertex.