Chapter 14.2, Problem 67E

To determine

$r(t0)$ is orthogonal to $r\text{'}(t0)$, if $\parallel r(t)\parallel$ takes a local minimum or maximum value at $t0$.

To determine

To prove: $r(t0)$.$r\text{'}(t0)=0$

Proof:

$\parallel r(t)\parallel$ has a local maximum or minimum value at $t=t0$. Therefore $\parallel r\left(t\right){\parallel }^2$ has also a local maximum or minimum value at $t=t0$. Hence the derivative of $\parallel r\left(t\right){\parallel }^2$ at $t=t0$ is 0.

$\frac{d}{dt}\parallel r\left(t0\right){\parallel }^2=\frac{d}{dt}r(t0).r(t0)$= 0

Using dot product rule

$\frac{d}{dt}r(t0).r(t0)=r(t0).r\text{'}(t0)+r\text{'}(t0).r(t0)=0$

Therefore, $r(t0)$.$r\text{'}(t0)=0$. Hence $r(t0)$ is orthogonal to $r\text{'}(t0)$.

In the figure the minimum is at $\parallel r(t0)\parallel$ the path intersect the sphere of radius $\parallel r(t0)\parallel$ at a point. Therefore $r\text{'}(t0)$ is tangent to the sphere at $t0$.