Suppose a particle moves in space with its position vector at time $t$ given by $\vec{r}(t) = f(t)\hat{i}+g(t)\hat{j}+h(t)\hat{k},\,\,$ $a < t < b$. Denote the particle's velocity at time $t$ by $\vec{v}(t)$. Assume that the particle does not pass through the origin. Let $t_0$ be the instant of closest approach to the origin. Show that $\vec{v}(t_0)$ is perpendicular to $\vec{r}(t_0)$.
Is it so easy that $\vec{v}(t)$ at $t_0$ is $0$? Hence $\vec{r}(t_0) \cdot \vec{v}(t_0) = 0$?
Best Answer
The square $S(t)$ of the distance of our point from the origin is given by $$S(t)=(f(t))^2 +(g(t))^2 +(h(t))^2.$$ Differeniate, and set the derivative equal to $0$. We get $$2f(t)f'(t)+2g(t)g'(t)+2h(t)h'(t)=0.$$ That says exactly what we want.
For a "vector" version of the same thing, the square of the distance from $\bar{v}(t)$ to the origin is $\bar{v}(t)\cdot \bar{v}(t)$. Differentiate. We get $2\bar{v}(t)\bar{v}'(t)=0$.