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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 $\vec{r}(t_0) \cdot \vec{v}(t_0) = 0$?

Cameron Williams
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John
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2 Answers2

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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 $2\bar{v}(t)\bar{v}'(t)=0$.

André Nicolas
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  • So if I got this right I'm supposed to create a vector from the origin to the point where t=t0.

    Differeniate, and set the derivative equal to 0. We get 2f(t)f′(t)+2g(t)g′(t)+2h(t)h′(t)=0. Why is the derivative = 0?

     – John Sep 21 '13 at 03:45
  • Remember that at a local min (or max) the derivative is $0$. – André Nicolas Sep 21 '13 at 03:49
  • Why is that point a min or max? Could it not be the closest to the origin without beeing a min or max? Really appreciate your help. – John Sep 21 '13 at 03:56
  • closes to the origin says the sum of the squares (aka $\bar{v}\cdot\bar{v}$) is a min. – André Nicolas Sep 21 '13 at 04:04
  • Sorry man, I don't get it – John Sep 21 '13 at 04:36
  • We are told that at time $t_0$, the distance of the point from the origin has reached a minimum. The distance is the square root of $W(t)=(f(t))^2+(g(t))^2+(h(t))^2$. Minimizing distance is equivalent to minimizing $W(t)$. We have a local (and absolute) min at $t_0$, so $W'(t_0)=0$. Usual beginning calculus max/min procedure. – André Nicolas Sep 21 '13 at 04:43
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It is not true that $v(t_0)=0$, but it is true that $\vec r(t_0)\cdot \vec v(t_0)=0$ If the velocity were not perpendicular to the radius, it would have a component along the radius. In that case, the radius would be increasing or decreasing and we would not be at the minimum.

Ross Millikan
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