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Let $r(t)$, where $t$ is a parameter $(t ∈R)$, be a position vector such that $r(t)\times \frac{dr(t)}{dt}=0$

I am asked to show that r(t) has a fixed direction. A hint says: Let $r(t) = f(t)\hat e(t)$ where $\hat e(t)$ is a unit vector).

Could someone tell me how to use the hint? May I ask for a proof?

EDIT: I did some searching and found this question Show fixed direction of a position vector But I cannot understand the answer. Could someone please use the hint to explicitly show that? Thanks in advance!

Community
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Y.X.
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  • Did you consider using the Frenet-Serret equations? – Mikhail Katz Mar 14 '17 at 12:26
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    Do you know the vectorial product of two vectors is a vector perpendicular to both original vectors? And thus, if the vect. product of two vectors is zero that means both vectors are parallel or else one of them is zero...and thus we're basically done? – DonAntonio Mar 14 '17 at 12:26
  • @MikhailKatz Sorry I have not learn about that. If it is relevant, could you please tell me how to consider about it? – Y.X. Mar 14 '17 at 12:31
  • @DonAntonio I think it makes sense geometrically, but here I need a proof using the hint. – Y.X. Mar 14 '17 at 12:33

2 Answers2

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If $r(t)=f(t)ê (t)$ then

$$ \frac{d \vec r}{dt} = f'(t)ê (t) + f(t)\frac{dê (t)}{dt} $$

$$ \vec r \times \frac{d \vec r}{dt} = \vec r \times f'(t)ê (t) + \vec r \times f(t)\frac{dê (t)}{dt} = \vec r \times f(t)\frac{dê (t)}{dt} = 0$$

Thus $\frac{dê (t)}{dt} = 0$ or $\vec r$ is parallel to $\frac{dê (t)}{dt}$

But $\vec r$ is perpendicular to $\frac{dê (t)}{dt}$ because of the property - "A vector of constant magnitude is perpendicular to it's derivative."

Therefore, $\frac{dê (t)}{dt} = 0$ i.e. there is no change in the direction vector.

yathish
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Tim The Enchanter
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  • May I ask how can we formally conclude that $\overrightarrow r\times f'(t)\hat e(t)=0$? – Y.X. Mar 14 '17 at 12:39
  • Why is $$ \vec r \times f'(t)ê (t) + \vec r \times f(t)\frac{dê (t)}{dt} = \vec r \times f(t)\frac{dê (t)}{dt}$$ true? Meaning, why $;r\times f'\hat e=0;$ ? – DonAntonio Mar 14 '17 at 12:40
  • @Y.X. $\vec r$ and $\hat e(t)$ are parallel from the assumption $\vec r = f(t) \hat e(t)$. Thus their cross product is zero. – Tim The Enchanter Mar 14 '17 at 12:42
  • @TimTheEnchanter But may I ask why can we assume it? That is, is that true that for all vector-valued function, we are able to express it in this form? How can I see this? – Y.X. Mar 14 '17 at 12:50
  • @Y.X. We can represent any vector in terms of it's magnitude and direction. All the expression $\vec r = f(t) \hat e(t)$ does is say that we have a time dependant vector $\vec r$ (denoting t by time) , whose magnitude is time dependant as $f(t)$ and its direction is time dependant as $\hat e(t)$ – Tim The Enchanter Mar 14 '17 at 12:54
  • @Y.X. It looks like it is going to be a little hard to avoid some basic geometry completely , as you seem to want. – DonAntonio Mar 14 '17 at 12:54
  • @Y.X. If you want to avoid geometry completely, you can call $\vec r$ an n-tuple and you can say the statement $\vec r = f(t) \hat e(t)$ is just splitting the n-tuple into a scalar f(t) and a normalised n-tuple $\hat e(t)$. Its a way of looking at it, just not a particularly appealing one. – Tim The Enchanter Mar 14 '17 at 12:58
  • Thanks a lot, I think I get it. – Y.X. Mar 14 '17 at 13:00
  • May I ask why we have$\frac{d\hat e(t)}{dt}=0$ instead of $r$ is parallel to $\frac{d\hat e(t)}{dt}$? – Y.X. Mar 16 '17 at 01:17
  • $\frac{d\hat e(t)}{dt}=0$ represents the tangent vector, and is hence always perpendicular to $\hat e(t)$ and hence $\vec r$ – Tim The Enchanter Mar 16 '17 at 01:30
  • Do you mean that it is the normal vector? Since I think the tangent vector is always parallel to the orginal vetor instead of perpendicular. – Y.X. Mar 16 '17 at 01:41
  • @Y.X. No $\frac{d\hat e(t)}{dt}$ is the tangent vector and the normal vector is $\hat e(t)$. Think of it this way. The vector $\hat e(t)$ occupies the unit sphere centred at origin. Clearly then the normal is actually $\hat e(t)$ and the tangent vector $\frac{d\hat e(t)}{dt}$ is perpendicular to $\hat e(t)$ ( from the properties of a sphere). – Tim The Enchanter Mar 16 '17 at 04:08
  • @PropositionX I've edited the answer to explain why $\frac{dê (t)}{dt} = 0$ instead of $\vec r$ is parallel to $\frac{dê (t)}{dt}$. No need to be confused thinking about geometrical interpretation in terms of tangent or normal.. – yathish Jul 22 '18 at 08:50
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This is not true. A counterexample:

$$ r(t) = \begin{cases} (0,t^2) & \text{for } t\le 0 \\ (t^2,0) & \text{for } t> 0 \end{cases} $$

If you have the additional assumption that $r(t)\ne 0$, then your property holds, and something like Tim the Enchanter's proof should work.

hmakholm left over Monica
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