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Consider an arbitrary vector real valued function smooth and continuous $\mathbf{r}$ and $\mathbf{r} \cdot \mathbf{r} = \| \mathbf{r} \|^2$

Given $ (\| \mathbf{r} \|^2)' = (\mathbf{r} \cdot \mathbf{r})' = \mathbf{r}' \cdot \mathbf{r} + \mathbf{r} \cdot \mathbf{r}' = 2\mathbf{r} \cdot \mathbf{r}' = 0$

Does it follow that $\|\mathbf{r}\|$ is a constant because $(\| \mathbf{r} \|^2)' = 0$?

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What is this prime supposed to represent? Differentiation with respect to...what? –  Muphrid Feb 3 '13 at 20:13
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What does the "Then" mean? When the function $t\mapsto{\bf r}(t)$ is "arbitrary" then you cannot expect that $2{\bf r}\cdot{\bf r}'\equiv0$. –  Christian Blatter Feb 3 '13 at 20:16
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I suspect the word "Then" should be replaced with, say, "Given"? –  Muphrid Feb 3 '13 at 20:18
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Assume ${\bf a}\ne{\bf 0}$ and put ${\bf r}(t):=t\>{\bf a}$. Then ${\bf r}'(t)\equiv{\bf a}$ and $2{\bf r}\cdot{\bf r}'=2 t\|{\bf a}\|^2\ne0$ $\ (t\ne0)$. –  Christian Blatter Feb 3 '13 at 20:35
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Okay clearly I've mispoken. I will make the corrections –  Hawk Feb 3 '13 at 21:46

3 Answers 3

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Use $$ f(t)=f(0)+\int_0^tf'(s)ds $$ with $f(s)=\|r(s)\|^2$.

So yes, $\|r\|$ is constant if and only if $(\|r\|^2)'=0$.

If you do not want to integrate, use the mean value theorem: $$ f(t)-f(0)=f'(s)(t-0)=f'(s)t $$ for some $s$ between $0$ and $t$.

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is there a proof where we can avoid integration? I want to do it soley in terms of vector products and derivatives. –  Hawk Feb 3 '13 at 20:11
    
Okay i can use MVT or integration, but what was wrong with my argument? –  Hawk Feb 3 '13 at 20:15
    
Nothing is wrong with your argument. I just recalled why a smooth function is constant (on an interval) if and only if its derivative is zero... –  1015 Feb 3 '13 at 20:17

Your argument is correct. The curve you depict is drawn in sphere. And since any radius and tangent of a sphere are perpendicular, the result you see with the dot product is no surprise.

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EDIT: I may have misread the question... if $\frac{d}{dt}\|r\|^2=0$ it follows that $\|r\|$ is constant. For arbitrary curves $r$ the former is not necessarily the case.

Yes. Let $f(t)$ be any differentiable scalar function with $\frac{d}{dt}(f^2) = 0$ on an interval $[a,b]$. By the chain rule, $2ff' = 0$, and $f$ is constant on every connected component where $f \neq 0$. It follows that $f$ must be constant on all of $[a,b]$.

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That's what I thought as well, but I wasn't sure if the same logic for scalar function will work on vector valued ones. Though to my previous knowledge, everything works exactly the same –  Hawk Feb 3 '13 at 20:23
    
Well, notice that in your case, $f(t) = \|r(t)\|$ is a scalar function. –  user7530 Feb 3 '13 at 20:28

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