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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 $(\| \mathbf{r} \|^2)' = 0$?

Lemon
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  • What is this prime supposed to represent? Differentiation with respect to...what? – Muphrid Feb 03 '13 at 20:13
  • Any parameter. I left it out because I thought it may create confusion. – Lemon Feb 03 '13 at 20:15
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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 03 '13 at 20:16
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    I suspect the word "Then" should be replaced with, say, "Given"? – Muphrid Feb 03 '13 at 20:18
  • @ChristianBlatter, are there times when it isn't true? What if I add the fact that $\mathbf{r}$ has constant speed? That is $| \mathbf{r}' | = \text{constant}$ – Lemon Feb 03 '13 at 20:22
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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 03 '13 at 20:35
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    Okay clearly I've mispoken. I will make the corrections – Lemon Feb 03 '13 at 21:46

3 Answers3

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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 $t$.

Julien
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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. – Lemon Feb 03 '13 at 20:11
  • Okay i can use MVT or integration, but what was wrong with my argument? – Lemon Feb 03 '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... – Julien Feb 03 '13 at 20:17
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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 $[a,b]$.

user7530
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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 – Lemon Feb 03 '13 at 20:23
  • Well, notice that in your case, $f(t) = |r(t)|$ is a scalar function. – user7530 Feb 03 '13 at 20:28
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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.

ncmathsadist
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