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Let $C$ be a plane curve parametrized by arc length by $\alpha(s)$, $T(s)$ (unit tangent vector) and $N(s)$ (unit normal vector). Prove that $$\frac{d}{ds} N(s)=-\kappa(s)T(s).$$

I know that since $C$ is a curve parametrized by arc length then the tangent vector $\alpha'(s)$ has unit length which equals $T(s) = \frac{\alpha'(s)}{||\alpha'(s)||}$, thus $N(s) = \frac{T'(s)}{||T'(s)||}$.

How can I prove this? .

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    $\begingroup$ Check your definitions! $\endgroup$
    – Ted Shifrin
    Sep 2, 2013 at 0:50
  • $\begingroup$ @TedShifrin Which definition do I need checking? $\endgroup$
    – Lays
    Sep 2, 2013 at 0:54
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    $\begingroup$ The definition of curvature. $\endgroup$
    – Ted Shifrin
    Sep 2, 2013 at 0:55
  • $\begingroup$ The way we defined it is. "The magnitude of rate of change of the unit tangent vector $\alpha$ with respect to the arc length $s$ along the curve". $\endgroup$
    – Lays
    Sep 2, 2013 at 1:00
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    $\begingroup$ It's fine. It was just missing when you wrote the question. You need all your definitions to attack a problem like this. $\endgroup$
    – Ted Shifrin
    Sep 2, 2013 at 1:47

2 Answers 2

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Since $N(s)$ and $T(s)$ are normal for all $s$, we have

$\langle T, N \rangle = 0$

for all $s$. Thus

$0 = \langle T, N \rangle' = \langle T', N \rangle + \langle T, N' \rangle$,

whence

$\langle T', N \rangle = -\langle T, N' \rangle$;

by definition, $T' = \kappa N$; therefore

$\langle T, N' \rangle = -\kappa$.

Now since $\langle N, N' \rangle = 0$ we must have

$N' = -\kappa T$,

since $T$ and $N$ form an orthorormal basis for the tangent space at any point they are defined. QED!

Thirteen minutes single-finger 'droid typing!

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    $\begingroup$ Thanks Robert! How come $\langle T, N \rangle' = \langle T', N \rangle + \langle T, N' \rangle$? And how did you get that, $\langle T, N' \rangle = -\kappa$ from $T' = \kappa N$? $\endgroup$
    – Lays
    Sep 2, 2013 at 1:46
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    $\begingroup$ Actually I see now. $\langle T, N \rangle' = \langle T', N \rangle + \langle T, N' \rangle$ is just the product rule. But I still dont understand how you got $\langle T, N' \rangle = -\kappa$ from $T' = \kappa N$? $\endgroup$
    – Lays
    Sep 2, 2013 at 1:57
  • $\begingroup$ @Lays: sorry about the delay. Use $\langle T, N \rangle' = 0$, since $\langle T, N \rangle = 0$. Then $\langle T', N \rangle = - \langle T, N' \rangle$; now plug in $T'= \kappa N$, to get $\langle T, N' \rangle = -\kappa \langle N, N \rangle = \kappa$, since $N$ is a unit vector. Cheers. $\endgroup$
    – Robert Lewis
    Sep 2, 2013 at 7:03
  • $\begingroup$ I see it now, thanks! From $\langle T, N' \rangle = -\kappa$ could you just plug in a $T$ instead of using $\langle N, N' \rangle = 0$ because I didnt understand that part too? $\endgroup$
    – Lays
    Sep 2, 2013 at 7:14
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    $\begingroup$ @Lays: since $N$ and $N'$ are orthogonal, $N'$ must be collinear with $T$, and the component is $-\kappa$! $\endgroup$
    – Robert Lewis
    Sep 2, 2013 at 7:40
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Let's express $\frac{d\mathbf{N}}{ds}$ as a linear combination of the tangent, normal, and binormal vectors $$\frac{d\mathbf{N}}{ds}=a(s)\mathbf{T}+b(s)\mathbf{N}+c(s)\mathbf{B}.$$ These are all unit vectors perpendicular to each other, and the scalar functions $a,b,c$ are arbitrary scalar functions. So if we take the dot product with respect to the tangent vector $\mathbf{T}$ $$\frac{d\mathbf{N}}{ds}\cdot\mathbf{T}=a(s)\mathbf{T}\cdot\mathbf{T}+b(s)\mathbf{N}\cdot\mathbf{T}+c(s)\mathbf{B}\cdot\mathbf{T}=a(s)+0+0=a(s)$$ Now we have that $$a(s)=\frac{d\mathbf{N}}{ds}\cdot\mathbf{T}=-\mathbf{N}\cdot\frac{d\mathbf{T}}{ds}=\mathbf{N}\cdot\kappa\mathbf{N}=-\kappa$$

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  • $\begingroup$ The OP is doing plane curves only. You should have $d/ds$, not partials. And I'm going to be picky and complain that if you throw in the notation of $a(s)$, etc., you should be consistent and put $\mathbf T(s)$, etc. and your last line has a confusing error/typo. $\endgroup$
    – Ted Shifrin
    Sep 2, 2013 at 1:28
  • $\begingroup$ you're right...I'm actually studying the exact same stuff and have been doing partials...will change... Thanks. I've been working with Colley's "Vector Calculus" and this is consistent with the notations she uses, but i understand where you are coming from. $\endgroup$
    – Eleven-Eleven
    Sep 2, 2013 at 1:32
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    $\begingroup$ Cool. :) Enjoy learning. I taught the author of your book when she was a graduate student. :) $\endgroup$
    – Ted Shifrin
    Sep 2, 2013 at 1:40
  • $\begingroup$ Thank you, Christopher! How did you get "$\frac{d\mathbf{N}}{ds}=a(s)\mathbf{T}+b(s)\mathbf{N}+c(s)\mathbf{B}.$"? $\endgroup$
    – Lays
    Sep 2, 2013 at 1:43
  • $\begingroup$ @TedShifrin, that is awesome! By the way, where is my typo? $\endgroup$
    – Eleven-Eleven
    Sep 2, 2013 at 1:44

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