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? .

  • 1
    $\begingroup$ Check your definitions! $\endgroup$ – Ted Shifrin Sep 2 '13 at 0:50
  • $\begingroup$ @TedShifrin Which definition do I need checking? $\endgroup$ – Lays Sep 2 '13 at 0:54
  • 1
    $\begingroup$ The definition of curvature. $\endgroup$ – Ted Shifrin Sep 2 '13 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 '13 at 1:00
  • 3
    $\begingroup$ 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 '13 at 1:47

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

$\langle T, N \rangle = 0$

for all $s$. Thus

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


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

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!

  • 1
    $\begingroup$ 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 '13 at 1:46
  • 1
    $\begingroup$ 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 '13 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 '13 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 '13 at 7:14
  • 1
    $\begingroup$ @Lays: since $N$ and $N'$ are orthogonal, $N'$ must be collinear with $T$, and the component is $-\kappa$! $\endgroup$ – Robert Lewis Sep 2 '13 at 7:40

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 $$a(s)=\frac{d\mathbf{N}}{ds}\cdot\mathbf{T}=-\mathbf{N}\cdot\frac{d\mathbf{T}}{ds}=\mathbf{N}\cdot\kappa\mathbf{N}=-\kappa$$

  • $\begingroup$ 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 '13 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 '13 at 1:32
  • 1
    $\begingroup$ Cool. :) Enjoy learning. I taught the author of your book when she was a graduate student. :) $\endgroup$ – Ted Shifrin Sep 2 '13 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 '13 at 1:43
  • $\begingroup$ @TedShifrin, that is awesome! By the way, where is my typo? $\endgroup$ – Eleven-Eleven Sep 2 '13 at 1:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.