# Curvature of curve: equivalence between tangent vector and angle definitions

I know that curvature for some curve $$C$$ defined parametrically is:

$$\kappa=\left\|{d\vec{T}\over ds}\right\|$$

Which basically is the rate at which the tangent vector to the curve changes, as the arclength of the curve changes.

In another source, I saw the definition of curvature as the following:

If $$P_1$$ and $$P_2$$ are two points on the curve, $$|P_1P_2|$$ is the arclength between those two points, and $$\Phi$$ is the limit of the angle between tangent vectors at the points $$P_1$$ and $$P_2$$ (as it goes to zero I assume), then the curvature is defined as:

$$\kappa=\lim_{|P_1P_2|\to 0}{\Phi\over |P_1P_2|}$$

Which basically means, the rate at which the angle of tangent vectors in global frame of reference changes, as the arclength of the curve changes.

I assume that this second definition can be rewritten using the notation from the first example as:

$$\kappa={d\phi\over ds}$$

Where $$\phi$$ is the angle between the vector tangent to the curve, and some constant global axis of reference (which could be the x axis, but realy it could be any line or vector on the same plane).

Given the second (weird in my opinion) definition of curvature, I can't see how those two definitions can be equivalent. Maybe they are not, I don't know. May be they are; if yes, how?

Also, here's a picture of the section from the book where the second definition appears in (it's not in English):

Note that the text agrees with yet another definition of curvature, which I am aware of: $$\kappa=\frac1r$$, where $$r$$ is radius of curvature.

• Note that all of these definitions except $\frac{d\phi}{ds}$ work just as well in three dimensions as in two dimensions. To use $\frac{d\phi}{ds}$ in three dimensions you need to find the plane in which the tangent and normal vectors lie, and then you can apply that definition too. – David K Dec 27 '18 at 12:45
• @DavidK thank you for the info, didn't even think about how that formula can be applied to three dimensions – KKZiomek Dec 27 '18 at 14:46

## 2 Answers

Let's work with the first definition. We have \begin{align} \kappa (s) &= \left\| \frac{d\vec T}{ds}(s)\right\| \\ &= \lim_{\delta s\to 0}\frac 1 {\delta s}\left\| \vec T(s + \delta s) - \vec T(s) \right\| \\ &= \lim_{\delta s \to 0} \frac 1 {\delta s} \sqrt{(\vec T(s + \delta s) - \vec T(s)).(\vec T(s + \delta s) - \vec T(s))} \\ &= \lim_{\delta s \to 0} \frac 1 {\delta s} \sqrt{\| \vec T(s + \delta s)\|^2 + \| \vec T(s) \|^2 - 2\vec T(s).\vec T(s + \delta s)}\end{align} But the curve is parameterised by arc length! So $$\| \vec T(s + \delta s)\|^2 = \| \vec T(s)\|^2 = 1$$ and $$\vec T(s).\vec T(s + \delta s) = \cos \Phi(s, s + \delta s),$$ where $$\Phi(s, s + \delta s)$$ is the angle between $$\vec T(s)$$ and $$\vec T(s + \delta s)$$.

Plugging this in, we get \begin{align} \kappa &= \lim_{\delta s \to 0} \frac 1 {\delta s} \sqrt{2 - 2 \cos \Phi(s, s + \delta s)} \\ &= \lim_{\delta s \to 0} \frac 1 {\delta s} 2 \sin \left( \frac { \Phi(s, s + \delta s) } {2}\right) \\ &= \lim_{\delta s \to 0} \frac {\Phi(s, s + \delta s)} {\delta s} \times \frac{\sin \left( \frac { \Phi(s, s + \delta s) } {2}\right)}{\frac{\Phi(s, s + \delta s)}{2}} \end{align} Clearly, $$\lim_{\delta s \to 0} \Phi(s, s + \delta s) = 0$$, so \begin{align} \kappa &= \lim_{\delta s \to 0} \frac {\Phi(s, s + \delta s)} {\delta s} \times \lim_{\Phi \to 0} \frac{\sin \left( \frac { \Phi } {2}\right)}{\frac{\Phi}{2}} \\ &= \lim_{\delta s \to 0} \frac {\Phi(s, s + \delta s)} {\delta s} \times 1 \\ &= \lim_{\delta s \to 0} \frac {\Phi(s, s + \delta s)} {\delta s}\end{align} which agrees with the second definition.

• How do you get from $\sqrt{2-2\cos\Phi}$ to $2\sin\left(\frac{\Phi}{2}\right)$? – Noble Mushtak Dec 27 '18 at 0:36
• @NobleMushtak I used $1 - \cos x = 2 \sin^2 (\tfrac x 2)$. – Kenny Wong Dec 27 '18 at 0:37

By your first statement, $$T'(s)=\kappa(s)N(s)$$ where $$\kappa$$ is the curvature and $$N$$ is the unit normal vector. Now, let's consider $$T(s)$$ and $$T(s+\Delta s)$$, so that we can compare the angles between tangent vectors. Now, $$T(s)$$, $$T(s+\Delta s)$$, and $$N(s)$$ are all unit vectors, so we can draw the following picture representing all of these vectors: Here, $$\Delta \theta$$ is the angle between the two tangent vectors and $$\Delta T=T(s+\Delta s)-T(s)$$. From the diagram, we can find that $$\Delta T\cdot T(s)=1-\cos \Delta \theta$$ and $$\Delta T\cdot N(s)=\sin d\theta$$. From the x-component, we get the following equation:

$$\frac{dT}{d\theta}\cdot T(s)=\lim_{\Delta \theta\rightarrow 0}\frac{\Delta T\cdot T(s)}{\Delta \theta}=\lim_{\Delta \theta\rightarrow 0}\frac{1-\cos(\Delta \theta)}{\Delta \theta}=0$$

This was pretty obvious since $$T'(s)=\kappa(s)N(s)$$ is orthogonal to $$T(s)$$, so nothing new there. However, from the y-component, we get:

$$\frac{dT}{d\theta}\cdot N(s)=\lim_{\Delta \theta\rightarrow 0}\frac{\Delta T\cdot N(s)}{\Delta \theta}=\lim_{\Delta \theta\rightarrow 0}\frac{\sin(\Delta \theta)}{\Delta \theta}=1$$

Now, let's use chain rule to figure out what $$\frac{dT}{d\theta}$$ is:

$$\frac{dT}{ds}=\kappa(s)N(s)=\frac{dT}{d\theta}\frac{d\theta}{ds} \rightarrow \frac{dT}{d\theta}=\frac{\kappa(s)N(s)}{\frac{d\theta}{ds}}$$

Finally, let's plug this value for $$\frac{dT}{d\theta}$$ into the equation with the dot product:

$$\frac{dT}{d\theta}\cdot N(s)=1\rightarrow \frac{\kappa(s)N(s)\cdot N(s)}{\frac{d\theta}{ds}}=1\rightarrow \frac{d\theta}{ds}=\kappa(s)N(s)\cdot N(s)=\kappa(s)$$

(Note that the last step uses $$N(s)\cdot N(s)=1$$ since $$N(s)$$ is a unit vector.)

At this point, we have shown $$\frac{d\theta}{ds}=\kappa(s)$$, which is what we originally set out to prove. Q.E.D.

Of course, this isn't exactly a formal proof since it assumes $$T(s+\Delta s)$$ is in the plane spanned by $$T(s)$$ and $$N(s)$$, which isn't necessarily true. However, I think this is a fair approximation since $$T'(s)=\kappa(s)N(s)$$, so $$T(s+\Delta s)\approx T(s)+\Delta sT'(s)=T(s)+[\Delta s\kappa(s)]N(s)$$. In any case, I feel like this geometric argument gives a better visual intuition for why $$|T'(s)|=\frac{d\theta}{ds}$$.

• Thank you for the answer. I accepted the other answer only because it was more straightforward, at least to me – KKZiomek Dec 27 '18 at 1:57