# Curvature of a space curve?

1. Nov 29, 2007

### scottie_000

So this book I have (Mathematical Methods for Physics and Engineering, Riley, Hobson, Bence) defines curvature as being:

$$\kappa = \left | \frac{d \hat{\textbf{t}}}{d s} \right | = \left | \frac{d^2 \hat{\textbf{r}}}{d s^2} \right |$$

where t hat is the unit tangent to the curve and r hat is the unit vector describing the curve.
Is the second equality correct? Surely you have to normalize the tangents and not the vectors....

2. Nov 29, 2007

### scottie_000

The above doesn't really seem answer my question.
I only want to know if the my formula is correct, and if it is not then why
Besides that reply is beyond me (for now :tongue2: )

3. Nov 30, 2007

### ObsessiveMathsFreak

The tangent is normalised in this equation. Well, that is to say, it is understood to be normalised.

The equation uses the letter "s" to denoted the parameter of the curve function. r=r(s). When used in this way, s is generally understood to be the arc length parameter, i.e. a parameter such that

$$\left | \frac{d \hat{\textbf{t}}}{d s} \right | = 1$$

In other words, if your curve is parameterized in terms of arc length "s", then the tangent to the curve will have unit length.

Hopefully you can see the reason that s is referred to as the arc length parameter. To calculated the length of the curve in this system we use
$$l(s)=\int_0^s \left| \frac{d \hat{\textbf{t}}}{d s} \right| ds = \int_0^s 1 ds = s$$

So the arc length parameter gives the length of the curve, starting from r(0).

If your parameter for the curve wasn't arc length, the correct formula would be:
$$\kappa = \left | \frac{\frac{d \hat{\textbf{r}}}{d t} }{\left|\frac{d \hat{\textbf{r}}}{d t}\right|}\right | = \left| \frac{1}{\left| \frac{d \hat{\textbf{r}}}{d t}}\right|} \left( \frac{d^2 \hat{\textbf{r}}}{d t^2} - \frac{d \hat{\textbf{r}}}{d t} \left(\frac{\frac{d \hat{\textbf{r}}}{d t} \cdot \frac{d^2 \hat{\textbf{r}}}{d t^2}}{\left| \frac{d \hat{\textbf{r}}}{d t}\right|^2} \right) \right) \right|$$

Which I think you'll agree is a lot more complicated than the arc length case.