Length of curve in Polar coordinate system

[ltd5241]

I want to caculate length of curve in Polar coordinate system like this: if r=r(a)
then length of the curve is ∫r(a)da Is this right? if not ,why ?
What's the right one ?
I konw the way in rectangular coordinate system,I just want to do it in Polar coordinate system .

 

[jeffreydk]

You can think of it like an infinitesimal form of the Euclidean distance formula. For a function $f(t)=\langle x_1(t),x_2(t),x_3(t),\ldots\rangle$

$$\sum_a^b \sqrt { \Delta x_1^2 + \Delta x_2^2+\Delta x_3^2+\ldots } \longrightarrow s=\int_{a}^{b} \sqrt { dx_1^2 + dx_2^2+dx_3^2+\ldots} = \int_{a}^{b} \sqrt { \left(\frac{dx_1}{dt}\right)^2 + \left(\frac{dx_2}{dt}\right)^2+\left(\frac{dx_3}{dt}\right)^2+\ldots}\text{ } dt$$

 

[ltd5241]

Is this ∫r(a)da wrong? Why?

 

[arildno]

You need to know what the appropriate infinitesemal length segments are!

Now, if you do polar coordinates, you may decompose a stretch of a curve into two parts:
1. The change in the radial position from the initial point on the curve to the final point.
Infinitesemally, this has length dr.

2. Here's the tricky part: The tiny arc by which the curve segment can be approximated by a circular arc, supported by a tiny angular change between the first point and the final point on the curve.
Clearly, that circular arc lies AT a radius of the value "r", and setting the angular change as [itex]d\theta[/tex], we get the expression [itex]rd\theta[/tex] for that length segment.<br /> <br /> 3. Now, we apply the Pythogorean theorem to these two length segment to gain the proper curve segment ds:<br /> $$ds=\sqrt{(dr)^{2}+(rd\theta)^{2}}$$<br /> <br /> 4. Assuming that the radial position of the point of the curve is describable as a function of the angular variable, we may rewrite this as:<br /> $$ds=\sqrt{(\frac{dr}{d\theta})^{2}+r^{2}}d\theta$$<br /> <br /> 5. This is then the proper infinitesemal form of the lengthn segment, and the length s of the curve can then be calculated as:<br /> $$s=\int_{\theta_{0}}^{\theta_{1}}\sqrt{(\frac{dr}{d\theta})^{2}+r^{2}}d\theta$$<br /> <br /> 6. Note that in the case of a CIRCLE, where r is a constant function of the angle, this reduces to:<br /> $$s=\int_{\theta_{0}}^{\theta_{1}}rd\theta$$<br /> as it should do.[/itex][/itex]

 

[ltd5241]

Thank you ! Especially you,Arildno.