0
$\begingroup$

Consider the function $r=f(\theta)$ in polar coordinates. The length of an arc of a circle is just $$S=\theta r$$ Where $r$ is the radius of the circle and $\theta$ is the angle that represents this arc. But since $r=f(\theta)$ and $\theta$ Should approach zero so that we can get the exact value of the arc, So $$\,dS=f(\theta)\,d\theta$$ Integrating from $\theta_1$ to $\theta_2$, we get : $$S=\int_{\theta_1}^{\theta_2} f(\theta)\,d\theta $$ But the actual formula for the length of a curve in polar coordinates is $$\int_{\theta_1}^{\theta_2} \sqrt{f^2(\theta)+f’(\theta)^2}\,d\theta.$$

I know that my approach isn’t rigorous enough, but it’s is still reasonable, so why it is different from the actual formula?

$\endgroup$
5
  • 1
    $\begingroup$ No contradiction ! The polar equation of a circle is $f(\theta)=R$ (constant), therefore $f'(\theta)=0$ disappears in the computation of the integral... $\endgroup$
    – Jean Marie
    Apr 11, 2021 at 22:14
  • $\begingroup$ No, i’ve tried to find a general equation for the length of a curve in polar, not just the length of a circle. The computations of the arc length of a circle were in order to approximate the arc length of a function in polar. @JeanMarie $\endgroup$
    – PNT
    Apr 11, 2021 at 22:20
  • $\begingroup$ I insist : I was right at the heart of your issue: you have infered "your" rule of arc lngth computation because it was working for the circle, but the other formula works also for the circle because terms $f'(\theta)$ is zero... $\endgroup$
    – Jean Marie
    Apr 11, 2021 at 22:26
  • 1
    $\begingroup$ Well, we can the difference between the two formulas just be computing the arc length of an arbitrary function in polar.@JeanMarie $\endgroup$
    – PNT
    Apr 11, 2021 at 22:29
  • 2
    $\begingroup$ You've noticed a polar coordinates version of the staircase paradox. $\endgroup$ Apr 12, 2021 at 17:53

1 Answer 1

1
$\begingroup$

Your equation for $dS$ is incorrect. In polar coordinates the differential length would be $f(\theta) d\theta$ only if $r$ is a constant. If you draw a little diagram you'll see that, in fact, the differential element of length is given by

$$dS=\sqrt{f^2+ \bigg(\frac{df}{d\theta}\bigg)^2}\ d\theta$$

$\endgroup$
2
  • $\begingroup$ I didn’t get it, why r should be a constant ?, $r=f(\theta)$ Then $r$ will change with respect to $theta$, ?? @CyeWaldman $\endgroup$
    – PNT
    Apr 12, 2021 at 22:45
  • $\begingroup$ What I said is that $dS=f(\theta)d\theta$ only when $r$ is constant. Otherwise it is as I indicated above. You can derive this by starting from $(dS/d\theta)^2=(dx/d\theta)^2+(dy/d\theta)^2$ and taking $x=r\cos\theta\ \text{&} \ y=r\sin\theta$. $\endgroup$ Apr 13, 2021 at 15:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .