Length of Arc in Rectangular Plane

Apply Pythagorean theorem to the triangular strip shown in the figure:
$(ds)^2 = (dx)^2 + (dy)^2$   ←   Equation (1)
 

length-of-arc-xy-plane.jpg

 

Divide both sides of Equation (1) by (dx)2
$\dfrac{(ds)^2}{(dx)^2} = \dfrac{(dx)^2}{(dx)^2} + \dfrac{(dy)^2}{(dx)^2}$

$\left( \dfrac{ds}{dx} \right)^2 = 1 + \left( \dfrac{dy}{dx} \right)^2$

$\dfrac{ds}{dx} = \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2}$

$ds = \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} \, dx$
 

Integrate both sides

$\displaystyle s = \int_{x_1}^{x_2} \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} \, dx$

 

Divide both sides of Equation (1) by (dy)2
$\dfrac{(ds)^2}{(dy)^2} = \dfrac{(dx)^2}{(dy)^2} + \dfrac{(dy)^2}{(dy)^2}$

$\left( \dfrac{ds}{dy} \right)^2 = \left( \dfrac{dx}{dy} \right)^2 + 1$

$\dfrac{ds}{dy} = \sqrt{\left( \dfrac{dx}{dy} \right)^2 + 1}$

$ds = \sqrt{\left( \dfrac{dx}{dy} \right)^2 + 1} ~ dy$

 

Integrate both sides

$\displaystyle s = \int_{y_1}^{y_2} \sqrt{\left( \dfrac{dx}{dy} \right)^2 + 1} ~ dy$

 

Length of Arc in Polar Plane

Recall the relationship between polar and rectangular coordinates:
$x = r \cos \theta$   ←   Equation (1)

$y = r \sin \theta$   ←   Equation (2)
 

length-of-arc-polar-plane-1st.jpg

 

Differentiate both sides of Equations (1) and (2) in terms of θ
$\dfrac{dx}{d\theta} = \dfrac{dr}{d\theta} \cos \theta - r\sin \theta$

$\dfrac{dy}{d\theta} = \dfrac{dr}{d\theta} \sin \theta + r\cos \theta$
 

From the triangular strip shon in the figure:
$(ds)^2 = (dx)^2 + (dy)^2$

$\left( \dfrac{ds}{d\theta} \right)^2 = \left( \dfrac{dx}{d\theta} \right)^2 + \left( \dfrac{dy}{d\theta} \right)^2$

$\left( \dfrac{ds}{d\theta} \right)^2 = \left( \dfrac{dr}{d\theta} \cos \theta - r\sin \theta \right)^2 + \left( \dfrac{dr}{d\theta} \sin \theta + r\cos \theta \right)^2$

$\left( \dfrac{ds}{d\theta} \right)^2 = \left[ \left( \dfrac{dr}{d\theta} \right)^2 \cos^2 \theta - 2r \sin \theta \cos \theta \dfrac{dr}{d\theta} + r^2 \sin^2 \theta \right] \\ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ + \left[ \left( \dfrac{dr}{d\theta} \right)^2 \sin^2 \theta + 2r \sin \theta \cos \theta \dfrac{dr}{d\theta} + r^2 \cos^2 \theta \right]$

$\left( \dfrac{ds}{d\theta} \right)^2 = \left( \dfrac{dr}{d\theta} \right)^2 \cos^2 \theta + r^2 \sin^2 \theta + \left( \dfrac{dr}{d\theta} \right)^2 \sin^2 \theta + r^2 \cos^2 \theta$

$\left( \dfrac{ds}{d\theta} \right)^2 = \left( r^2 \sin^2 \theta + r^2 \cos^2 \theta \right) + \left[ \left( \dfrac{dr}{d\theta} \right)^2 \sin^2 \theta + \left( \dfrac{dr}{d\theta} \right)^2 \cos^2 \theta \right]$

$\left( \dfrac{ds}{d\theta} \right)^2 = r^2 \left( \sin^2 \theta + \cos^2 \theta \right) + \left( \dfrac{dr}{d\theta} \right)^2 \left( \sin^2 \theta + \cos^2 \theta \right)$

$\left( \dfrac{ds}{d\theta} \right)^2 = r^2 + \left( \dfrac{dr}{d\theta} \right)^2$

$\dfrac{ds}{d\theta} = \sqrt{r^2 + \left( \dfrac{dr}{d\theta} \right)^2}$

$ds = \sqrt{r^2 + \left( \dfrac{dr}{d\theta} \right)^2} ~ d\theta$
 

Integrate both sides:

$\displaystyle s = \int_{\theta_1}^{\theta_2} \sqrt{r^2 + \left( \dfrac{dr}{d\theta} \right)^2} ~ d\theta$

 

Another way to derive the formula
dsr = arc of a circle of radius r and central angle dθ
$\dfrac{ds_r}{d\theta} = \dfrac{2\pi r}{2\pi}$

$\dfrac{ds_r}{d\theta} = r$
 

length-of-arc-polar-plane.jpg

 

From the figure above:
$(ds)^2 = (ds_r)^2 + (dr)^2$

$\left( \dfrac{ds}{d\theta} \right)^2 = \left( \dfrac{ds_r}{d\theta} \right)^2 + \left( \dfrac{dr}{d\theta} \right)^2$

$\left( \dfrac{ds}{d\theta} \right)^2 = r^2 + \left( \dfrac{dr}{d\theta} \right)^2$

$\dfrac{ds}{d\theta} = \sqrt{r^2 + \left( \dfrac{dr}{d\theta} \right)^2}$

$ds = \sqrt{r^2 + \left( \dfrac{dr}{d\theta} \right)^2} \, d\theta$
 

Integrate both sides:
$\displaystyle s = \int_{\theta_1}^{\theta_2} \sqrt{r^2 + \left( \dfrac{dr}{d\theta} \right)^2} ~ d\theta$           okay!
 

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