# What is the arc length of the polar curve f(theta) = cos(3theta-pi/2) +thetacsc(-theta)  over theta in [pi/12, pi/8] ?

Feb 10, 2016

L = int_(alpha1)^(alpha2) sqrt(f'^" 2"(alpha) + f^2(alpha)) " "dalpha
$L \approx .2037$

#### Explanation:

The Arc Length, L is given:
from parametric length equation we have the familiar:
$s = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$ it is just the aggregate distance formula that you may have seen many time. If you think the polar coordinate as a parametric representation in $r , \alpha$ then:
$r = f \left(\alpha\right)$
$x = r \cos \alpha = f \left(\alpha\right) \cos \alpha$
$y = r \sin \alpha = f \left(\alpha\right) \sin \alpha$
${\left[\frac{\mathrm{dx}}{\mathrm{da} l p h a}\right]}^{2} = {\left[f ' \left(\alpha\right) \cos \alpha - f \left(\alpha\right) \sin \left(\alpha\right)\right]}^{2}$
${\left[\frac{\mathrm{dy}}{\mathrm{da} l p h a}\right]}^{2} = {\left[f ' \left(\alpha\right) \sin \alpha + f \left(\alpha\right) \cos \left(\alpha\right)\right]}^{2}$
${\left[\frac{\mathrm{dx}}{\mathrm{da} l p h a}\right]}^{2} + {\left[\frac{\mathrm{dy}}{\mathrm{da} l p h a}\right]}^{2} = f {'}^{\text{ 2}} \left(\alpha\right) + {f}^{2} \left(\alpha\right)$
L = int_(alpha1)^(alpha2) sqrt(f'^" 2"(alpha) + f^2(alpha)) " "dalpha
$f {'}^{\text{ 2}} \left(\alpha\right) = {\left[\frac{d}{\mathrm{da} l p h a} \left(\cos \left(3 \alpha - \frac{\pi}{2}\right) + \alpha \csc \left(- \alpha\right)\right)\right]}^{2}$
${f}^{2} \left(\alpha\right) = {\left[\cos \left(3 \alpha - \frac{\pi}{2}\right) + \alpha \csc \left(- \alpha\right)\right]}^{2}$
Integrand, I =sqrt( f'^" 2"(alpha) + f^2(alpha)); alpha = t
I =sqrt(3cos3t + ((t cott - 1)csct)^2 + (sin3t - tcsct)^2
$L = {\int}_{\frac{\pi}{12}}^{\frac{\pi}{8}} \left(I\right) \mathrm{dt}$
$L \approx .2037$