What is the arclength of the polar curve #f(theta) = 2tan^2theta+cos3theta # over #theta in [0,(5pi)/12] #?

Question

What is the arclength of the polar curve #f(theta) = 2tan^2theta+cos3theta  # over #theta in [0,(5pi)/12]  #?

Emma Aldridge · Accepted Answer

#int_0^((5 pi)/12) sqrt((3 sin(3 t)-4 sec^2(t) tan(t))^2+(cos(3 t)+2 tan^2(t))^2) dt~~27.41370155...#

The Arc Length L in polar is given by: #L = int_a^b sqrt(r^2 + ((dr)/(d theta ))^2 d theta# where #a, b in [0, 2pi] # For this case we need evaluate #L = int_0^((5pi)/12) sqrt(f^2(theta ) + ((df(theta))/(d theta))^2 ) d theta# This a a nasty integral indeed: #f^2(theta)= 4tan^4theta + 4tan^2thetacos3theta +cos^2theta# #(df(theta))/(d theta) = 4sec^2theta tantheta-3sin3theta # #[(df(theta))/(d theta)]^2 = 16(sec^2theta tantheta)^2-24sec^2theta tanthetasin3theta + 9sin(3theta)# Now add take the square root and compute using one of the online integral calculators.

#int_0^((5 pi)/12) sqrt((3 sin(3 t)-4 sec^2(t) tan(t))^2+(cos(3 t)+2 tan^2(t))^2) dt~~27.41370155...#

I use Wolframalpha

Emilia Aguilar · Answer

undefined To find the arc length of the polar curve $ f(\theta) = 2\tan^2(\theta) + \cos(3\theta) $ over $ \theta $ in $ \left[0, \frac{5\pi}{12}\right] $, we use the formula for arc length:

$$ L = \int_{\alpha}^{\beta} \sqrt{[f(\theta)]^2 + [f'(\theta)]^2} \, d\theta $$

where $ \alpha $ and $ \beta $ are the limits of integration.

First, we need to find $ f'(\theta) $, the derivative of $ f(\theta) $ with respect to $ \theta $:

$$ f'(\theta) = \frac{d}{d\theta} (2\tan^2(\theta) + \cos(3\theta)) $$

$$ = 4\tan(\theta)\sec^2(\theta) - 3\sin(3\theta) $$

Now, we can plug $ f(\theta) $ and $ f'(\theta) $ into the arc length formula and integrate:

$$ L = \int_{0}^{\frac{5\pi}{12}} \sqrt{(2\tan^2(\theta) + \cos(3\theta))^2 + (4\tan(\theta)\sec^2(\theta) - 3\sin(3\theta))^2} \, d\theta $$

This integral can be complex to solve analytically, and it might be more suitable to use numerical methods or software to approximate the value of the arc length.