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

1 Answer
Mar 9, 2016

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...#

Explanation:

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