Question

What is the arclength of the polar curve `f(theta) = sin4theta-cos3theta` over `theta in [pi/3,pi/2]`?

Answer 1

The arclength is approximately `0.46371`

Explanation:

The equation for arclength of a polar curve is:

`int_a^b sqrt(r^2 + ((dr)/(d theta))^2) d theta`

In this case,

`r^2 = (sin4theta-cos3theta)^2`

`(dr)/(d theta) = d/(d theta) (sin4theta-cos3theta)`

`((dr)/(d theta))^2 = (4cos4theta + 3sin3theta)^2`

So, the arclength can be found by:

`int_(pi/3)^(pi/2) sqrt( (sin(4theta)-cos(3theta))^2 + (4cos4theta + 3sin3theta)^2) d theta`

`~~ 0.46371`

Final Answer

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I am not entirely sure how to solve this integral manually; if anyone else has a valid method of integration, feel free to extend this answer, but as far as I can see, the only way to solve this problem is with a graphing calculator or similar interface.