What is the arc length of the polar curve #f(theta) = 5sintheta-4theta # over #theta in [pi/8, pi/3] #?

1 Answer
Jul 30, 2017

L #approx# 0.47235219096 #approx# 0.472 (3 decimal places)

Explanation:

The derivation of the arc length of a polar curve over the interval #[a, b]# could be found in another one of my posts, the link to which is here: https://socratic.org/questions/what-is-the-arclength-of-the-polar-curve-f-theta-3sin-3theta-2cot4theta-over-the#455941

Applying to this question, we must first find the derivative of the function with respect to #theta#:

#f'(theta) = d/(d theta)(5 sin(theta) - 4 theta)#

Since #d/(d theta) (sin (theta)) = cos(theta)#,

#f'(theta) = 5cos(theta) - 4#

The formula for the arc length of a polar curve over the interval #[a, b]# is given by:

#L = int_a^bsqrt(r^2 + ((dr)/(d theta))^2) d theta#

where #r = f(theta)# and therefore it follows that #(dr)/(d theta) = f'(theta)#

Therefore, the arc length for the given function over the given interval would be:

#L = int_(pi/8)^(pi/3)sqrt((5 sin(theta) - 4 theta)^2 +( 5cos(theta) - 4)^2) d theta#

Using a graphing utility (for the sake of simplicity), the integral equals about #0.47235219096#, or to three decimal places, it is equal to #0.472#