Slove the math step by step

Question

.Find the entire arc length of the curve r=2acos³(θ/3)

Answer 1

Given, r = 2a cos³(θ/3)

Using chain rule while differentiating with respect to θ,

dr / dθ = d / dθ * [2a cos³(θ/3)]

= 2a * d / dθ [cos³(θ/3)

= 2a * 3 cos²(θ/3) * d/dθ [cos(θ/3)]

= 6a cos³(θ/3) * [-sin(θ/3)] * d/dθ [θ/3]

= 6a cos³(θ/3) * [-sin(θ/3)] * 1/3

= -2a*cos²(θ/3)*sin(θ/3)

Now, As Tom K mentioned in his comment,

cos(3π + θ) = - cos(θ)

(cos (3 π + θ), sin(3 π + θ)) = - (cos (θ), sin( θ))

This means values of x and y repeat every 3 π

Now, for the arc length L, integrate sqrt [r²+(dr/dθ)²dθ] with [0,3π].

Note: I am new to Wyzant so couldn't use the integration symbol properly. Everything below this line is with an definite integral with [0,3π]

= sqrt [ 4a²cos⁶(θ/3) + 4a²cos⁴(θ/3) sin²(θ/3) ] dθ

= sqrt [ 4a²cos⁴(θ/3) (cos²(θ/3)+sin²(θ/3)) ] dθ

We know, cos²(θ/3)+sin²(θ/3) = 1

And, cos (2θ) = 2 cos²(θ) - 1

= 2a cos²(θ/3) dθ

= 2a * 1/2 * [ 1 + cos (2θ/3) ] dθ

= a [1 + cos (2θ/3)] dθ

In the following steps, we will drop the integration sign as we are plugging the values,

= a [ θ + (3/2) sin(2θ/3) ] on [3π,0]

= a [ (3π + 0) - (0 +0) ]]

= 3πa

Therefore, the required length of the arc is 3πa.

Answer 2

see https://tutorial.math.lamar.edu/classes/calcii/polararclength.aspx

The formula is I[0, 2 π] sqrt(r^2 + (dr/dθ)^2) dθ

Note that our immediate thought should be, as we have cos(θ/3), we need to integrate from 0 to 6 π. However, as cos(3 π + θ) = - cos(θ) and (cos (3 π + θ), sin(3 π + θ)) = - (cos (θ), sin( θ)), the values of x and y repeat every 3 π

If r = 2acos³(θ/3), dr/dθ = 2acos²(θ/3)sin(θ/3)

I[0, 2 π] sqrt(r^2 + (dr/dθ)^2) dθ =

2a I[0, 3 π] sqrt(cos⁶(θ/3)+cos⁴(θ/3)sin²(θ/3)) dθ = as sin²(θ/3) + cos²(θ/3) = 1, and sqrt(cos⁴(θ/3)) = cos²(θ/3)

2a I[0, 3 π] cos²(θ/3) dθ = as cos²(θ/3) = 1/2 + 1/2 cos (2θ/3)

2a I[0, 3 π] 1/2 + 1/2 cos (2θ/3) dθ =

a (θ + 3/2 sin (2θ/3)) E[0, 3π] =

3π a