# Thread: length of polar curve

1. ## length of polar curve

i'm going to use x in place of theta since i don't know how to make the symbol

find the length of the cardioid r=1 + cos(x).

so first i take the derivative, which is -sin(x). that squared would be sin(x)^2

(1+cos(x))^2 is 1 +2cos(x) +cos(x)^2.

put these both under the square root symbol and take the integral of the whole thing. i apologize for my lack of proper formatting.

so we have int(sqrt(1+2cos(x)+cos(x)^2 +sin(x)^2))
the cos(x)^2 + sin(x)^2 =1, which gives:
int(sqrt(2+2cos(x)). there i am stuck, i don't know how to solve that integral. can anyone help? and did i at least make it to this point without any mistakes? please let me know wherever i messed up. also i have the limits of integration from 0 to 2pi, but i'm not very sure that is correct.

2. Originally Posted by isuckatcalc
i'm going to use x in place of theta since i don't know how to make the symbol

find the length of the cardioid r=1 + cos(x).

so first i take the derivative, which is -sin(x). that squared would be sin(x)^2

(1+cos(x))^2 is 1 +2cos(x) +cos(x)^2.

put these both under the square root symbol and take the integral of the whole thing. i apologize for my lack of proper formatting.

so we have int(sqrt(1+2cos(x)+cos(x)^2 +sin(x)^2))
the cos(x)^2 + sin(x)^2 =1, which gives:
int(sqrt(2+2cos(x)). there i am stuck, i don't know how to solve that integral. can anyone help? and did i at least make it to this point without any mistakes? please let me know wherever i messed up. also i have the limits of integration from 0 to 2pi, but i'm not very sure that is correct.
everything you've done up to here is correct.

pull out the 2 ...

$\displaystyle \sqrt{1+\cos{t}} \cdot \frac{\sqrt{1-\cos{t}}}{\sqrt{1-\cos{t}}}$

$\displaystyle \frac{\sqrt{1 - \cos^2{t}}}{\sqrt{1-\cos{t}}}$

$\displaystyle \frac{|\sin{t}|}{\sqrt{1-\cos{t}}}$

since $\displaystyle \sin{t} \ge 0$ in quads I and II, use symmetry ...

$\displaystyle L = 2\sqrt{2} \int_0^{\pi} \frac{\sin{t}}{\sqrt{1-\cos{t}}} \, dt$

now use a simple substitution to evaluate $\displaystyle L$