Length of the cardioid r = 5(1 - cos?) between ?=0, ?=?

[MAC-A-TAC]

Please help. I am stuck between the following steps in the soulution.
S = ? 5?2-2cos? d?

S = ? 10?1- cos? d? /2

What am I missing?

Thank you for your help.

 

[galactus]

arc length in polar is $\displaystyle L=\int_{\alpha}^{\beta}\sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}d{\theta}$

Let's do the general case: $\displaystyle a(1-cos{\theta})$

$\displaystyle r^{2}+\left(\frac{dr}{d\theta}\right)^{2}=[a(1-cos{\theta})]^{2}+[asin{\theta}]^{2}=4a^{2}sin^{2}(\frac{\theta}{2})$

$\displaystyle \int_{0}^{\pi}[2asin(\frac{\theta}{2})]d{\theta}$

After you integrate and have the solution(it is very basic), note that you have a=5

Here is the region you want the arc length of:

 

[MAC-A-TAC]

Thank you for the detailed graph.
I am still not understanding how to manipulate the trig identities to come up with the second equation in my original post.... :?
Thank you.

 

[galactus]

There is no identities to use. Just factor out a 2.

Just do as the formula says.

I am going to use t instead of theta. No biggie, just easier.

$\displaystyle r^{2}=25(1-cos(t))^{2}$

$\displaystyle \left(\frac{dr}{dt}\right)^{2}=25sin^{2}(t)$

$\displaystyle \sqrt{r^{2}+\left(\frac{dr}{dt}\right)^{2}}=5\sqrt{2(1-cos(t))}$

There it is. But what you have does not correspond to what you have in your heading, $\displaystyle 5\sqrt{(1-cos(t))}$

The first one is correct. You forgot the 2 in the second one. See?.

But, $\displaystyle \sqrt{1-cos(t)}$ is difficult to integrate. That is why we manipulate it so we can integrate as I showed in my first post.

BTW, it would help to use proper grouping symbols. What you have is $\displaystyle 10\sqrt{1}-\frac{cos{\theta}}{2}$

 

[MAC-A-TAC]

Got it.
Thank you.