1. ## Length of Cycloid

Okay everyone, I'm sure this is a simple fix but let's see if I'm on the right track:

We have been given a "project" to "prove" that the length of one complete arc of a cycloid generated by a circle of radius a, equals the perimeter of a square that circumscribes the same circle. I know that the answer to my integral should be 8|a|, but it is not coming out that way. I know that I have made an error somewhere but I just can't seem to find it!

Assume a is a constant:

$\displaystyle f(t)=x=at-a\sin{t}$

$\displaystyle g(t)=y=a-a\cos{t}$

$\displaystyle 0\leq{t}\leq{2\pi}$

$\displaystyle f'(t)=a-a\cos{t}$

$\displaystyle g'(t)=a\sin{t}$

$\displaystyle L=\int_{0}^{2\pi}\sqrt{(f'(t))^2+(g'(t))^2}$ $\displaystyle dt$

$\displaystyle L=\int_{0}^{2\pi}\sqrt{(a-a\cos{t})^2+(a\sin{t})^2}$ $\displaystyle dt$

$\displaystyle L=\int_{0}^{2\pi}\sqrt{a^2-2a^2\cos^2{t}+a^2\cos^2{t}+a^2\sin^2{t}}$ $\displaystyle dt$

$\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2-2a^2\cos^2{t}}$ $\displaystyle dt$

$\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos^2{t})}$ $\displaystyle dt$

$\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(1-\cos^2{t})}$ $\displaystyle dt$

$\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(sin^2{t})}$ $\displaystyle dt$

$\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}|\sin{t}|$ $\displaystyle dt$

$\displaystyle |\sin{t}|$ $\displaystyle =$ $\displaystyle \sin{t}$ for $\displaystyle t$ $\displaystyle \epsilon$ $\displaystyle [0,\pi]$

$\displaystyle |\sin{t}|$ $\displaystyle =$ $\displaystyle -\sin{t}$ for $\displaystyle t$ $\displaystyle \epsilon$ $\displaystyle [\pi,2\pi]$

$\displaystyle L=\sqrt{2}|a|\int_{0}^{\pi}\sin{t}$ $\displaystyle dt$ + $\displaystyle \sqrt{2}|a|\int_{\pi}^{2\pi}-\sin{t}$ $\displaystyle dt$

$\displaystyle L=\sqrt{2}|a|(-\cos{t})]_0^\pi$ + $\displaystyle \sqrt{2}|a|(\cos{t})]_\pi^{2\pi}$

$\displaystyle L=\sqrt{2}|a|(-(-1)-(-1))+\sqrt{2}|a|(1-(-1))$

$\displaystyle L=4\sqrt{2}|a|$

It's evident that I should not have a $\displaystyle \sqrt{2}$, instead I should have just 2 outside the integral. Right?

2. Originally Posted by elizsimca
Okay everyone, I'm sure this is a simple fix but let's see if I'm on the right track:

We have been given a "project" to "prove" that the length of one complete arc of a cycloid generated by a circle of radius a, equals the perimeter of a square that circumscribes the same circle. I know that the answer to my integral should be 8|a|, but it is not coming out that way. I know that I have made an error somewhere but I just can't seem to find it!

Assume a is a constant:

$\displaystyle f(t)=x=at-a\sin{t}$

$\displaystyle g(t)=y=a-a\cos{t}$

$\displaystyle 0\leq{t}\leq{2\pi}$

$\displaystyle f'(t)=a-a\cos{t}$

$\displaystyle g'(t)=a\sin{t}$

$\displaystyle L=\int_{0}^{2\pi}\sqrt{(f'(t))^2+(g'(t))^2}$ $\displaystyle dt$

$\displaystyle L=\int_{0}^{2\pi}\sqrt{(a-a\cos{t})^2+(a\sin{t})^2}$ $\displaystyle dt$

$\displaystyle L=\int_{0}^{2\pi}\sqrt{a^2-2a^2\cos^2{t}+a^2\cos^2{t}+a^2\sin^2{t}}$ $\displaystyle dt$

$\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2-2a^2\cos^2{t}}$ $\displaystyle dt$

$\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos^2{t})}$ $\displaystyle dt$

$\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(1-\cos^2{t})}$ $\displaystyle dt$

$\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(sin^2{t})}$ $\displaystyle dt$

$\displaystyle L=\sqrt{2}|a|\int_{0}^{2\pi}|\sin{t}|$ $\displaystyle dt$

$\displaystyle |\sin{t}|$ $\displaystyle =$ $\displaystyle \sin{t}$ for $\displaystyle t$ $\displaystyle \epsilon$ $\displaystyle [0,\pi]$

$\displaystyle |\sin{t}|$ $\displaystyle =$ $\displaystyle -\sin{t}$ for $\displaystyle t$ $\displaystyle \epsilon$ $\displaystyle [\pi,2\pi]$

$\displaystyle L=\sqrt{2}|a|\int_{0}^{\pi}\sin{t}$ $\displaystyle dt$ + $\displaystyle \sqrt{2}|a|\int_{\pi}^{2\pi}-\sin{t}$ $\displaystyle dt$

$\displaystyle L=\sqrt{2}|a|(-\cos{t})]_0^\pi$ + $\displaystyle \sqrt{2}|a|(\cos{t})]_\pi^{2\pi}$

$\displaystyle L=\sqrt{2}|a|(-(-1)-(-1))+\sqrt{2}|a|(1-(-1))$

$\displaystyle L=4\sqrt{2}|a|$

It's evident that I should not have a $\displaystyle \sqrt{2}$, instead I should have just 2 outside the integral. Right?
No time now. If no-one else replies I'll take a look later.

3. It's okay Mr. F.
It would have been up 45 mins ago, but I decided to try my hand at LaTeX so everyone could see the work I had already done.

4. ## Errors

Elizsimca
You were correct up to here:
$\displaystyle L=\int_{0}^{2\pi}\sqrt{(a-a\cos{t})^2+(a\sin{t})^2}$
but you made an error in the next line:
$\displaystyle L=\int_{0}^{2\pi}\sqrt{a^2-{\color{red}2a^2\cos{t}}+a^2\cos^2{t}+a^2\sin^2{t} }$
So
$\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2-2a^2\cos{t}}$
$\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos{t})}$
$\displaystyle L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos{t})}$
$\displaystyle L=|a|\int_{0}^{2\pi}\sqrt{2(1-\cos{t})}$
$\displaystyle L=|a|\int_{0}^{2\pi}\sqrt{2(2\sin^2(\frac{t}{2}))}$
$\displaystyle L=2|a|\int_{0}^{2\pi}\sqrt{\sin^2(\frac{t}{2})}$
And from there, you should be able to get the correct answer.

--Kevin C.

5. Oh what a dumb mistake!! Thank you I super appreciate it!

6. Originally Posted by elizsimca
Oh what a dumb mistake!! Thank you I super appreciate it!
And we appreciate you being so dilligent in showing all your working. Latex can be tricky. Well done - as you can see, it made it very easy to get a good answer.