## TomLikesPhysics 3 years ago I want to find the length of a curve which is given by the equation: r=(t-sint,1-cost, 2). To calculate the length I first calculate the first derivative of r which is v=(1-cost, sint, 0). Than I went on to integrate the absolulte value of v. This gave me the integral of the squareroot of (2-2cost). Than I ended up with -2*squareroot of (2+2cost). Which is the same as Wolfram Alpha got. However, if I want to integrate this curve from 0 to 2pi than I end up with 0, but it should be 8. If I plug in 0 and 2pi I get -4-(-4) which is 0, but the curve can not have a lenght of 0.

1. TomLikesPhysics

In a nutshell: I ended up with the right integral but if I plug in my boundaries (0 and 2pi) I get 0 as the lenght of the curve. So somehow the lenghts eat each other up? So, how do I "use" the integral in the right way to get 8 as a lenght?

2. dumbcow

hmm to make the math work maybe you should split it into 2 integrals with limits (0,pi) and (pi,2pi)

3. Fellowroot

This is very strange, wolfram does indeed give the answer as 8 yet whenever i try and evaluate the integral I get zero just think 0 and 2pi are the same value when evaluated at cos(t) no wonder its zero, yet wolfram says otherwise.

4. TomLikesPhysics

Changing the boundaries from before 0 to 2pi to 0 to pi and pi to 2pi did not help at all. I only started learning about calculating the length of a curve so I have no clue what this is supposed to tell me. The Curce must clearly have some length.

5. TomLikesPhysics

Here is my (wrong) calculation so far:

6. TomLikesPhysics

The second integral goes from pi/2 to what?

7. sirm3d

correction: it's $\int\limits_{0}^{\pi}$ and $\int\limits_{\pi}^{2\pi}$

8. sirm3d

here's the corrected equation: $\sqrt{2-2\cos t}=2\cos \frac{ t }{ 2 }$ when $0 \le t \le \pi$ and $\sqrt{2-2\cos t}=-2\cos \frac{ t }{ 2 }$ in $\pi \le t \le 2\pi$

9. TomLikesPhysics

So I did the integral wrong?

10. sirm3d

so that $\int\limits_{0}^{2\pi}\sqrt{2-2\cos t}dt = \int\limits_{0}^{\pi}2\cos \frac{ t }{ 2 }dt + \int\limits_{\pi}^{2 \pi} -2\cos \frac{ t }{ 2 }dt$

11. TomLikesPhysics

Yeah, if I rewrite that Integral I get 8. However, I could nowhere find that Trig. Identity that you used.

12. TomLikesPhysics

Ah... now I see where you got that from. If you don´t mind could you check where my substitution went wrong?

13. TomLikesPhysics

Thank you so far for your help sirm3d.

14. sirm3d

$\cos^2 A=\frac{ 1 }{ 2 }(1+\cos 2A)$ that's half-angle identity

15. sirm3d

16. sirm3d

here's the mistake: $\sqrt{2-2\cos t}$ $=\sqrt{\frac{ (2-2\cos t)(2+2\cos t) }{ (2 + 2\cos t) }}$$=\frac{ \sqrt{4-4\cos^2 t} }{ \sqrt{2+2\cos t} }$ up to this point it is still correct.

17. TomLikesPhysics

I guess it might be the point where I set dt=du/2sint - sint could be zero so I am dividing by zero!?!?!?!?

18. sirm3d

$\sqrt{4-4\cos^2 t} \neq 2\sin t$ due to the fact that $2\sin$ is negative in $\pi < t < 2\pi$

19. sirm3d

$\sqrt{4-4\cos^2t}=2\sin t$ in $0<t<\pi$ and $\sqrt{4-4\cos^2 t}=-2 \sin t$ in $\pi < t < 2 \pi$

20. sirm3d

a common mistake in definite integrals involving trig function is the choice of + over minus. be careful next time.

21. TomLikesPhysics

Thank you for your help, sirm3d. I try to be more carefull next time. :) Thx a lot.