ambigus question

[Dr.Peterson]

OP, why do you bother yourself to start with that angle when you are interested in a positive length in a full revolution? Just do the [imath](0,2\pi)[/imath] interval. I don't blame you to choose that complicated initial angle as the point of interest was at the origin [imath]\left(\displaystyle \theta = -\frac{\pi}{2}\right)[/imath].

I'm confused. I see no [imath]\theta[/imath] anywhere in this thread (or in the reference), and the point is at the origin for [imath]t=0[/imath]. I would absolutely use 0 to [imath]2\pi[/imath] for the limits.

the angle start at \(\displaystyle -\frac{\pi}{2}\) and for one red curve it end at \(\displaystyle -\frac{\pi}{2} - 2\pi = -\frac{5\pi}{2}\)

Can you explain where you got these angles? I never pointed out this error because there were other issues to deal with. My guess is that you made up your own (unstated) definition for "the angle", and never noticed that it doesn't agree with [imath]t[/imath].

 

[mario99]

This is only true for certain intervals. [math]\sqrt{2-2\cos t} = 2\sin\!\left(\frac{t}{2}\right)[/math]It'd be wrong to omit the absolute value without good reasons. Your suggestion of picking a different starting angle is essentially applying a phase shift.

See your messy calculations and compare them with my calculations. Frankly, were there good reasons for me to do that?

I'm confused. I see no [imath]\theta[/imath] anywhere in this thread (or in the reference), and the point is at the origin for [imath]t=0[/imath]. I would absolutely use 0 to [imath]2\pi[/imath] for the limits.

Well BigBeachBanana does not agree with you unless you have good reasons to omit the absolute value! And oops, I meant [imath]\displaystyle t = -\frac{\pi}{2}[/imath]. I don't see any mistake in choosing [imath]t[/imath] like that. It only complicates things.

 

[BigBeachBanana]

See your messy calculations and compare them with my calculations. Frankly, were there good reasons for me to do that?


Well BigBeachBanana does not agree with you unless you have good reasons to omit the absolute value! And oops, I meant [imath]\displaystyle t = -\frac{\pi}{2}[/imath]. I don't see any mistake in choosing [imath]t[/imath] like that. It only complicates things.

I do not see anything that you deem as messy but rather added details. You seem to disagree with my statement that equality does not hold for all domains
[math]\sqrt{2-2\cos t} \neq 2\sin\!\left(\frac{t}{2}\right)[/math]

 

[mario99]

I do not see anything that you deem as messy but rather added details. You seem to disagree with my statement that equality does not hold for all domains
[math]\sqrt{2-2\cos t} \neq 2\sin\!\left(\frac{t}{2}\right)[/math]

I am not against your idea. You are correct that the two integrals are not equal in all domains, but in our domain they are equal and that what really matters.

By the way, I love the messy and complicated approach to solve a problem. When I have given the OP that solvable integral, I expected him to wonder about: how did you do that, why did you do that, what about the domain? But he did not say anything, so I could not have the chance to alert him about any restrictions.

For professor Dave, [imath]t = 0[/imath] is the correct and logical initial angle. For me, I don't care what is the initial angle as long as it will end in a full revolution because whatever I get from the integral, I will take the absolute value of that because the length of an arc is positive.

In simple words, this was my intention:

[imath]\displaystyle \bigg| \int_{-\frac{\pi}{2}}^{-\frac{5\pi}{2}} \sqrt{2 - 2\cos t} \ dt \bigg| = 2\int_{0}^{2\pi} \sin\left(\frac{t}{2}\right) \ dt[/imath]