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TomLikesPhysics
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I want to find the length of a curve which is given by the equation: r=(tsint,1cost, 2).
To calculate the length I first calculate the first derivative of r which is v=(1cost, sint, 0).
Than I went on to integrate the absolulte value of v.
This gave me the integral of the squareroot of (22cost). 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.
 one year ago
 one year ago
TomLikesPhysics Group Title
I want to find the length of a curve which is given by the equation: r=(tsint,1cost, 2). To calculate the length I first calculate the first derivative of r which is v=(1cost, sint, 0). Than I went on to integrate the absolulte value of v. This gave me the integral of the squareroot of (22cost). 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.
 one year ago
 one year ago

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TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
hmm to make the math work maybe you should split it into 2 integrals with limits (0,pi) and (pi,2pi)
 one year ago

Fellowroot Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Here is my (wrong) calculation so far:
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
The second integral goes from pi/2 to what?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
correction: it's \[\int\limits_{0}^{\pi} \] and \[\int\limits_{\pi}^{2\pi}\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
here's the corrected equation: \[\sqrt{22\cos t}=2\cos \frac{ t }{ 2 }\] when \[0 \le t \le \pi\] and \[\sqrt{22\cos t}=2\cos \frac{ t }{ 2 }\] in \[\pi \le t \le 2\pi\]
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
So I did the integral wrong?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
so that \[\int\limits_{0}^{2\pi}\sqrt{22\cos t}dt = \int\limits_{0}^{\pi}2\cos \frac{ t }{ 2 }dt + \int\limits_{\pi}^{2 \pi} 2\cos \frac{ t }{ 2 }dt\]
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Yeah, if I rewrite that Integral I get 8. However, I could nowhere find that Trig. Identity that you used.
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Ah... now I see where you got that from. If you don´t mind could you check where my substitution went wrong?
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Thank you so far for your help sirm3d.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
\[\cos^2 A=\frac{ 1 }{ 2 }(1+\cos 2A)\] that's halfangle identity
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
im still checking your solution.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
here's the mistake: \[\sqrt{22\cos t}\] \[=\sqrt{\frac{ (22\cos t)(2+2\cos t) }{ (2 + 2\cos t) }}\]\[=\frac{ \sqrt{44\cos^2 t} }{ \sqrt{2+2\cos t} }\] up to this point it is still correct.
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
I guess it might be the point where I set dt=du/2sint  sint could be zero so I am dividing by zero!?!?!?!?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
\[\sqrt{44\cos^2 t} \neq 2\sin t\] due to the fact that \[2\sin \] is negative in \[\pi < t < 2\pi\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
\[\sqrt{44\cos^2t}=2\sin t\] in \[0<t<\pi\] and \[\sqrt{44\cos^2 t}=2 \sin t\] in \[\pi < t < 2 \pi\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
a common mistake in definite integrals involving trig function is the choice of + over minus. be careful next time.
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Thank you for your help, sirm3d. I try to be more carefull next time. :) Thx a lot.
 one year ago
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