A community for students.
Here's the question you clicked on:
 0 viewing
TomLikesPhysics
 3 years ago
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.
TomLikesPhysics
 3 years ago
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.

This Question is Closed

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0In 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?

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

Fellowroot
 3 years ago
Best ResponseYou've already chosen the best response.0This 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.

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0Changing 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.

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0Here is my (wrong) calculation so far:

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0The second integral goes from pi/2 to what?

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

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.1here'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 \[\pi \le t \le 2\pi\]

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0So I did the integral wrong?

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

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

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

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you so far for your help sirm3d.

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

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.1im still checking your solution.

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.1here'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.

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

sirm3d
 3 years ago
Best 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 \[\pi < t < 2\pi\]

sirm3d
 3 years ago
Best 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 \[\pi < t < 2 \pi\]

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

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you for your help, sirm3d. I try to be more carefull next time. :) Thx a lot.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.