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Parametrization of plane curves

  1. Mar 10, 2013 #1
    1. The function f(x,y) = x + y

    2. The area A is formed by the lines : x = 0 and x = pi/4 and by the graphs : x + cos(x) and x + sin(x)

    3. I have to parametrize A

    4. 'Formula' : r(u,v) = (u,v*f(u)+(1-v)*g(u))


    Could this be a parameterization of A :


    assuming f(u) = u+ cos(u) and g(u) = u + sin(u)

    r(u,v) = (u, v*(u+cos(u))+(1-v)*(u+sin(u))) <->

    r(u,v) = (u, v*cos(u)+u+sin(u)-v*sin(u)) and

    http://mathhelpforum.com/attachments/calculus/27437d1362743750-integration-flat-area-parameterization-billede-2.png


    ? ? ?

    I would really appreciate it if someone could look at this and give me some feedback because I'm not sure if the parametrization is correct :/ and I cannot solve the rest of the problem if the parametrization is incorrect.
     
  2. jcsd
  3. Mar 10, 2013 #2

    LCKurtz

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    You haven't made it very clear what you are trying to do. Are you trying to calculate the volume under the surface ##f(x,y)=x+y## above the described area in the ##xy## plane? If so, try the parameterization of your surface$$
    \vec r(x,y) =\langle x,y,x+y\rangle$$
     
  4. Mar 10, 2013 #3
    I have to find the area of the plane curve B.

    And to do that I need to parametrize B.

    After that I can find the Jacobi function and then integrate it and find the area of B.
     
  5. Mar 10, 2013 #4

    LCKurtz

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    Your original post uses f for two different things, f(x,y) and f(u). You also said you have to parameterize A, whatever A is. You didn't say. So I took a guess what it was. But now you say you need to parameterize B. We aren't mind readers here. What the heck is B? And you say need the area of plane curve B. Plane curves don't have areas anyway. I'm guessing that English isn't your first language, but you still need to state the problem completely enough so we know what it is.
     
  6. Mar 10, 2013 #5
    I'm sorry.

    I'm given a function f(x,y) = y + x and A.

    A is the set of points in the (x,y) plane that is formed by the lines : x = 0 and x = pi/4 and by the graphs : x + cos(x) and x + sin(x).

    I need to find the area of A.


    And I have to use a specific formula :
    tumblr_mjgvwuEsKm1qbhqjfo1_r1_400.png


    r(u,v) is the parameterization of A.


    I hope you understand the problem now ?
     
  7. Mar 10, 2013 #6

    LCKurtz

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    What does ##f(x,y) = x+y## have to do with finding the area of A? And if you want the area of A, what's wrong with just doing the integral$$
    \int_0^{\frac \pi 4}\int_{x+\sin x}^{x+\cos x} 1\, dydx$$Why would you want to introduce a change of variables in this problem?
     
  8. Mar 10, 2013 #7
    The function is given because I need to use it to another part of the problem. So it is not relevant in this case.

    I don't think there is anything wrong with solving the problem like that but I just think that the purpose of the problem is to learn parameterisizing set of points like A.
     
  9. Mar 10, 2013 #8

    LCKurtz

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    ##\vec r(x,y) = \langle x,y,0\rangle,\ x+\sin x\le y \le x+\cos x,\ 0\le x \le \frac \pi 4##. There is no reason to parameterize that region any other way that I can see. I don't see how to help you further.
     
  10. Mar 10, 2013 #9
    Alright

    Thank you
     
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