1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Setting up parametrization

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Compute the line integral of the scalar function.
    f(x,y,z) = x[itex]e^{z^2}[/itex], piecewise linear path from (0,0,1) to (0,2,0) to (1,1,1)

    2. Relevant equations



    3. The attempt at a solution
    In this problem, all I need is a parametrization. First I drew the line from (0,0,1) to (0,2,0) xyz-plane. I got the slope as z = 1 - [itex]\frac{y}{2}[/itex]. So I set y = t then z will be 1 - [itex]\frac{t}{2}[/itex]. I got parametrization as c(t) = <0,t,1-[itex]\frac{y}{2}[/itex]>. But it's wrong. It's c(t) = <0,2t,1-2t>. Would anyone help me how to get parametrization ??
     
  2. jcsd
  3. Dec 2, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, the line from (0, 0, 1) to (0, 2, 0) can be written as z= 1- y/2. Since you are using t as parameter, that would be <0, t, 1- t/2>. Why do you say that is wrong?
    When t= 0, your parameterization gives (0, 0, 1) and when t= 2, it gives (0, 2, 1). A line is determined by two points so it gives the correct line.

    It is c(t)= <0, 2t, 1- 2t> that is wrong. In order that y= 2t= 2, t must be 1. But then z= 1- 2= -1, not 0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Setting up parametrization
  1. Volume integral set up (Replies: 9)

  2. Setting up a DE (Replies: 3)

Loading...