September 6th, 2017, 09:53 AM  #1 
Member Joined: May 2017 From: Slovenia Posts: 87 Thanks: 0  Line integral, help pls :(
I found this solved but I don't understand the parameterization. Why x = 1t and y = 0+2t and why are the bounds from 0 to 1? 
September 6th, 2017, 02:02 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,410 Thanks: 715 
If you don't know this, learn it now: a point on the line segment between two points $A,~B$ can be represented as $p = A + (BA)t,~t \in [0,1]$ Think about that until you understand why in your bones. Now apply the values of the two points you are given, $(1, 0),~(0,2)$ then examine the two components of that vector. You'll see they are exactly the equations for $x$ and $y$ you are given, with $t$ ranging from $0 \to 1$. Last edited by skipjack; September 6th, 2017 at 03:01 PM. 
September 6th, 2017, 04:03 PM  #3 
Senior Member Joined: Nov 2015 From: Alabama Posts: 140 Thanks: 17 Math Focus: Geometry, Trigonometry, Calculus  Couldn't be more true. Not knowing this made me screw up a question on a cal 3 exam. Lesson learned... lol.

September 7th, 2017, 04:08 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,649 Thanks: 681 
Do you understand that there is no one "correct" parameterization? There exist an infinite number of "correct" parameterizations for the same figure. The parameter goes from 0 to 1 because the parameterization was chosen that way that makes it particularly easy. For this problem, I would have argued that, since the figure is a straight line, the equations for x, y, and z can be linear ("can be" not "must be"). That is, I can take x= at+ b, y= ct+ d. The line must go through (1, 0) so, taking that point correspond to t= 0, we must have x= a(0)+ b= b= 1, y= c(0)+ d= d= 0. The line must go through (0, 2) so, taking that point to correspond to t= 1, we must have x= a(1)+ b= a+ b= a+ 1= 0, a= 1, and y= c(1)+ d= c+ d= c= 2. The parameterization is x= t+ 1, y= 2t. But I could, just as well, have decided to take t= 1 at (1, 0) and t= 5 at (0, 2). Then I would have x= a(1)+ b= a+ b= 1, c(1)+ d= c+ d= 0 and x= a(5)+ b= 5a+ b= 0, y= c(5)+ d= 5c+ d= 2. I still have two equations to solve for a, b and two equations to solve for c, d. They happen to give a= 1/4, b= 5/4, c= 1/2, d= 1/2 so the parametric equations are x= (1/4)t+ 5/4, y= (1/2)t 1/2. This is a valid parameterization because (1) both equations are linear so it describes a straight line, (2) when t= 1, x= 1/4+ 5/4= 1 and y= 1/2 1/2= 0, (3) when t= 5, x= 5/4+ 5/4= 0, y= 5/2 1/2= 2. Taking the parameter to be 0 at one point and 1 at the other just makes the equations easier. Last edited by Country Boy; September 7th, 2017 at 04:12 AM. 
September 7th, 2017, 05:00 AM  #5  
Member Joined: May 2017 From: Slovenia Posts: 87 Thanks: 0  Quote:
Will it be correct if: find the equation between those two points, that will be y = 2x+2 take x = t and y = 2t+2 and take the bounds from 1 to 0. I get the same result.  

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integral, line, pls 
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