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September 6th, 2017, 10:53 AM   #1
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Line integral, help pls :(

I found this solved but I don't understand the parameterization.
Why x = 1-t and y = 0+2t
and why are the bounds from 0 to 1?
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September 6th, 2017, 03:02 PM   #2
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If you don't know this, learn it now:

a point on the line segment between two points $A,~B$ can be represented as

$p = A + (B-A)t,~t \in [0,1]$

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 04:01 PM.
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September 6th, 2017, 05:03 PM   #3
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Quote:
Originally Posted by romsek View Post
If you don't know this, learn it now:

a point on the line segment between two points $A,~B$ can be represented as

$p = A + (B-A)t,~t \in [0,1]$

Think about that until you understand why in your bones.
Couldn't be more true. Not knowing this made me screw up a question on a cal 3 exam. Lesson learned... lol.
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September 7th, 2017, 05:08 AM   #4
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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 05:12 AM.
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September 7th, 2017, 06:00 AM   #5
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Quote:
Originally Posted by Country Boy View Post
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.

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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