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If we parameterize curve by x=t , y=t , what is the range of t ? Is it 0<= t <=1? why?

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- Thread starter nothGing
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- #1

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If we parameterize curve by x=t , y=t , what is the range of t ? Is it 0<= t <=1? why?

- #2

HallsofIvy

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A "line integral" is along a one-dimensional line and a "surface integral" is over a two dimensional surface.what is the different between line integral and surface integral?

There are no bounds. if you are talking about a portion of a curve, say from [itex](x_0, y_0)[/itex] to [itex](x_1, y_1)[/itex], the range of t is from whatever value of t gives [itex]x_0[/itex] and [itex]y_0[/itex] to whatever value of t gives [itex]x_1[/itex] and [itex]y_1[/itex].If we parameterize curve by x=t , y=t , what is the range of t ? Is it 0<= t <=1? why?

- #3

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Let say, integral ( x^2 + y + z)ds , the line segment of curve is from (0,0,0) to (1,2,1).

We parameterize curve C by x=t , y=2t , z=t.

As you said, t should be between 0 and 2.

but according to my reference book, 0<=t<=1.

Why?

- #4

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With your paramterization, its 0<=t<=1. But I could choose:

x=t/2, y=t, z=t/2

Which still describe a line with the same orientation, instead now t must range from 0 to 2 to give the desired segment.

- #5

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When

Let say, integral ( x^2 + y + z)ds , the line segment of curve is from (0,0,0) to (1,2,1).

We parameterize curve C by x=t , y=2t , z=t.

As you said, t should be between 0 and 2.

but according to my reference book, 0<=t<=1.

Why?

- #6

HallsofIvy

Science Advisor

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No, I did NOT say that! I said " if you are talking about a portion of a curve, say from (xErm.. ok..

Let say, integral ( x^2 + y + z)ds , the line segment of curve is from (0,0,0) to (1,2,1).

We parameterize curve C by x=t , y=2t , z=t.

As you said, t should be between 0 and 2.

We must have x(t)= t= 0, y(t)= 2t= 0, and z(t)= t= 0. Obviously t= 0 satisfies all of those.

We must also have x(t)= t= 1, y(t)= 2t= 2, and z(t)= t= 1. Obviously t= 1, NOT t= 2, satisfies all of those.

but according to my reference book, 0<=t<=1.

Why?

- #7

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Well, now i understand it.. thanks to elibj123,Redbelly98, HallsofIvy for helping..

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