Wislar
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Line integral: \[\int\limits_{C}^{}x ^{2}dx+y ^{2}dy\] of the line segment from (0,2) to (4,3)
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Wislar
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Is it:
\[\int\limits_{0}^{4}16t ^{2}(4)+(2+t)^{2}(1)dt\]
Wislar
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The parameterized curve is r(t)=(4t, 2+t) right?
Wislar
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@phi @amistre64 @Zarkon @TuringTest
amistre64
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id have a little reading to do to catch up on this :)
Wislar
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I think I have it down correctly, I'm just not sure if the limit is from 0 to 4
amistre64
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we need to parametrize the line from 0,2 to 4,3 right?
Wislar
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Yes
amistre64
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4,3
-0-2
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<4,1> is the direction vector, and we can apply it to either point given so one parametric could be:
x = 0 + 4t
y = 2 + t
Wislar
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That's what I got! I'm not sure about the limits of integration though
abb0t
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sketch out the graph.
amistre64
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the dx and dy parts in your post got me confuddled. im used to seeing a ds in it
amistre64
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\[\int_CPdx+Qdy=\int_CPdx+\int_CQdy\]
amistre64
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x = 0 + 4t
dx = 4 dt
y = 2 + t
dy = dt
Wislar
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I have that down @amistre64 I'm just really not sure about the limits. I just used the x coordinates.
amistre64
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what are the limits of t for the given line?
Wislar
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1 and 4?
amistre64
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when t=0 we are at (0,2)
when t=? we are at (4,3)
Wislar
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I'm not sure?
amistre64
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0 = 0 + 4t
2 = 2 + t
t=0
4 = 0 + 4t
3 = 2 + t
t = 1
so the limit of our t is 0 to 1
Wislar
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Wait how did you do that?
amistre64
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we are changing x and y into functions of t correct?
Wislar
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Yes
amistre64
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then "with respect to t" the line is from t=0 to t=1 since (0,2) is t=0; and (4,3) is t=1
Wislar
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I see!
amistre64
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we are simply defining all the terms with respect to t; x,y and movement
Wislar
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I have another question that asks for the line integral that consists of the arc of the circle x^2+y^2=4 from (2,0 to (0,2)
I know how to find the line integral, but the limits are what I'm not sure about again
Wislar
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The parameterized curve is (2cost,2sint)
amistre64
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is it asking you to define it by the line that goes from 2,0 to 0,2? or from the arc that goes from 2,0 to 0,2 ?
Wislar
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It says the arc of the circle from (2,0) to (0,2)
amistre64
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x = r cos(t)
y = r sin(t)
and from the equation r = 2
x = 2 cos(t)
dx = -2sin(t)
y = 2 sin(t)
dy = 2 cos(t)
what is the value of t for the point x=2, y=0?
what is the value of t for the point x=0, y=2?
Wislar
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cos(t)=1 and sin(t)=1?
amistre64
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2 = 2 cos(t)
0 = 2 sin(t)
cos(t) = 1 and sin(t) = 0 when t = 0
0 = 2 cos(t)
2 = 2 sin(t)
cos(t) = 0 and sin(t) = 1 when t = pi/2
Wislar
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So it's from 0 to pi/2 right?
amistre64
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yes
Wislar
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Thank you so much!!
amistre64
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youre welcome, and good luck :)