anonymous
  • anonymous
Please Check! Line integral: \[\int\limits_{C}^{}x ^{2}dx+y ^{2}dy\] of the line segment from (0,2) to (4,3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Is it: \[\int\limits_{0}^{4}16t ^{2}(4)+(2+t)^{2}(1)dt\]
anonymous
  • anonymous
The parameterized curve is r(t)=(4t, 2+t) right?
anonymous
  • anonymous
@phi @amistre64 @Zarkon @TuringTest

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amistre64
  • amistre64
id have a little reading to do to catch up on this :)
anonymous
  • anonymous
I think I have it down correctly, I'm just not sure if the limit is from 0 to 4
amistre64
  • amistre64
we need to parametrize the line from 0,2 to 4,3 right?
anonymous
  • anonymous
Yes
amistre64
  • amistre64
4,3 -0-2 ----- <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
anonymous
  • anonymous
That's what I got! I'm not sure about the limits of integration though
abb0t
  • abb0t
sketch out the graph.
amistre64
  • amistre64
the dx and dy parts in your post got me confuddled. im used to seeing a ds in it
amistre64
  • amistre64
here we go http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtII.aspx
amistre64
  • amistre64
\[\int_CPdx+Qdy=\int_CPdx+\int_CQdy\]
amistre64
  • amistre64
x = 0 + 4t dx = 4 dt y = 2 + t dy = dt
anonymous
  • anonymous
I have that down @amistre64 I'm just really not sure about the limits. I just used the x coordinates.
amistre64
  • amistre64
what are the limits of t for the given line?
anonymous
  • anonymous
1 and 4?
amistre64
  • amistre64
when t=0 we are at (0,2) when t=? we are at (4,3)
anonymous
  • anonymous
I'm not sure?
amistre64
  • amistre64
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
anonymous
  • anonymous
Wait how did you do that?
amistre64
  • amistre64
we are changing x and y into functions of t correct?
anonymous
  • anonymous
Yes
amistre64
  • amistre64
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
anonymous
  • anonymous
I see!
amistre64
  • amistre64
we are simply defining all the terms with respect to t; x,y and movement
anonymous
  • anonymous
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
anonymous
  • anonymous
The parameterized curve is (2cost,2sint)
amistre64
  • amistre64
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 ?
anonymous
  • anonymous
It says the arc of the circle from (2,0) to (0,2)
amistre64
  • amistre64
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?
anonymous
  • anonymous
cos(t)=1 and sin(t)=1?
amistre64
  • amistre64
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
anonymous
  • anonymous
So it's from 0 to pi/2 right?
amistre64
  • amistre64
yes
anonymous
  • anonymous
Thank you so much!!
amistre64
  • amistre64
youre welcome, and good luck :)

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