Please Check! 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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Please Check! Line integral: \[\int\limits_{C}^{}x ^{2}dx+y ^{2}dy\] of the line segment from (0,2) to (4,3)

Mathematics
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Is it: \[\int\limits_{0}^{4}16t ^{2}(4)+(2+t)^{2}(1)dt\]
The parameterized curve is r(t)=(4t, 2+t) right?

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

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