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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)

- anonymous

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- anonymous

Is it:
\[\int\limits_{0}^{4}16t ^{2}(4)+(2+t)^{2}(1)dt\]

- anonymous

The parameterized curve is r(t)=(4t, 2+t) right?

- anonymous

@phi @amistre64 @Zarkon @TuringTest

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## More answers

- amistre64

id have a little reading to do to catch up on this :)

- anonymous

I think I have it down correctly, I'm just not sure if the limit is from 0 to 4

- amistre64

we need to parametrize the line from 0,2 to 4,3 right?

- anonymous

Yes

- 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

That's what I got! I'm not sure about the limits of integration though

- abb0t

sketch out the graph.

- amistre64

the dx and dy parts in your post got me confuddled. im used to seeing a ds in it

- amistre64

here we go
http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtII.aspx

- amistre64

\[\int_CPdx+Qdy=\int_CPdx+\int_CQdy\]

- amistre64

x = 0 + 4t
dx = 4 dt
y = 2 + t
dy = dt

- anonymous

I have that down @amistre64 I'm just really not sure about the limits. I just used the x coordinates.

- amistre64

what are the limits of t for the given line?

- anonymous

1 and 4?

- amistre64

when t=0 we are at (0,2)
when t=? we are at (4,3)

- anonymous

I'm not sure?

- 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

Wait how did you do that?

- amistre64

we are changing x and y into functions of t correct?

- anonymous

Yes

- 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

I see!

- amistre64

we are simply defining all the terms with respect to t; x,y and movement

- 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

The parameterized curve is (2cost,2sint)

- 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

It says the arc of the circle from (2,0) to (0,2)

- 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

cos(t)=1 and sin(t)=1?

- 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

So it's from 0 to pi/2 right?

- amistre64

yes

- anonymous

Thank you so much!!

- amistre64

youre welcome, and good luck :)

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