Question

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

Answer 1

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

Answer 2

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

Answer 3

@phi @amistre64 @Zarkon @TuringTest

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

Yes

Answer 8

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

Answer 9

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

Answer 10

sketch out the graph.

Answer 11

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

Answer 12

here we go

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

Answer 13

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

Answer 14

x = 0 + 4t
dx = 4 dt

y = 2 + t
dy = dt

Answer 15

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

Answer 16

what are the limits of t for the given line?

Answer 17

1 and 4?

Answer 18

when t=0 we are at (0,2)

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

Answer 19

I'm not sure?

Answer 20

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

Answer 21

Wait how did you do that?

Answer 22

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

Answer 23

Yes

Answer 24

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

Answer 25

I see!

Answer 26

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

Answer 27

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

Answer 28

The parameterized curve is (2cost,2sint)

Answer 29

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 ?

Answer 30

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

Answer 31

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?

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

yes

Answer 36

Thank you so much!!

Answer 37

youre welcome, and good luck :)