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

1. anonymous

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

2. anonymous

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

3. anonymous

@phi @amistre64 @Zarkon @TuringTest

4. amistre64

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

5. anonymous

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

6. amistre64

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

7. anonymous

Yes

8. 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

9. anonymous

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

10. abb0t

sketch out the graph.

11. amistre64

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

12. amistre64
13. amistre64

$\int_CPdx+Qdy=\int_CPdx+\int_CQdy$

14. amistre64

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

15. anonymous

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

16. amistre64

what are the limits of t for the given line?

17. anonymous

1 and 4?

18. amistre64

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

19. anonymous

I'm not sure?

20. 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

21. anonymous

Wait how did you do that?

22. amistre64

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

23. anonymous

Yes

24. 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

25. anonymous

I see!

26. amistre64

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

27. 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

28. anonymous

The parameterized curve is (2cost,2sint)

29. 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 ?

30. anonymous

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

31. 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?

32. anonymous

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

33. 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

34. anonymous

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

35. amistre64

yes

36. anonymous

Thank you so much!!

37. amistre64

youre welcome, and good luck :)