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

1. Wislar Group Title

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

2. Wislar Group Title

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

3. Wislar Group Title

@phi @amistre64 @Zarkon @TuringTest

4. amistre64 Group Title

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

5. Wislar Group Title

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

6. amistre64 Group Title

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

7. Wislar Group Title

Yes

8. amistre64 Group Title

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. Wislar Group Title

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

10. abb0t Group Title

sketch out the graph.

11. amistre64 Group Title

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

12. amistre64 Group Title
13. amistre64 Group Title

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

14. amistre64 Group Title

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

15. Wislar Group Title

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

16. amistre64 Group Title

what are the limits of t for the given line?

17. Wislar Group Title

1 and 4?

18. amistre64 Group Title

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

19. Wislar Group Title

I'm not sure?

20. amistre64 Group Title

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. Wislar Group Title

Wait how did you do that?

22. amistre64 Group Title

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

23. Wislar Group Title

Yes

24. amistre64 Group Title

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. Wislar Group Title

I see!

26. amistre64 Group Title

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

27. Wislar Group Title

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. Wislar Group Title

The parameterized curve is (2cost,2sint)

29. amistre64 Group Title

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. Wislar Group Title

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

31. amistre64 Group Title

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. Wislar Group Title

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

33. amistre64 Group Title

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. Wislar Group Title

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

35. amistre64 Group Title

yes

36. Wislar Group Title

Thank you so much!!

37. amistre64 Group Title

youre welcome, and good luck :)