## anonymous one year ago Medal****Please Help Given the vectorfield F(x,y)= (x+2xsiny) i + (x^2cosy +2y) j Determine whether F is conservative. If it is, find a potential function

1. anonymous

@ganeshie8

2. ganeshie8

Hint : curl must be 0 for the vector field to be conservative

3. anonymous

what formula would i apply here?

4. ganeshie8

$$F = Mi + Nj$$ $$\text{curl(F)} = N_x - M_y$$

5. ganeshie8

F(x,y)= (x+2xsiny) i + (x^2cosy +2y) j M = x + 2xsiny N = x^2cosy + 2y N_x = ? M_y = ?

6. anonymous

i take the derivative?

7. ganeshie8

N_x means the partial of N with respect to x

8. anonymous

2xcosy?

9. ganeshie8

what is 2xcosy ?

10. anonymous

Nx

11. ganeshie8

OK, work the curl

12. anonymous

2xcosy-My....i dont know the My

13. Loser66

.

14. ganeshie8

You need to work M_y too

15. anonymous

-2xsiny+2 for My?

16. ganeshie8

M = x + 2xsiny M_y = ?

17. anonymous

1+2cosy

18. ganeshie8

try again

19. anonymous

2xcosy

20. ganeshie8

Yes, curl = ?

21. anonymous

2xcosy-2xcosy=0

22. ganeshie8

Since the curl is 0, the given vector field is conservative and a potential function exists

23. ganeshie8

Let $$f(x,y)$$ be the potential function, then this must satisfy : $$f_x = M$$ $$f_y = N$$

24. anonymous

what if it wasnt 0, it would be non-conservative but still a potential function?

25. ganeshie8

If the curl is not 0, then the vector field is not conservative and consequently there will not be a potential function. You cannot find a potential function.

26. ganeshie8

potential function for a vector field exists if and only if the curl of vector field is 0

27. anonymous

oki! from fx=m fy=n....whats the next step?

28. ganeshie8

plug in m and n

29. ganeshie8

$f_x = x + 2x\sin y$ simply integrate both sides with respect to $$x$$ to get $$f$$

30. anonymous

x^2/2 +2x^3/3siny?

31. ganeshie8

try again

32. ganeshie8

and what about the integration constant ?

33. anonymous

x.... x^2/2+x^2siny

34. anonymous

x^2/2+2x^2/2siny

35. ganeshie8

Easy, $f_x = x + 2x\sin y$ integrating both sides with respect to $$x$$ gives $f = \frac{x^2}{2} + x\sin y + \color{red}{g(y)}$

36. ganeshie8

that $$\color{red}{g(y)}$$ is the arbitrary constant, it shows up everytime you integrate

37. ganeshie8

you need to find $$\color{red}{g(y)}$$

38. anonymous

xsiny why isnt it x^2siny

39. ganeshie8

Oops! my mistake... see if below looks fine $f_x = x + 2x\sin y$ integrating both sides with respect to $$x$$ gives $f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)}$

40. anonymous

i take the third derviative to find g(Y)

41. ganeshie8

There is an easy way take the derivative of $$f$$ with respect to $$y$$ and compare it with the equation $$f_y = N$$

42. ganeshie8

$f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)}$ $f_y = ?$

43. anonymous

Y^2

44. ganeshie8

how

45. anonymous

nevermind...is it x^2cosy

46. ganeshie8

$f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)}$ $f_y = x^2\cos y + \color{red}{g'(y)}$

47. ganeshie8

compare this with the other equation $$f_y = N$$

48. anonymous

g'(y) is 2y?

49. ganeshie8

Yes, integrate and solve $$g(y)$$

50. anonymous

y^2

51. ganeshie8

Looks good! plug that in the potential function and you're done!

52. ganeshie8

$f = \frac{x^2}{2} + x^2\sin y + \color{red}{g(y)} = \frac{x^2}{2} + x^2\sin y + \color{red}{y^2}$

53. anonymous

thank you so much! x^2/2+x^2siny+Y^2

54. ganeshie8

just a sanity check : find the partials and see if you really get M and N

55. ganeshie8

you must get $$f_x = M$$ and $$f_y = N$$

56. anonymous

yes i do

57. anonymous

thank you again ganeshie8!!

58. IrishBoy123

for completeness, there's a really handy way around the final step that avoids fiddling around with the integration constants. because the field is conservative, it is path independent so you can do a line integral. $$W = \int \vec F \bullet \vec dr = \int <x+2xsiny, x^2cosy +2y> \bullet <dx, dy>$$ from (0,0) to (x,y) using, for example, these steps: |dw:1436724170339:dw| the line integral is straightforward

59. ganeshie8

I prefer the line integral too (ofcourse when eyeballing method fails :))