Question

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

Answer 1

@ganeshie8

Answer 2

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

Answer 3

what formula would i apply here?

Answer 4

\(F = Mi + Nj\)

\(\text{curl(F)} = N_x - M_y\)

Answer 5

F(x,y)= (x+2xsiny) i + (x^2cosy +2y) j

M = x + 2xsiny
N = x^2cosy + 2y

N_x = ?
M_y = ?

Answer 6

i take the derivative?

Answer 7

N_x means the partial of N with respect to x

Answer 8

2xcosy?

Answer 9

what is 2xcosy ?

Answer 10

Nx

Answer 11

OK, work the curl

Answer 12

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

Answer 13

.

Answer 14

You need to work M_y too

Answer 15

-2xsiny+2 for My?

Answer 16

M = x + 2xsiny

M_y = ?

Answer 17

1+2cosy

Answer 18

try again

Answer 19

2xcosy

Answer 20

Yes, curl = ?

Answer 21

2xcosy-2xcosy=0

Answer 22

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

Answer 23

Let \(f(x,y)\) be the potential function, then this must satisfy :

\(f_x = M\)
\(f_y = N\)

Answer 24

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

Answer 25

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.

Answer 26

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

Answer 27

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

Answer 28

plug in m and n

Answer 29

\[f_x = x + 2x\sin y\]

simply integrate both sides with respect to \(x\) to get \(f\)

Answer 30

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

Answer 31

try again

Answer 32

and what about the integration constant ?

Answer 33

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

Answer 34

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

Answer 35

Easy,

\[f_x = x + 2x\sin y\]

integrating both sides with respect to \(x\) gives

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

Answer 36

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

Answer 37

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

Answer 38

xsiny why isnt it x^2siny

Answer 39

Oops! my mistake... see if below looks fine

\[f_x = x + 2x\sin y\]

integrating both sides with respect to \(x\) gives

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

Answer 40

i take the third derviative to find g(Y)

Answer 41

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

Answer 42

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

\[f_y = ?\]

Answer 43

Y^2

Answer 44

how

Answer 45

nevermind...is it x^2cosy

Answer 46

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

\[f_y = x^2\cos y + \color{red}{g'(y)}\]

Answer 47

compare this with the other equation \(f_y = N\)

Answer 48

g'(y) is 2y?

Answer 49

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

Answer 50

y^2

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

you must get
\(f_x = M\) and \(f_y = N\)

Answer 56

yes i do

Answer 57

thank you again ganeshie8!!

Answer 58

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:



the line integral is straightforward

Answer 59

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