Determine whether or not F is a conservative vector field. If it is, find a function f such that F = ∇f. (If the vector field is not conservative, enter DNE.) F(x, y) = (y2 − 4x)i + 2xyj

f = xy²-2x² satisfies that  F = ∇f.

Step-by-step explanation:

F(x,y) = (y² - 4x) i + 2xy j

We want f(x,y) such that

Lets find a primitive of y²-4x respect to the variable x. We need to think y (and y²) as constants here, so a primitive of y² would be xy² the same way that a primitive of k is xk (because we treat y² as constant). A primitive of x is x²/2, thus a primitive of 4x is 2x². Thus, a primitive of y²-4x is xy² - 2x². We can obtain any other primitive by summing a constant, however since we treated y as constant, then we have that

where c(y) only depends on y (thus, it is constant repsect with x).

We will derivate the expression in terms of y to obtain information about c(y)

Thus, is constant. We can take f(x,y) = xy²-2x². This function f satisfies that  F = ∇f.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.