Question

THEOREM. Suppose that F(x, y) = (P(x, y), Q(x, y)) is a vector-valued function of two variables and that the domain of P(x,y) and Q(x,y) is all of R2. Then it is possible to find a function f(x,y) satisfying Vf = F if and only if Py = Q. Instructions: Use this Theorem to test whether or not each of the following vector-valued functions F(x,y) has a function f(x, y) that satisfies VS = F (that is, if there is an "anti- gradient" of F). If so, then find all possible such functions. 1. F(x, y) = (4.xºy- 3r”, 3x"y? + 2y) 2. F(x,y) = (3ry",y) 3. F(x,y) = (2x sin y + 2r?cos y + sec? y) 4. F(x,y) = (xy?, ²y + 2) 5. F(x, y) = (10.3e" + 2,5x?e+ 2x) 6. F(x,y) = VY y V + 2.r, "257 -2V7+5) *Note: The domain D of the function F(,y) in Example 6 is not R², it is the first quadrant D = {(x,y) € R2 > 0.y >0}. However, the Theorem holds as stated above for this example as well. We will discuss these issues in class next week.

Answer 1

© Give, f(@y)= (42205-392, 32442 +245 Play) = 42 343 . 3 x² .8 324 y² + 245 123843 127 %y? Py: Bx - of af holds. . 20 ap ay 02 12 Let 마 - APR at 42 343 32² ax = + = 9943- 13.23 8, (y) af 324 y² + 2ys oy 5 to 2479 + 492 +0,0) Take a = 4 ½ d₂ = -23 ::f=7943-93 +4% F(1.4) (22y2, 43) Q = 4² B x = 0 . Pyta = F doesn't hold MAY 7 . 2 JUN NOTES P: 3 x y Py bay of