Math 127(3) WD Smith, HW#6 ANSWERS ------------------------------------- 7 problems total (1) 2 2 F(x,y) = 5xy exp(-x - y ), to find all flat spots and decide if saddles, maxes, or mins. 2 2 2 2 Grad F = ( 5 y [1-2 x ], 5 x [1-2 y ] ) exp(-x -y ) 2 2 = (0,0) if x=y=0 or x = y = 1/2. So the 5 flat spots are (0,0) and (+-SquareRoot(1/2), +-SquareRoot(1/2)). x=y=0: saddle. WHY (EASY WAY): Behaves just like F(x,y)=xy, the classic saddle example. (+SquareRoot(1/2), +SquareRoot(1/2)) and (-SquareRoot(1/2), -SquareRoot(1/2)): Clearly by sym, these two points behave the same as each other. Both MAX. WHY (EASY WAY): The easiest way to attack this seems to be to use new variables s=x+y and m=x-y. These form a 45-degree rotated coordinate system. In that caseF is proportional to 2 2 2 2 2 2 (s-m) (s+m) exp( -(1/2) (s + m ) = (s -m ) exp( -(1/2) (s + m ) and at our point s=SquareRoot(2), m=0, you can see that the behavior is concave-down in s since 2 2 s exp(-s/ 2) has negative 2nd deriv at s=SquareRoot(2) and meanwhile is concave-down in m (obviously), hence behaves like a max. (By sym, in s,m, the mixed derivative is 0, which was one of the points of doing this.) HARD WAY: One could also use the brute force method in the book: let A = F B = F C = F where the subscripts denote partial derivatives, xx xy yy and find, after a lot of work, 2 discriminant = B - AC = 2 2 2 2 2 2 2 4 4 2 4 4 2 25 exp(-x - y ) (1 - 4 y - 4 x - 20 x y + 4 y + 8 y x + 4 x + 8 x y ) which when (x,y) = (+SquareRoot(1/2), +SquareRoot(1/2)) is discriminant = -100 exp(-2) < 0 hence it must be a min or max, NOT a saddle. Great. Now, it is a MAX since A = -10 exp(-1) < 0. whew! (+SquareRoot(1/2), +SquareRoot(1/2)) (-SquareRoot(1/2), -SquareRoot(1/2)) both MIN. Why: Clearly by sym, these two points behave the same as the above two, only with overall sign changed. ------------------------------------ (2) 2 F(x,y) = x + sin(y) to find all flat spots and decide if saddles, maxes, or mins. Grad F = (2x, cos(y)) = (0,0) if x=0 and y=pi/2 + pi n (n integer) So infinite number of flat spots. If n odd (minimizes sin(y)), min, since min with respect to y, and also with respect to x. 2 If n even (maximizes sin(y), but x=0 is minimizing x ), saddle. SANITY CHECKS: you can try numbers near the point in question with calculator. If always get F(x,y) > F(at the min) then that supports the conclusion it was a min! (Otherwise you know insane!) If always < ..(ditto).. max. If both ways ..(ditto).. saddle. ------------------------------------ (3) F(x,y) = (x+y)(xy+5) to find all flat spots and decide if saddles, maxes, or mins. 2 2 Grad F = (2xy + 5 + y , 2xy + 5 + x ) 2 2 = (0,0) if and only if x = y . Hence one of the following 2 cases: 2 2 x=y and 3x+ 5 = 0 x = -y and 5-x = 0 no solution so x = +-SquareRoot(5) and y = -+SquareRoot(5). So we have these two points: (+SquareRoot(5), -SquareRoot(5)) (-SquareRoot(5), +SquareRoot(5)) which by symmetry both behave the same. To apply the test in the book (instead of ad hoc tricks like above), let A = F B = F C = F where the subscripts denote partial derivatives, xx xy yy and A = 2y B = 2x+2y C=2x, then 2 2 2 discriminant = B - AC = 4 (x + y - xy) which is always >0, in particular at our points. Hence, SADDLE. ------------------------------------ 2 2 2 2 (4) F(x,y) = (x + y ) exp( x - y ) 2 2 2 2 gradF = 2 (x + y + 1) exp(x- y ) ( x, -y ) --- a scaling factor --- vector so gradF=0 iff x=y=0. 2 2 This is a MIN since behaves like x + y near 0 (note is always >=0 for all x,y). ---------------------------------------- (5) div(x,y,z) = dx/dx + dy/dy + dz/dz = 1+1+1 = 3. curl(x,y,z) = (0,0,0). The first 0 in this curl is dz/dy-dy/dz = 0-0. The others are similar. ---------------------------------------- (6) div(3x+y+z, 5y+z, x+y) = d(3x+y+z)/dx + d(5y+z)/dy + d(x+y)/dz = 3+5+0=8. curl(3x+y+z, 5y+z, x+y) = (0, 0, -1). The first-coord 0 is d(x+y)/dy - d(5y+z)/dz = 1-1 = 0. The second-coord 0 is d(x+y)/dx - d(3x+y+z)/dz = 1-1 = 0. The third-coord -1 is d(5y+z)/dx - d(3x+y+z)/dy = 0-1 = -1. ---------------------------------------- (7) To show that the Newton gravity force field 2 2 2 3 F = -GmM (x,y,z) / SquareRoot(x +y +z ) has curlF=(0,0,0) and divF=0: First, we write out in terms of x,y,z as we just did. Second, we can now find the div and curl by using the epxressions in terms of derivatives... simplified due to the symmetry of the expression with x,y,z playing the same role. Thus we only need to find d(F1)/dx and then d(F2)/dy and d(F3)/dz will be the same if letters permuted. We get 2 2 2 2 2 2 5 d(F1)/dx = (-GmM) (y + z - 2x ) / SquareRoot(x +y +z ) so they all (that is, divF = dF1/dx + dF2/dy + dF3/dz) add up to 0 (the -2 cancels +1 and +1...) showing divF = d(F1)/dx + d(F2)/dy + d(F3)/dz = 0. Similarly we really only need to find 1 coordinate of the curl and the other 2 will be the same if letters permuted: 2 2 2 5 d(F2)/dz - d(F3)/dy = (Gmm) (3xy - 3xy) / SquareRoot(x +y +z ) = 0 so the curl is (0,0,0). ---------------------------------------- (8) To show the identities (capital letters will denote vectors, lowercase=scalars) div(f V) = gradf . V + f (div V) This is essentially the product rule for ordinary deriviatives, (a b)' = a' b + a b' used component by component. Note our formula has the same structure as the product rule, but with vector-types of "derivatives" and vector-types of "multiplication." curl(f V) = gradf X V + f (curl V) This too is essentially the product rule for ordinary deriviatives, used component by component. You can work out a component to see this (due to symmetry, for all components, proof works the same). (c) curl(curl V) = grad(div V) - laplacian V You have to work out the components... The known vector identity A . (A X C) = 0 might make you think you get 0, but that would be wrong! The right analogy is to the known vector identity A X (A X C) = A (A . C) - (A . A) C and the proof goes thru exactly the same only using psuedovector del, not vector A. ---------done.