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1. Let w = f (x, y) where x = u + v and y = uv. a) Find ∂w ∂u b) Find ∂2 w ∂u∂v . and ∂w ∂v in terms of partial derivatives of w with respect to x and y. a) First use the chain rule for 1 variable namely ∂w ∂x ∂w ∂y ∂w ∂w ∂w = + = + v ∂u ∂x ∂u ∂y ∂u ∂x ∂y (1) ∂w ∂x ∂w ∂y ∂w ∂w ∂w = + = + u ∂v ∂x ∂v ∂y ∂v ∂x ∂y (2) and also b) For this part what we need to do is realize that must differentiate each object carefully. Start with this is h(u, v) = ∂w ∂x (x, y). Now we must take ∂h ∂u ∂2w ∂u∂v ∂w ∂x . = ∂ ∂w ∂u ∂v thus we We know that really and use the chain rule for this. Using the chain rule this equals ∂ ∂w ∂ 2 w ∂x ∂ 2 w ∂y ∂h = = + . ∂u ∂u ∂x ∂x2 ∂u ∂y∂x ∂u (3) Then evaluating the first partials we get that this equals ∂ 2w ∂2w ∂ ∂w + = v 2 ∂u ∂x ∂x ∂y∂x Now if we take the right hand side of (2) if we want to take (4) ∂ ∂u ∂w ∂y u then we must use both the product rule and the chain rule. Using the product rule we get that ∂w ∂ ∂w ∂ ∂w u+ u = ∂u ∂y ∂u ∂y ∂y (5) Now to get the last ingredient in our recipe for chain rule fun we must use the 1 chain rule again, This time on the function g(u, v) = ∂w ∂y (x, y). This gives us the following ∂g ∂ ∂w ∂ 2 w ∂x ∂ 2 w ∂y = = + ∂u ∂u ∂y ∂x∂y ∂u ∂y 2 ∂u (6) then adding in the first partials we get that this equals ∂2w ∂2w v + ∂x∂y ∂y 2 (7) Taking all of the ingredients together and mixing gives us that ∂2w ∂w ∂2w ∂2w ∂2w ∂2w + uv + = v + u + 2 2 ∂u∂v ∂x ∂y∂x ∂x∂y ∂y ∂y If conditions on f are nice enough we have that ∂2 w ∂x∂y = ∂2 w ∂y∂x (8) and thus that ∂2w ∂w ∂2w ∂2w ∂2w + uv + = (u + v) + 2 ∂u∂v ∂x ∂x∂y ∂y 2 ∂y (9) Now let us do some double integrals Problem 26 asks us to calculate the mass and the center of mass for the lamina bounded by x = y 2 and x = 4 with the given density ρ(x, y) = y + 3. We are in the following region enclosed in red 2 y 2 1 x 1 2 3 4 5 6 7 8 9 −1 −2 Let us take M the center of mass given on this region by M= ZZ ρ(x, y)dA = R 4 Z Z √ x √ − x 0 (y + 3)dxdy = Z 4 0 √ y2 x + 3y|−√x 2 (10) Taking (10) we get that this equals Z 4 Z 4 √ √ √ 3 x x 6 xdx = 4x 2 |40 = 32. ( + 3 x) − ( − 3 x) dx = M= 2 2 0 0 (11) Then we have that for Mx we have that Mx = ZZ R yρ(x, y)dA = Z 4 0 Z √ x √ − x (y 2 + 3y)dydx = Z 4 3 3 √x y + y 2 |−√x dx (12) 3 2 0 reducing (12) we find that Mx = Z 0 4 4 5 4 128 2 3 x 2 dx = x 2 |0 = 3 15 15 3 (13) For My we then similarly compute that My = ZZ xρ(x, y)dA = R Z 0 4 Z √ x √ − x (xy + 3x)dydx = Z 4 x 0 √ y2 x + 3xy|−√x dx (14) 2 expanding (14) we find that it equals My = Z 4 0 Z (x + 3x ) − (x − 3x ) dx = 2 3 2 2 3 2 0 4 3 6x 2 dx = 12 5 4 384 x 2 |0 = 5 5 M (15) x Putting (11), (13) and (15) we have that the center of mass is ( My , M M ) = 4 ( 12 5 , 5 ). Hopefully we all understand lamina and double integrals better now. 4