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Unformatted text preview: Math 53 Homework 12 Solutions 16.5 # 1: a) curl vector F = vector F = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle k /x /y /z xyz x 2 y vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = ( x 2 0) ( 2 xy xy ) + (0 xz ) k = x 2 + 3 xy xz k. b) div vector F = vector F = x ( xyz ) + y (0) + z ( x 2 y ) = yz + 0 + 0 = yz . 16.5 # 9: If we write vector F = P + Q , then P = 0, while Q is independent of x (and z ), and decreases as y increases, so Q/y < 0. a) div vector F = P x + Q y = 0 + Q y < 0. b) since we are considering a 2D vector field, the x and y components of the curl vanish, and curl vector F = parenleftbigg Q x P y parenrightbigg k = (0 0) k = 0. (This illustrates the interpretation of divergence for velocity fields: the field vector F corresponds to a ow that compresses areas, hence div < 0. On the other hand, there is no spinning motion, hence curl is zero). 16.5 # 11: If we write vector F = P + Q , then Q = 0, while P is independent of x (and z ), and increases as y increases, so P/y > 0. a) div vector F = P x + Q y = 0 + 0 = 0. b) since we are considering a 2D vector field, the x and y components of the curl vanish, and curl vector F = parenleftbigg Q x P y parenrightbigg k = P y k points in the negative z direction. (This illustrates the interpretation of curl for velocity fields in terms of rotation, see also exercise #37. The ow described by vector F is neither compressing nor expanding, hence diver gence is zero; however the shearing motion causes a particle placed in this velocity field to spin clockwise). 16.5 # 15: vector F = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle k /x /y /z 2 xy x 2 + 2 yz y 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = (2 y 2 y ) (0 0) + (2 x 2 x ) k = 0. Since vector F is defined everywhere and curl vector F = 0, we conclude that vector F is conservative. f ( x,y,z ) is a potential function if it satisfies: f x = 2 xy , f y = x 2 + 2 yz , f z = y 2 . Integrating f x = 2 xy , we get that f ( x,y,z ) = x 2 y + g ( y,z ) for some function g ( y,z ). Thus f y = x 2 + g y = x 2 + 2 yz , which implies that g y = 2 yz , hence g ( y,z ) = y 2 z + h ( z ), and f ( x,y,z ) = x 2 y + y 2 z + h ( z ). Finally, f z = y 2 + h ( z ) = y 2 , so h ( z ) = 0, so h ( z ) = c for...
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 Fall '07
 Hutchings
 Math

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