hw5_solutions

Hw5_solutions - February 6th 2009 Math 20e Assignment#5 Solutions Problem 1(page 310#2 Find the divergence of the vector eld V(x y z = zy i xz j xy

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February 6th, 2009 Math 20e - Assignment #5 Solutions Problem 1 (page 310 #2) . Find the divergence of the vector field V ( x,y,z ) = zy i + xz j + xy k . Solution. Let V ( x,y,z ) = h yz,xz,xy i = h V 1 ( x,y,z ) ,V 2 ( x,y,z ) ,V 3 ( x,y,z ) i . Then we com- pute the divergence by taking div V = ∇ · V = ∂V 1 ∂x + ∂V 2 ∂y + ∂V 3 ∂z = ∂x ( yz ) + ∂y ( xz ) + ∂z ( xy ) = 0 + 0 + 0 = 0 . Problem 2 (page 311 #7) . Sketch a few flow lines for F ( x,y ) = y i . Calculate ∇ · F and explain why your answer is consistent with your sketch. Solution. ∇ · F = ∂F 1 ∂x + ∂F 2 ∂y = ∂x ( y ) + ∂y (0) = 0 . 1
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F = ∇ · F = 0 is consistent with the graph because zero divergence means the vector field should have no expansions or contractions, and the flow lines do not spread out or get closer together. Problem 3 (page 311 #8) . Sketch a few flow lines for F ( x,y ) = - 3 x i - y j . Calculate ∇· F and explain why your answer is consistent with your sketch. Solution. ∇ · F = ∂F 1 ∂x + ∂F 2 ∂y = ∂x ( - 3 x ) + ∂y ( - y ) = - 4 . div F = ∇ · F = - 4 is consistent with the graph because negative divergence means the vector field should contract, and the flow lines do not spread out or get closer together. Problem 4 (page 312 #26) . Show that F = ( x 2 + y 2 ) i - 2 xy j is not a gradient field. Solution. If F was a gradient field, then we could write F = f for some scalar function f . But then by Theorem 1 on page 303 we know that the curl of a gradient is always zero, which means 0 = ∇ × ∇ f = ∇ × F. 2
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This note was uploaded on 01/30/2010 for the course MATH 20E 20E taught by Professor Enright during the Fall '09 term at UCSD.

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Hw5_solutions - February 6th 2009 Math 20e Assignment#5 Solutions Problem 1(page 310#2 Find the divergence of the vector eld V(x y z = zy i xz j xy

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