ch20 - 20.1 SOLUTIONS 1339 CHAPTER TWENTY Solutions for...

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20.1 SOLUTIONS 1339 CHAPTER TWENTY Solutions for Section 20.1 Exercises 1. The frst vector feld appears to be diverging more at the origin, since both felds are zero at the origin and the vectors near the origin are larger in feld (I) than they are in feld (II). 2. div ~ F = ∂x ( - x ) + ∂y ( y ) = - 1 + 1 = 0 3. div ~ F = ∂x ( - y ) + ∂y ( x ) = 0 4. div ~ F = ∂x ( x 2 - y 2 ) + ∂y (2 xy ) = 2 x + 2 x = 4 x 5. div ~ F = ∂x ( - x + y ) + ∂y ( y + z ) + ∂z ( - z + x ) = - 1 + 1 - 1 = - 1 6. Using the Formula For ~a × ~ r in Cartesian coordinates, we get div ~ F = ∂x ( a 2 z - a 3 y ) + ∂y ( a 3 x - a 1 z ) + ∂z ( a 1 y - a 2 x ) = 0 7. Taking partial derivatives, we get div ~ F = ∂x µ - y ( x 2 + y 2 ) + ∂y µ x ( x 2 + y 2 ) = 2 xy ( x 2 + y 2 ) 2 - 2 yx ( x 2 + y 2 ) 2 = 0 . 8. In coordinates, we have ~ F ( x, y, z ) = ( x - x 0 ) p ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ~ i + ( y - y 0 ) p ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ~ j + ( z - z 0 ) p ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ~ k . So iF ( x, y, z ) 6 = ( x 0 , y 0 , z 0 ) , then div ~ F = 1 p ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 - ( x - x 0 ) 2 (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) 3 / 2 ! + 1 p ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 - ( y - y 0 ) 2 (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) 3 / 2 ! + 1 p ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 - ( z - z 0 ) 2 (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) 3 / 2 ! = µ ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) 3 / 2 - ( x - x 0 ) 2 (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) 3 / 2 + µ ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) 3 / 2 - ( y - y 0 ) 2 (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) 3 / 2 + µ ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) 3 / 2 - ( z - z 0 ) 2 (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) 3 / 2 = 3(( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) - (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) (( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 ) 3 / 2 = 2 p ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 = 2 k ~ r - ~ r 0 k .

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1340 Chapter Twenty /SOLUTIONS 9. (a) Positive. The infow From the lower leFt is less than the outfow From the upper right. The net outfow is positive. (b) Zero. The infow on the right side is equal to outfow on the leFt. (c) Negative. The infow From above is greater than the outfow below. The net outfow is negative. 10. Two vector ±elds that have positive divergence everywhere are in ²igures 20.1 and 20.2. x y Figure 20.1 x y Figure 20.2 11. Two vector ±elds that have negative divergence everywhere are in ²igures 20.3 and 20.4. x y Figure 20.3 x y Figure 20.4 12. Two vector ±elds that have zero divergence everywhere are in ²igures 20.5 and 20.6. x y Figure 20.5 x y Figure 20.6
20.1 SOLUTIONS 1341 Problems 13. Since div F (1 , 2 , 3) is the fux density out oF a small region surrounding the point (1 , 2 , 3) , we have div ~ F (1 , 2 , 3) ±lux out oF small region around (1 , 2 , 3) Volume oF region .

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This note was uploaded on 08/26/2009 for the course MATH 1ABCD taught by Professor All during the Winter '08 term at Foothill College.

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ch20 - 20.1 SOLUTIONS 1339 CHAPTER TWENTY Solutions for...

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