15_2_10 - = z 2-x 2-y 2 2 xz = z 2-( 1 + A 2 ) x 2 2 zx ....

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10. F = 2 x z i + 2 y z j - x 2 + y 2 z 2 k = G + k , where G is the vector field F of Exercise 9. Since G is conservative (except on the plane z = 0), so is F , which has scalar potential φ( x , y , z ) = x 2 + y 2 z + z = x 2 + y 2 + z 2 z , since x 2 + y 2 z is a potential for G and z is a potential for the vector k . The equipotential surfaces of F are φ( x , y , z ) = C , or x 2 + y 2 + z 2 = Cz which are spheres tangent to the xy -plane having centres on the z -axis. The field lines of F satisfy dx 2 x z = dy 2 y z = dz 1 - x 2 + y 2 z 2 . As in Exercise 9, the first equation has solutions y = Ax , representing vertical planes containing the z -axis. The remaining equations can then be written in the form dz dx
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Unformatted text preview: = z 2-x 2-y 2 2 xz = z 2-( 1 + A 2 ) x 2 2 zx . This first order DE is of homogeneous type (see Section 9.2), and can be solved by a change of dependent variable: z = x v( x ) . We have v + x d v dx = dz dx = x 2 v 2-( 1 + A 2 ) x 2 2 x 2 v x d v dx = v 2-( 1 + A 2 ) 2 v-v = -v 2 + ( 1 + A 2 ) 2 v 2 v d v v 2 + ( 1 + A 2 ) = -dx x ln ( v 2 + ( 1 + A 2 ) ) = -ln x + ln B v 2 + 1 + A 2 = B x z 2 x 2 + 1 + A 2 = B x z 2 + x 2 + y 2 = Bx ....
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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