1211s4 - THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1211 Multivariable Calculus 2010-11 1st Semester: Solution to Assignment 4 1. Let x ( t ) = ( x ( t ) , y ( t ) , z ( t )). Since z ( t ) = z 3 ( t ), we have z 3 dz = dt . Integrating both sides, z 2 ( t ) = (2 t + 2 C ) 1 for some constant C . Substituting z (0) = 7, we get 49 = 1 / (2 C ) or equivalently, C = 1 / 98. Therefore, z ( t ) = (1 / 49 2 t ) 1 2 = 7(1 98 t ) 1 2 . For the y- component, we have y ( t ) = 3 y ( t ), and the solution is y ( t ) = y (0) e 3 t = 5 e 3 t . Since x ( t ) satisfies x ( t ) = 2 y ( t ) = 10 e 3 t , integrating both sides and we get x ( t ) = x (0) + 10 3 (1 e 3 t ) = 19 3 10 3 e 3 t . Combining, we have x ( t ) = ( 19 3 10 3 e 3 t , 5 e 3 t , 7 1 98 t ) . 2. (a) Guess f ( x, y, z ) = x 2 y 2 3 z . Then we have f = f x i + f y j + f z k = 2 x i 2 y j 3 k = F and it shows that F is a gradient field. (Remark: the function f is found by solving f x = 2 x , f y = 2 y and f z = 3.) (b) The equipotential surfaces of F ares the level sets { ( x, y, z ) | f ( x, y, z ) = c } , or equivalently, { ( x, y, z ) | 3( z + c/ 3) = x 2 y 2 } , where c is a constant. Referring to Page 90 in the textbook, the equipotential surfaces are hyperbolic paraboloids. 3. By the chain rule, we have G ( t ) = d dt f ( x ( t )) = f ( x ( t )) · x ( t ) = f ( x ( t )) · ∇ f ( x ( t )) 0 , and hence G is an increasing function of t . 4. Write F = ( F 1 , F 2 , F 3 ). Using curl F = ∇ × F = ( ∂F 3 ∂y ∂F 2 ∂z ) i + ( ∂F 1 ∂z ∂F 3 ∂x ) j + ( ∂F 2 ∂x ∂F 1 ∂y ) k , we have div (curl F ) = ∇ · ( ∇ × F ) = ∂x ( ∂F 3 ∂y ∂F 2 ∂z ) + ∂y ( ∂F 1 ∂z ∂F 3 ∂x ) + ∂z ( ∂F 2 ∂x ∂F 1 ∂y ) . By the equality of mixed partials,
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This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.

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1211s4 - THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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