M25008SelSol6

M25008SelSol6 - Mathematics 250 Assignment 6 Selected...

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Fall 2008 Assignment # 6 Selected Soltuions Exercises; MVM 3.6.3, 3.6.6, 3.6.7, 3.6.8. MVM 3.6.6: Let f be a scalar-valued function on R 2 , and let g be a mapping from R 2 to itself. Write g ±² u v ³´ = x ±² u v ³´ y ±² u v ³´ . Set F = f g . Compute 2 F ∂u∂v . Response: The Chain Rule tells us that ∂F ∂v = ∂f ∂x ∂x ∂v + ∂f ∂y ∂y ∂v . To compute 2 F ∂u∂v , we should differentiate the right hand side with respect to u . We can begin by using the sum rule and the product rule. This gives us 2 F ∂u∂v = ∂u ± ∂f ∂x ´ ∂x ∂v + ∂f ∂x ∂u ± ∂x ∂v ´ + ∂u ± ∂f ∂y ´ ∂x ∂v + ∂f ∂y ∂u ± ∂y ∂v ´ In this expression, the terms ∂x ∂v and ∂y ∂v are presumably expressed directly in terms of u and v , and so can be differentiated directly. However, the terms ∂f ∂x and ∂f ∂y are presumably expressed in terms of x and y - they are implicitly composite functions with g . Hence to diffeentiate them with respect to u , we again need the Chain Rule. Implmenting the indicated differentiation according to this scheme gives the result 2 F ∂u∂v = ± 2 f ∂x 2 ∂x ∂u + 2 f ∂y∂x ∂y ∂v ´ ∂x ∂v + ∂f ∂x ± 2 x ∂u∂v ´ + ± 2 f ∂x∂y ∂x ∂u + 2 f ∂y 2 ∂y ∂v ´ ∂x ∂v + ∂f ∂y ± 2 y ∂u∂v ´ = ∂f ∂x ± 2 x ∂u∂v ´ + ∂f ∂y ± 2 y ∂u∂v ´ + 2 f ∂x 2 ± ∂x ∂u ∂x ∂v ´ + 2 f ∂x∂y ± ∂x ∂u ∂y ∂v + ∂y ∂u ∂x ∂v ´ + 2 f ∂y 2 ± ∂y ∂u ∂y ∂v ´ . MVM 3.6.7: Let f be a smooth scalar-valued function on R 2 , and let P ( ² r θ ³ ) = ² r cos θ r sin θ ³ be polar coordi- nates. If F = f P , show that 2 f ∂x 2 + 2 f ∂y 2 = 2 F ∂r 2 + 1 r ∂F ∂r + 1 r 2 2 F ∂θ 2 . (3 . 6 . 7 . 1) Response: Although it may not provide meximum insight, perhaps the simplest approach to establishing this identity is to compute the right hand side following the procedure used for problem MVM 3.6.6. According to that scheme, we first compute ∂F ∂r = ∂f ∂x ∂x ∂r + ∂f ∂y ∂y ∂r = ∂f ∂x cos θ + ∂f ∂y sin θ. (3 . 6 . 7 . 2) We differentiate again with respect to r , keeping in mind that ∂f ∂x and ∂f ∂y are still functions of x and y . The result is slightly simpler than the general case, because ∂x ∂r = cos θ and ∂y ∂r = sin θ do not depend on r . The formula is 2
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This note was uploaded on 09/24/2009 for the course MATH 250 taught by Professor Rogerhowe during the Fall '06 term at Yale.

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M25008SelSol6 - Mathematics 250 Assignment 6 Selected...

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