Solutions3

Solutions3 - 1 Mathematics 1c: Solutions, Homework Set 3...

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1 Mathematics 1c: Solutions, Homework Set 3 Due: Monday, April 19th by 10am. 1. (10 Points) Section 3.1, Exercise 16 Let w = f ( x,y ) be a function of two variables, and let x = u + v, y = u - v. Show that 2 w ∂u∂v = 2 w ∂x 2 - 2 w ∂y 2 . Solution. By the chain rule, ∂w ∂v = ∂w ∂x · ∂x ∂v + ∂w ∂y · ∂y ∂v = w x - w y . Thus, 2 w ∂u∂v = ∂u ± ∂w ∂v ² = ∂u ( w x - w y ) = ∂u w x - ∂u w y = ∂w x ∂x · ∂x ∂u + ∂w x ∂y · ∂y ∂u - ± ∂w y ∂x · ∂x ∂u + ∂w y ∂y · ∂y ∂u ² = w xx + w xy - ( w yx + w yy ) = w xx - w yy i.e., 2 w ∂u∂v = 2 w ∂x 2 - 2 w ∂y 2 . 2. (10 Points) Section 3.1, Exercise 22 (a) Show that the function g ( x,t ) = 2 + e - t sin x satisfies the heat equation: g t = g xx . [ Here g ( x,t ) represents the temper- ature in a metal rod at position x and time t . ] (b) Sketch the graph of g for t 0 . ( Hint: Look at sections by the planes t = 0 ,t = 1 , and t = 2 . ) (c) What happens to g ( x,t ) as t → ∞ ? Interpret this limit in terms of the behavior of heat in the rod. Solution. (a) Since g ( x,y ) = 2 + e - t sin x , then g t = - e - t sin x,g x = e - t cos x, and g xx = - e - t sin x . Therefore, g t = g xx . (b) The graph of g and the times t = 0 , 1 , and 2 is shown in the figure—try this yourself on the computing site.
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2 –4 –2 0 2 4 0 1 2 3 4 0 1 2 3 4 t x z t = 0 t = 1 t = 2 (c) Note that lim t →∞ g ( x,t ) = lim t →∞ (2 + e - t sin x ) = 2 This means that the temperature in the rod at position x tends to be a constant (= 2) as the time t is large enough. 3. (10 Points) Section 3.2, Exercise 6 Determine the second-order Taylor for- mula for the function f ( x,y ) = e ( x - 1) 2 cos y expanded about the point x 0 = 1 ,y 0 = 0. Solution.
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This note was uploaded on 07/19/2010 for the course MA 1C taught by Professor Ramakrishnan during the Spring '08 term at Caltech.

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Solutions3 - 1 Mathematics 1c: Solutions, Homework Set 3...

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