ps3-solns-2009

ps3-solns-2009 - 18.02A pset 3, part II solutions, fall...

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Unformatted text preview: 18.02A pset 3, part II solutions, fall 2009 Problem 1 a) C =- 2 x 10 4 e- ( x 2 + y 2 ) / 10 4 ,- 4 y 10 4 e- ( x 2 + y 2 ) / 10 4 =- 2 C ( x,y ) 10 4 h x, 2 y i . The radial direction at ( x,y ) is b u = h x,y i p x 2 + y 2 . dC ds b u = C b u =- 2 C ( x,y ) 10 4 ( x 2 + 2 y 2 ) p x 2 + y 2 . b) The path must have the same slope as the gradient vector at each point. That is, dy dx = 2 y x . In 18.01 you learned to separate variables and integrate: dy y = 2 dx x ln y = ln( x 2 ) + C y = Kx 2 . Using the starting position to determine K , we get the sharks path is along y = y x 2 x 2 . Problem 2 a) The surface is the level surface w = z 2 + x 2 y 2 + y 3 + x 2 = 4. The normal is w = h 2 xy 2 + 2 x, 2 yx 2 + 3 y 2 , 2 z i . At (1 , 1 , 1) we get w = h 4 , 5 , 2 i the equation of the tangent plane is 4 x + 5 y + 2 z = 11. b) We want to minimize the distance squared = f ( x,y,z ) = x 2 + y 2 + z 2 subject to the constraint 4 x + 5 y + 2 z = 11. Lagrange multipliers gives: 2 x = 4 , 2 y = 5 , 2 z = 2 , 4 x + 5 y + 2 z = 11. Substituting for x , y , z in terms of in the constraint gives 8 + 25 2 + 2 = 11 = 22 45 the one critical point is 11 45 (4 , 5 , 2) minimum distance = 11 45 45 . Problem 3 We want to minimize distance squared = x 2 + ( y- b ) 2 , subject to the constraint y = x 2 ....
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ps3-solns-2009 - 18.02A pset 3, part II solutions, fall...

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