Problem 8a spring 2008 given the function f x y x 2 y

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Problem 8(a) - Spring 2008 Given the function f ( x , y ) = x 2 y + ye xy . Find the linearization L ( x , y ) of f at the point (0 , 5) and use it to approximate the value of f at the point ( . 1 , 4 . 9). (An unsupported numerical approximation to f ( . 1 , 4 . 9) will not receive credit.)
Problem 8(b) - Spring 2008 Given the function f ( x , y ) = x 2 y + ye xy . Suppose that x ( r , θ ) = r cos θ and y ( r , θ ) = r sin θ . Calculate f θ at r = 5 and θ = π 2 .
Problem 8(b) - Spring 2008 Given the function f ( x , y ) = x 2 y + ye xy . Suppose that x ( r , θ ) = r cos θ and y ( r , θ ) = r sin θ . Calculate f θ at r = 5 and θ = π 2 .
Problem 8(c) - Spring 2008 Given the function f ( x , y ) = x 2 y + ye xy . Suppose a particle travels along a path ( x ( t ) , y ( t )), and that F ( t ) = f ( x ( t ) , y ( t )) where f ( x , y ) is the function defined above. Calculate F 0 (3), assuming that at time t = 3 the particle’s position is ( x (3) , y (3)) = (0 , 5) and its velocity is ( x 0 (3) , y 0 (3)) = (3 , - 2).
Problem 8(c) - Spring 2008 Given the function f ( x , y ) = x 2 y + ye xy . Suppose a particle travels along a path ( x ( t ) , y ( t )), and that F ( t ) = f ( x ( t ) , y ( t )) where f ( x , y ) is the function defined above. Calculate F 0 (3), assuming that at time t = 3 the particle’s position is ( x (3) , y (3)) = (0 , 5) and its velocity is ( x 0 (3) , y 0 (3)) = (3 , - 2).
Problem 9(a) - Spring 2008 Consider the function f ( x , y ) = 2 p x 2 + 4 y . Find the directional derivative of f ( x , y ) at P = ( - 2 , 3) in the direction starting from P pointing towards Q = (0 , 4).
Problem 9(a) - Spring 2008 Consider the function f ( x , y ) = 2 p x 2 + 4 y . Find the directional derivative of f ( x , y ) at P = ( - 2 , 3) in the direction starting from P pointing towards Q = (0 , 4).
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Problem 9(b) - Spring 2008 Consider the function f ( x , y ) = 2 p x 2 + 4 y . Find all unit vectors u for which the directional derivative D u f ( - 2 , 3) = 0.