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math226fall11mt2_10s

# math226fall11mt2_10s - xy 1(a Determine whether f(x y z = 1...

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1 . (a) Determine whether f ( x, y, z ) = xy 1+ y 2 z is increasing or decreasing at (1 , 1 , 0) toward the point (3 , - 1 , 1) . At what rate? What is the maximal increase rate and in what direction it occurs? (b) Write the equation of the tangent plane to the surface S : xy 1+ y 2 z = 1 at (1 , 1 , 0). Answer. (a) A vector pointing from (1 , 1 , 0) toward (3 , - 1 , 1) is ~a = h 2 , - 2 , 1 i . The rate of change of f at (1 , 1 , 0) in ~a -direction is the number D ~u f (1 , 1 , 0) = f (1 , 1 , 0) · ~u , where ~u = ~a/ | ~a | = h 2 , - 2 , 1 i | ~a | = 2 3 , - 2 3 , 1 3 . So we find f x = y 1 + y 2 z , f y = x (1 + y 2 z ) - 2 xy 2 z (1 + y 2 z ) 2 , f z = - xy (1 + y 2 z ) 2 y 2 , f (1 , 1 , 0) = h 1 , 1 , - 1 i and D ~u f (1 , 1 , 0) = - 1 3 < 0 . The function f is decreasing at (1 , 1 , 0) in the direction of ~u at the rate 1 3 . The maximal increase rate is |∇ f (0 , 1 , 1) | = 3 and it occurs in the direction of f (0 , 1 , 1) = h 1 , 1 , - 1 i . (b) The tangent plane passes through (1 , 1 , 0) and is perpendicular to f (1 , 1 , 0) = h 1 , 1 , - 1 i : (1)( x - 1) + y - 1 + ( - 1)( z ) = 0 or x + y - z - 2 = 0 . 2. (a) Determine the set of points at which the function is continuous: f ( x, y ) = ( x 2 y 2 x 2 + y 4 , ( x, y ) 6 = (0 , 0) , 0 , ( x, y ) = (0 , 0) .

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math226fall11mt2_10s - xy 1(a Determine whether f(x y z = 1...

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