# ex8 - Klaus Schmitt WeBWorK problems 1(1 pt Compute the...

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Klaus Schmitt Math 2210-2 Fall 2002 WeBWorK problems. WeBWorK assignment 8 due 11/26/02 at 11:59 PM. 1. (1 pt) Compute the gradient vector fields of the following functions: A. f ( x , y ) = 6 x 2 + 10 y 2 f ( x , y ) = 12 x i + 20 y j B. f ( x , y ) = x 2 y 9 , f ( x , y ) = 2 xy 9 i + 9 x 2 y 8 j C. f ( x , y ) = 6 x + 10 y f ( x , y ) = 6 i + 10 j D. f ( x , y , z ) = 6 x + 10 y + 2 z f ( x , y ) = 6 i + 10 j + 2 k E. f ( x , y , z ) = 6 x 2 + 10 y 2 + 2 z 2 f ( x , y , z ) = 12 x i + 20 y j + 4 z k . 2. (1 pt) Match the following vector fields with the verbal descriptions of the level curves or level surfaces to which they are perpendicular by putting the letter of the verbal description to the left of the number of the vector field. Solution: If we can find, for the given vector field F a scalar field f such that f = F , then the vector field will be normal to the surface f ( x , y , z ) = constant . That is, we need to see whether the given vector field is conservative. 1. F = 2 x i + y j + z k . Using Theorem C, page 744, we see that this is a conservative vector field. To find f w eproceed as follows: Since f x = 2 x , f ( x , y , z ) = x 2 + k ( y , z ) . Therefore k y = y , i . e ., k ( y , z ) = 1 2 y 2 + g ( z ) and g prime ( z ) = z . The surface is hence given by f ( x , y , z ) = x 2 + 1 2 y 2 + z 2 = constant , which is the surfcae of an ellipsoid. 2. F = x i + y j - z k . We proceed in a similar manner to the above and obtain the surface x 2 + y 2 - z 2 = constant , which is the surface of a hyperboloid (circular in planes parallel to the x , y plane). 3. F = x i + y j - k . Proceeding as above, we obtain the sur- faces f ( x , y , z ) = x 2 + y 2 - z = constant which is a collection of paraboloids. 4. F = 2 i + j . The levels in this case are a family of curves given by 2 x + y = constant , which are parallel straight lines. 5. F = x i + y j + z k . Here f ( x , y , z ) = x 2 + y 2 + z 2 = constant , which is a family of concentric spheres cen- tered at the origin. 6. F = y i + x j . Here M = y , N = x , so M y = 1 = N x from which we conclude that f ( x , y ) = xy = constant which is a family of hyperbolas. 7. F = 2 i + j + k . In this case we obtain f ( x , y , z ) = 2 x + y + z = constant which is a family of parallel planes.

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• Fall '07
• RickRugangYe
• Vector Calculus, Vector field, hyperbolas, xI + yJ, Klaus Schmitt

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