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Unformatted text preview: 11/15/2007 SECOND HOURLY THIRD PRACTICE Math 21a, Fall 2007 Name: MWF 9 ChenYu Chi MWF 10 Oliver Knill MWF 10 Corina Tarnita MWF 11 Veronique Godin MWF 11 Stefan Hornet MWF 11 Jay Pottharst MWF 12 ChenYu Chi MWF 12 MingTao Chuan TTH 10 Thomas BarnetLamb TTH 10 Rehana Patel TTH 11:30 Thomas BarnetLamb TTH 11:30 Thomas Lam Start by printing your name in the above box and check your section in the box to the left. Do not detach pages from this exam packet or unstaple the packet. Please write neatly. Answers which are illeg ible for the grader can not be given credit. No notes, books, calculators, computers, or other electronic aids can be allowed. You have 90 minutes time to complete your work. The hourly exam itself will have space for work on each page. This space is excluded here in order to save printing resources. 1 20 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 Total: 110 1 Problem 1) True/False questions (20 points) Mark for each of the 20 questions the correct letter. No justifications are needed. 1) T F The directional derivative D u f is a vector normal to a level surface of f . Solution: The directional derivative is a scalar 2) T F At a critical point of a function f , the gradient vector has length 1. Solution: The gradient vector is the zero vector there. 3) T F At a critical point ( x, y ) of a function f , the tangent plane to the graph of f does not exist. Solution: The tangent plane is horizontal there. 4) T F For any point ( x, y ) which is not a critical point, there is a unit vector vectoru for which D vectoru f ( x, y ) is nonzero. Solution: Take a vector vector is perpendicular to the gradient, the directional derivative D vectoru f = f vectoru is zero. 5) T F If f xx (0 , 0) = 0 , D = f xx f yy f 2 xy negationslash = 0, and f (0 , 0) = ( , ) , then (0 , 0) is a saddle point. Solution: Because f xx = 0, we have D = f xx f yy f 2 xy = f 2 xy which can not be positive. Because D negationslash = 0, we must have D < 0. By the second derivative test, the critical point is a saddle point. 6) T F A continuous function defined on the closed unit disc x 2 + y 2 1 has an absolute maximum inside the disc or on the boundary. 2 Solution: The maximum can be either in the interior or at the boundary. 7) T F The function f ( x, y ) = x 2 y 2 has a neither a local maximum nor a local minimum at (0 , 0). Solution: It is a saddle point. 8) T F If ( x, y ) is a maximum of f ( x, y ) under the constraint g ( x, y ) = 5 then it is also a maximum of f ( x, y ) + g ( x, y ) under the constraint g ( x, y ) = 5. Solution: Indeed, on the constraint curve, the function f + g is just f + 5, which has the same maxima and minima as f on that curve....
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This test prep was uploaded on 04/06/2008 for the course MATH 21a taught by Professor Knill during the Fall '07 term at Harvard.
 Fall '07
 Knill
 Math, Multivariable Calculus

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