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math 21a midterm 2 solution1

# math 21a midterm 2 solution1 - SECOND HOURLY FIRST PRACTICE...

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11/15/2007 SECOND HOURLY FIRST PRACTICE Math 21a, Fall 2007 Name: MWF 9 Chen-Yu Chi MWF 10 Oliver Knill MWF 10 Corina Tarnita MWF 11 Veronique Godin MWF 11 Stefan Hornet MWF 11 Jay Pottharst MWF 12 Chen-Yu Chi MWF 12 Ming-Tao Chuan TTH 10 Thomas Barnet-Lamb TTH 10 Rehana Patel TTH 11:30 Thomas Barnet-Lamb 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

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Problem 1) TF questions (20 points) Mark for each of the 20 questions the correct letter. No justifications are needed. 1) T F (1 , 1) is a local maximum of the function f ( x, y ) = x 2 y x + cos( y ). Solution: (1 , 1) is not even a critical point. 2) T F If f is a smooth function of two variables, then the number of critical points of f inside the unit disc is finite. Solution: Take the example f ( x, y ) = x . Then the entire y axes inside the unit disc consists of critical points. 3) T F In polar coordinates, the equation r = cos( θ ) , 0 θ < π is equivalent to x 2 + y 2 = x so that this describes a circle. Solution: Indeed, r 2 = r cos( θ ) is x 2 + y 2 = x and completion of the square shows that this is a circle of radius 1 / 2 centered at (1 / 2 , 0). Solution: Take f ( x, y ) = x 2 for example. Every point on the y axes { x = 0 } is a critical point. 4) T F If (1 , 1) is a critical point for the function f ( x, y ) then (1 , 1) is also a critical point for the function g ( x, y ) = f ( x 2 , y 2 ). Solution: If f (1 , 1) = ( f x (1 , 1) , f y (1 , 1)) = (0 , 0) then also g (1 , 1) = ( f x (1 , 1)2 x, f y (1 , 1)2 y ) = (0 , 0). 5) T F There is no function f ( x, y, z ) of three variables, for which every point on the unit sphere is a critical point. 2
Solution: Take a function like g ( t ) = te t with a maximum at t = 1 and define f ( x, y, z ) = g ( x 2 + y 2 + z 2 ). 6) T F If ( x 0 , y 0 ) is a maximum of f ( x, y ) under the constraint g ( x, y ) = g ( x 0 , y 0 ), then ( x 0 , y 0 ) is a maximum of g ( x, y ) under the constraint f ( x, y ) = f ( x 0 , y 0 ). Solution: Assume you have a situation f, g , where this is true and where the constraint is g = 0, produce a new situation f, h = g , where the first statement is still true but where the extrema of h under the constraint of f is a minimum. 7) T F If vectoru is a unit vector tangent at ( x, y, z ) to the level surface of f ( x, y, z ) then D u f ( x, y, z ) = 0. Solution: The directional derivative measures the rate of change of f in the direction of u . On a level surface, in the direction of the surface, the function does not change (because f is constant by definition on the surface).

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math 21a midterm 2 solution1 - SECOND HOURLY FIRST PRACTICE...

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