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math 21a midterm 2 solution4

# math 21a midterm 2 solution4 - SECOND HOURLY FORTH PRACTICE...

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11/15/2007 SECOND HOURLY FORTH 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) True/False questions (20 points), no justifications needed 1) T F The point (1 , 1 , 1) in Cartesian coordinates is the same as ( ρ, φ, θ ) = ( 2 , π/ 4 , 1) in spherical coordinates. Solution: Already θ = π/ 4 negationslash = 1. 2) T F Assume f satisfies the PDE f x = f y . If g = f x , then g x = g y . Solution: Use Clairots theorem tells g x = f yx = f xy = g y . 3) T F The equation φ = π/ 4 in spherical coordinates ( ρ 0 , 0 φ π, 0 θ 2 π as usual) and the surface x 2 + y 2 = z 2 (with no further restrictions on x, y, z ) are the same surface. Solution: The first equation is a single cone, the second equation is a double cone. 4) T F Even with f x ( a, b ) = 0 and f y ( a, b ) = 0, it is possible that some directional derivative D vectorv ( f ) of f ( x, y ) at ( a, b ) is non-zero. Solution: We have D v f ( x, y ) = f x v 1 + f y v 2 . 5) T F There exists a pair of different points on a sphere, for which the tangent planes are parallel. Solution: Take the antipodes. 6) T F The equation given in cylindrical coordinates as r = 2 | z | is the cone x 2 + y 2 = 4 z 2 . Solution: Yes it is x 2 + y 2 = 4 z 2 . 2
7) T F Assume we have a smooth function f ( x, y ) for which the lines x = 0 , y = 0 and x = y are level curves f ( x, y ) = 0. Then (0 , 0) is a critical point with D < 0. Solution: It can not be a saddle point, nor can it be a local maximum or local minimum. If it would be a saddle point, the level curves through the point consist of two lines. A concrete example, where three level curves pass through the same point is a monkey saddle. This is a case, where D = 0. 8) T F The parametrized surface vector r ( s, t ) = ( s cos( t ) , s sin( t ) , t ) is a surface of revo- lution. Solution: The surface vector r ( s, t ) = ( s cos( t ) , s sin( t ) , s ) would be a surface of revolution.

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