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Unformatted text preview: 18.02 PRACTICE FINAL EXAM BJORN POONEN December 17, 2009, 9:00am12:00 (3 hours) Please turn cell phones off completely and put them away. No books, notes, or electronic devices are permitted during this exam. Generally, you must show your work to receive credit. Name: Student ID number: Recitation instructor’s last name: Recitation time (e.g., 10am): (Do not write below this line.) 1 out of 20 2 out of 10 3 out of 10 4 out of 10 5 out of 10 6 out of 10 7 out of 15 8 out of 10 9 out of 10 10 out of 15 11 out of 15 12 out of 10 13 out of 20 14 out of 20 15 out of 10 16 out of 10 17 out of 10 18 out of 10 19 out of 25 Total out of 250 2 1) For each of (a)(d) below: If the statement is true, write TRUE. If the statement is false, write FALSE. (Please do not use the abbreviations T and F.) No explanations are required in this problem. (a) The set of points ( x,y ) in R 2 satisfying x 2 + y 2 > 9 is simply connected. Solution. FALSE. The circle of radius 4 centered at (0 , 0) is in this region, but its interior is not. (b) If T is any region in R 2 , and F is a continuously differentiable 2 D vector field on T such that curl F = 0 at each point of T , then F is the gradient of some function on T . Solution. FALSE. To deduce this conclusion, one needs T to be simply connected. (c) If A is a 3 × 3 matrix such det A 6 = 0, and b ∈ R 3 , then there is exactly one vector x such that A x = b . Solution. TRUE. Multiplying both sides of the equation by A 1 on the left shows that x = A 1 b , and conversely, x = A 1 b actually is a solution. (d) If the position vectors h a,b i and h c,d i form two sides of a parallelogram, then the area of the parallelogram equals ad bc . Solution. FALSE. The sign could be wrong, because ad bc could be negative, while area is always positive. 3 2) Let A = (0 , 1 , 1), let B = ( 1 , 1 , 1), and let C = (1 , , 1). What is the radian measure of angle A of triangle ABC ? Solution. The answer is the angle θ between the vectors u := B A = h 1 , , i and v := C A = h 1 , 1 , i . We have u · v =  u  v  cos θ 11 √ 2cos θ so cos θ = 1 / √ 2, and θ = 3 π/ 4. 4 3) Find the equation of the plane through the points (2 , 3 , 5), (1 , 2 , 3), and ( 2 , , 0), in the form ax + by + cz = d . Solution. Let A , B , C be the three points. The vector n := ( B A ) × ( C A ) = ( 1 , 1 , 2) × ( 4 , 3 , 5) = i j k 1 1 2 4 3 5 = i + 3 j k is a normal vector to the plane, so the equation has the form x + 3 y z = d for some constant d . Plugging in ( 2 , , 0) shows that d = 2. Thus the answer is x + 3 y z = 2 . 5 4) Let A be the matrix 2 2 2 7 , and let I be the 2 × 2 identity matrix, For which real numbers c is A cI not invertible?...
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 Fall '08
 Auroux
 Multivariable Calculus, Vector Calculus, Cos, Trigraph

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