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Unformatted text preview: 18.02 MIDTERM #1 BJORN POONEN September 30, 2010, 1:05pm to 1:55pm (50 minutes). 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 leader’s last name: Recitation time (e.g., 10am): (Do not write below this line.) 1 out of 20 2 out of 15 3 out of 20 4 out of 20 5 out of 25 Total out of 100 1) Let f ( x, y ) = 2 x 2 3 xy + y 2 + 6 x 4 y . (a) (10 pts.) Find all critical points (if any) of f ( x, y ). (b) (10 pts.) For each critical point, determine whether it is a local minimum, local maximum, saddle point, or none of these. 2) Find an equation of the plane through (1 , , 1) and (0 , 1 , 1) and parallel to 2 i j + 2 k . 3) Let A be the 2 × 2 matrix whose associated function R 2 → R 2 sends each point to its reflection in the line y = x . (For example, A 2 1 = 1 2 .) (a) (12 pts.) What is the matrix A ? (b) (8 pts.) What is the matrix A 2 ? 4) Use the approximation formula to estimate the hypotenuse of a right triangle with sides of length 2 . 9 and 4 . 1. Give your answer to two places after the decimal point. 5) A hamster is running clockwise around a circle of radius 2 centered at (0 , 0). It runs with constant speed 3, and stays in the xyplane (hamsters do not fly). At time t = 0, it is at (0 , 2). Find a formula for its position vector r ( t ) as a function of t . This is the end! Scratch work, or continuation of work on problem number (Detach and recycle this page if it is not part of your solutions.) Scratch work, or continuation of work on problem number (Detach and recycle this page if it is not part of your solutions.) 18.02 MIDTERM #1 SOLUTIONS BJORN POONEN September 30, 2010, 1:05pm to 1:55pm (50 minutes). 1) Let f ( x, y ) = 2 x 2 3 xy + y 2 + 6 x 4 y . (a) (10 pts.) Find all critical points (if any) of f ( x, y ). Solution: Calculate the partial derivatives: f x = 4 x 3 y + 6 f y = 3 x + 2 y 4 . Critical points are points where both of these are 0: 4 x 3 y + 6 = 0 3 x + 2 y 4 = 0 . Multiply these by 2 and 3, respectively, to obtain 8 x 6 y + 12 = 0 9 x + 6 y 12 = 0 . Adding the first equation to the second yields 8 x 6 y + 12 = 0 x = 0 , which is equivalent to ( x, y ) = (0 , 2). Thus (0 , 2) is the only critical point. (b) (10 pts.) For each critical point, determine whether it is a local minimum, local maximum, saddle point, or none of these. Solution: We’ll use the second derivative test. First we calculate f xx = 4 , f xy = 3 , f yy = 2 . Evaluating these at (0 , 2) yields A = 4, B = 3, C = 2. Then AC B 2 = 1 < 0, so (0 , 2) is a saddle point....
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 Fall '11
 Preston
 Vector Calculus, Vector field, BJORN POONEN

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