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Fall 2007 - Hohnhold's Class - Quiz 3 (Version A)

Fall 2007 - Hohnhold's Class - Quiz 3 (Version A) - x = 1 3...

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Quiz 3 for Math 20B Fall Quarter 2007, UCSD Henning Hohnhold Name: Section: #1: #2: Total: (1) (a) Give the polar form of the complex number (4 - 4 i ) 4 . (2 points) The argument of 4 - 4 i is θ = 7 π 4 (the easiest way to see this is to look at the point 4 - 4 i in the complex plane, but you can also use the formula for the argument) and its modulus is p 4 2 + ( - 4) 2 = 4 2. Hence the polar form we are looking for is given by (4 - 4 i ) 4 = (4 2 · (cos(7 π/ 4) + i sin(7 π/ 4)) 4 = 4 5 · (cos(7 π ) + i sin(7 π )) Note that the argument is only unique up to adding multiples of 2 π , for example, we could replace 7 π by π . Note also that computing the cos and sin we see that (4 - 4 i ) 4 = - 4 5 = - 1024. (b) Find all solutions of the equation 3 x 2 - 2 x + 10 = 0 . (2 points) Just apply the quadratic formula to see that
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Unformatted text preview: x = 1 3 ± r-29 9 = 1 3 ± √ 29 3 i. (2) (a) Please sketch the area enclosed by the polar curve r ( θ ) = | sin θ | . (3 points)-2-1.5-1-0.5 0.5 1 1.5 2-1.5-1-0.5 0.5 1 (b) Please compute the area of the ‘flower’ described by the polar curve r ( θ ) = 5-sin(7 θ ). (3 points)-10-7.5-5-2.5 2.5 5 7.5 10-5-2.5 2.5 5 We use the formula for the area enclosed by a polar curve: 1 2 Z 2 π (5-sin(7 θ )) 2 dθ = 1 2 Z 2 π 25-10 sin(7 θ ) + sin 2 (7 θ ) dθ = 1 2 ± 25 · 2 π + 10 7 cos(7 θ ) | 2 π + 1 2 Z 2 π 1-cos(14 θ ) dθ ² = 1 2 ± 50 π + 1 2 · 2 π-1 28 sin(14 θ ) | 2 π ² = 51 2 π 2 Good Luck! 3...
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