Quiz_1_20C_Solution

Quiz_1_20C_Solution - Math 20C (Shenk). Quiz 1 Solution....

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Unformatted text preview: Math 20C (Shenk). Quiz 1 Solution. August 9, 2010. 1. 2. 3. 3 F = F1 + F2 = 7 − 11, −1 + 5 = −4, 4 (√ pounds) • [Angle of inclination] = 3 π or 4 π + 2nπ 4 √ with any integer n • [Magnitude] = |F| = 42 + 42 = 4 2 (pounds) − → − → P Q = 2 − 1, 4 − 1, 3 − 2 = 1, 3, 1 • P R = 0 − 1, 1 − 1, 5 − 2 = −1, 0, 3 • − →− → P Q · P R = 1(−1) + 3(0) + 1(3) = 2 • √ √ √ √ − → − → |P Q| = 12 + 32 + 12 = 11 • |P R| = 12 + 02 + 32 = 10 • Let θ be the angle at P with − →− → 2 2 PQ · PR • θ = cos–1 √ √ 0 ≤ θ ≤ π . • cos θ = − − → → =√ √ 11 10 11 10 |P Q||P R| i 4. 5. 6. x = 1 + 4t y=2 z = 3 − 2t − → P Q = 5 − 1, 2 − 2, 1 − 3 = 4, 0, −2 is parallel to the line. • L: j k 1 13 23 j+ i− (a) 1, 2, 3 × 2, −1, 0 = 1 2 3= 2 20 −1 0 2 −1 0 = [(2)(0) − (3)(−1)]i − [(1)(0) − (3)(2)]j + [(1)(−1) − (2)(2)]k √ √ = 3i + 6j − 5k = 3, 6, −5 (b) [Area] = 1 32 + 62 + 52 = 1 70 2 2 2 k −1 n =< 4, 3, 2 > is perpendicular to the plane. • Plane: 4(x − 3) + 3(y − 2) + 2(z − 1) = 0 Figure A5 x = x( t ) y = y (t ) 0≤t≤3 y 5 −5 5 x Figure A5 7. (a) x′ = 3 − 2t • y ′ = 3t2 − 6t • x′ (1) = 1 • y ′ (1) = −3 • v(1) = 1, −3 • (x(1), y (1)) = (2, −1) • Draw an arrow from (2, −1) to (2 + 1, −1 − 3) = (3 − 4) • Figure A7 3.25 (b) [Length] = −0.8 (3 − 2t)2 + (3t2 − 6t)2 dt y 4 −4 Figue A6 4x −4 1 ...
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This note was uploaded on 08/31/2011 for the course MATH 20C taught by Professor Helton during the Spring '08 term at UCSD.

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