Exam_1_20C_Solution

Exam_1_20C_Solution - Math 20C(Shenk Summer 2010 Exam 1...

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Unformatted text preview: Math 20C (Shenk). Summer, 2010. Exam 1 Solution. 1. 2. 3. 2 A ยท B = 2, 1, 3 ยท โˆ’1, 4, k = โˆ’2 + 4 + 3k = 2 + 3k โ€ข A ยท B = 0 โ‡โ‡’ 2 + 3k = 0 โ€ข k = โˆ’ 3 โˆš C ยท D = 1, 2, 2 ยท 1, 2, โˆ’2 = 1(1) + 2(2) + 2(โˆ’2) = 1 โ€ข |C| = 12 + 22 + 22 = 3 โ€ข โˆš CยทD 1 1 = = 1 โ€ข ฮธ = cosโ€“1 ( 9 ) |D| = 12 + 22 + 22 = 3 โ€ข cos ฮธ = 9 |C||D| 3(3) โˆ’ โ†’ โˆ’ โ†’ One solution: OQ = 3, 3, 3 โ€ข P S = โˆ’1 โˆ’ 1, 4 โˆ’ 1, 2 โˆ’ 1 = โˆ’2, 3, 1 โˆ’ โ†’โˆ’ โ†’โˆ’ โ†’โˆ’ โ†’โˆ’ โ†’ OR = OQ + QR = OQ + P S = 3, 3, 3 + โˆ’2, 3, 1 = 1, 6, 4 โ€ข R = (1, 6, 4) x =3+t y =1โˆ’t z = 2 + 2t 4. One answer: Normal vector: n =< 1, โˆ’1, 2 > โ€ข Line: 5. โˆ’ โ†’ One answer: P Q = โˆ’2 โˆ’ 1, 3 โˆ’ 4, 0 โˆ’ 2 = โˆ’3, โˆ’1, โˆ’2 โ€ข Line: 6. โˆ’ โ†’ One solution: P Q = 4 โˆ’ 2, 3 โˆ’ 3, 5 โˆ’ 3 = 2, 0, 3 i jk x = 1 โˆ’ 3t y =4โˆ’t z = 2 โˆ’ 2t โˆ’ โ†’ โ€ข P R = 2 โˆ’ 2, 6 โˆ’ 3, 4 โˆ’ 2 = 0, 3, 2 โˆ’ โ†’โˆ’ โ†’ 03 23 20 iโˆ’ [Normal vector ] = n = P Q ร— P R = 2 0 3 = j+ k 32 02 03 032 = [0(2) โˆ’ 3(3))]i โˆ’ [(2(2) โˆ’ 3(0)]j + [2(3) โˆ’ 0(0)]k = โˆ’9i โˆ’ 4j + 6k = โˆ’9, โˆ’4, 6 โ€ข โˆ’ โ†’ โˆ’ โ†’ Check: P Q ยท n = 2(โˆ’9) + 0(โˆ’4) + 3(6) = 0 โ€ข P R ยท n = 0(โˆ’9) + 3(โˆ’4) + 2(6) = 0 โ€ข Plane (using P ): โˆ’9(x โˆ’ 2) โˆ’ 4(y โˆ’ 3) + 6(z โˆ’ 2) = 0 7. 8. โ€ข 123 01 05 15 +3 โˆ’2 1, 2, 3 ยท [ 0, 1, 5 ร— 4, 0, 1 ] = 0 1 5 = 1 40 41 01 401 = 1[(1)(1)โˆ’(5)(0)]โˆ’2[(0)(1)โˆ’(5)(4)]+3[(0)(0)โˆ’(1)(4)] = 1(1)โˆ’2(โˆ’20)+3(โˆ’4) = 1+40โˆ’12 = 29 Figure A8 y 3 2 1 1 Figure A6 1 2 3 x Math 20C, Summer, 2010 9. Exam 1 Solution, p. 2 d โ€ข v (t ) = t โˆ’ t2 , t2 + t = 1 โˆ’ 2t, 2t + 1 โ€ข [Speed at time t] dt โˆš โˆš = |v(t)| = (1 โˆ’ 2t)2 + (2t + 1)2 = 1 โˆ’ 4t + 4t2 + 4t2 + 4t + 1 = 2 + 8t2 (b) The speed is a minimum at t = 0. โ€ข v(0) = 1, 1 โ€ข Figure A9 (a) r(t) = t โˆ’ t2 , t2 + t y 3 v(0) = 1, 1 Figure A9 โˆ’3 x ...
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