Calculus with Analytic Geometry by edwards & Penney soln ch12

# Calculus with Analytic Geometry

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y x R S P (2,3) R S x y P (–10, –20) (1,–2) (4,2) u v x y w = u + v Section 12.1 C12S01.001: v = RS = 3 1 , 5 2 = 2 , 3 . The position vector of the point P (2 , 3) and RS are shown next. C12S01.002: v = RS = 1 ( 2) , 4 ( 3) = 3 , 7 . C12S01.003: v = RS = 5 5 , 10 10 = 10 , 20 . The position vector of the point P and RS are shown next. C12S01.004: v = RS = 15 ( 10) , 25 20 = 25 , 45 . C12S01.005: w = u + v = 1 , 2 + 3 , 4 = 1 + 3 , 2 + 4 = 4 , 2 . The next figure illustrates this computation in the form of the triangle law for vector addition (see Fig. 12.1.6 of the text). C12S01.006: u + v = 4 , 2 + 2 , 5 = 4 2 , 2 + 5 = 2 , 7 . C12S01.007: Given: u = 3 i + 5 j , v = 2 i 7 j : 1

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(5,–2) (3,5) u v u + v u + v = 3 i + 5 j + 2 i 7 j = (3 + 2) i + (5 7) j = 5 i 2 j . The next figure illustrates the triangle law for vector addition using u and v . C12S01.008: u + v = 7 10 , 5 + 0 = 3 , 5 = 3 i + 5 j . C12S01.009: Given: a = 1 , 2 and b = 3 , 2 . Then: | a | = (1) 2 + ( 2) 2 = 5 , | − 2 b | = | − 6 , 4 | = 36 + 16 = 2 13 , | a b | = | 1 ( 3) , 2 2 | = 16 + 16 = 4 2 , a + b = 1 3 , 2 + 2 = 2 , 0 , 3 a 2 b = 3 , 6 6 , 4 = 3 ( 6) , 6 4 = 9 , 10 . C12S01.010: Given: a = 3 , 4 and b = 4 , 3 . Then: | a | = 9 + 16 = 25 = 5 , | − 2 b | = | − 8 , 6 | = 64 + 36 = 10 , | a b | = | 3 ( 4) , 4 3 | = 49 + 1 = 5 2 , a + b = 3 4 , 4 + 3 = 1 , 7 , 3 a 2 b = 9 , 12 8 , 6 = 9 ( 8) , 12 6 = 17 , 6 . C12S01.011: Given: a = 2 , 2 and b = 3 , 4 . Then: | a | = 4 + 4 = 8 = 2 2 , | − 2 b | = | 6 , 8 | = 36 + 64 = 10 , | a b | = | − 2 ( 3) , 2 ( 4) | = | 1 , 2 | = 1 + 4 = 5 , a + b = 2 3 , 2 4 = 5 , 6 , 3 a 2 b = 6 , 6 6 , 8 = 6 ( 6) , 6 ( 8) = 0 , 2 . 2
C12S01.012: Given: a = 2 4 , 7 = 8 , 14 and b = 3 4 , 2 = 12 , 6 . Then: | a | = 64 + 196 = 260 = 2 65 , | − 2 b | = | − 24 , 12 | = 576 + 144 = 720 = 12 5 , | a b | = | − 8 12 , 14 6 | = | − 20 , 20 | = 400 + 400 = 20 2 , a + b = 8 + 12 , 14 + 6 = 4 , 8 , 3 a 2 b = 24 , 42 24 , 12 = 24 24 , 42 12 = 48 , 54 . C12S01.013: Given: a = i + 3 j and b = 2 i 5 j . Then: | a | = | i + 3 j | = 1 + 9 = 10 , | − 2 b | = | 4 i 10 j | = 16 + 100 = 2 29 , | a b | = | − i + 8 j | = 1 + 64 = 65 , a + b = i + 3 j + 2 i 5 j = 3 i 2 j , 3 a 2 b = 3 i + 9 j 4 i + 10 j = i + 19 j . C12S01.014: Given: a = 2 i 5 j and b = i 6 j . Then: | a | = | 2 i 5 j | = 4 + 25 = 29 , | − 2 b | = | − 2 i + 12 j | = 4 + 144 = 2 37 , | a b | = | i + j | = 2 , a + b = 2 i 5 j + i 6 j = 3 i 11 j , 3 a 2 b = 6 i 15 j 2 i + 12 j = 4 i 3 j . C12S01.015: Given: a = 4 i and b = 7 j . Then: | a | = | 4 i | = 16 = 4 , | − 2 b | = | 14 j | = (14) 2 = 14 , | a b | = | 4 i + 7 j | = 16 + 49 = 65 , a + b = 4 i 7 j , 3 a 2 b = 12 i + 14 j . C12S01.016: Given: a = i j and b = 2 i + 2 j . Then: | a | = 1 + 1 = 2 , | − 2 b | = | − 4 i 4 j | = 32 = 4 2 , 3

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| a b | = | − 3 i 3 j | = 18 = 3 2 , a + b = i j + 2 i + 2 j = i + j , 3 a 2 b = 3 i 3 j 4 i 4 j = 7 i 7 j . C12S01.017: Because | a | = 9 + 16 = 5, u = 1 5 a = 3 5 i 4 5 j and v = 1 5 a = 3 5 i + 4 5 j . C12S01.018: Because | a | = 25 + 144 = 13, u = 1 13 a = 5 13 i 12 13 j and v = 1 13 a = 5 13 i + 12 13 j .
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