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CEGEP CHAMPLAIN - ST. LAWRENCE 201-103-RE: Differential Calculus Patrice Camiré Problem Sheet #21 Linear Approximation 1. (a) Give the equation of the tangent line to the curve y = f (x) at the point xo . (b) State the main principle behind the method of linear approximation. 2. Use the method of linear approximation to estimate the following numbers by hand. √ √ 3 5 1 (a) (1.0004)12 (g) 131 (j) 249 (d) 5.01 √ √ 4 √ 6 √ (b) 66 (h) 78 60 (k) (e) 120 √ 4 2/3 √ √ (c) (8.05) (i) 632 3 (f) 62 (l) 80 3. Use the method of linear approximation to estimate the following numbers by hand. √ √ (a) (0.997)6 (k) 144.1 (f) 99.9 √ √ 3 (b) (2.0003)7 (g) 65 (l) 139.97 √ √ 5 (c) 99.99 (h) (2.01)10 (m) 0.99 1 √ (i) ln(1.02) 5 (d) (n) 1.1 1002 0.03 √ √ (j) e (o) 82 (e) 25.1 Answers 1. (a) Txo (x) = f (xo ) + f 0 (xo )(x − xo ) (b) Near the point of tangency, the tangent line is very close to the curve. Mathematically, this can be stated as f (x) ≈ f (xo ) + f 0 (xo )(x − xo ) , if x ≈ xo . 2. (a) f (1.0004) = (1.0004)12 ≈ 1.0048 f (x) = x12 and xo = 1 √ 65 (b) f (66) = 66 ≈ = 8.125 8 √ f (x) = x and xo = 64 (c) f (8.05) = (8.05)2/3 ≈ 4.016̄ f (x) = x2/3 and xo = 8 (d) f (5.01) = 1 ≈ 0.1996 5.01 1 f (x) = and xo = 5 x √ 241 (e) f (120) = 120 ≈ = 10.954 22 √ f (x) = x and xo = 121 √ 95 3 (f) f (62) = 62 ≈ = 3.9583̄ 24 √ f (x) = 3 x and xo = 64 3. (a) (0.997)6 ≈ 0.982 (b) (2.0003)7 ≈ 128.1344 √ (c) 99.99 ≈ 9.9995 (d) (e) (f) (g) 1 1002 √ √ ≈ 0.000998 25.1 ≈ 5.01 99.9 ≈ 9.995 √ 3 65 ≈ 193 = 4.02083̄ 48 √ 127 3 (g) f (131) = 131 ≈ = 5.08 25 √ f (x) = 3 x and xo = 125 √ 107 4 (h) f (78) = 78 ≈ = 2.972̄ 36 √ f (x) = 4 x and xo = 81 √ 2507 4 (i) f (632) = 632 ≈ = 5.014 500 √ f (x) = 4 x and xo = 625 √ 407 5 (j) f (249) = 249 ≈ = 3.0148 135 √ f (x) = 5 x and xo = 243 √ 95 6 = 1.97916̄ (k) f (60) = 60 ≈ 48 √ f (x) = 6 x and xo = 64 √ (l) f (80) = 80 ≈ 8.94̄ √ f (x) = x and xo = 81 (h) (2.01)10 ≈ 1075.2 (i) ln(1.02) ≈ 0.02 (j) e0.03 ≈ 1.03 √ (k) 144.1 ≈ 12.00416̄ √ (l) 139.97 ≈ 11.832083̄ √ 5 (m) 0.99 ≈ 0.998 √ 5 (n) 1.1 ≈ 1.02 √ (o) 82 ≈ 9.05̄