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Unformatted text preview: Chapter 12 Approximation methods 12.1 An approximation for the square root Use a linear approximation to find a rough estimate of the following functions at the indicated points. (a) y = √ x at x = 10. (Use the fact that √ 9 = 3.) (b) y = 5 x 2 at x = 1. (c) y = sin( x ) at x = 0 . 1 and at x = π + 0 . 1 Detailed Solution: (a) We have y prime = 1 2 √ x . Therefore our approximation is √ x ≈ √ x + 1 2 √ x ( x x ). Plugging in x = 9, we have √ 10 ≈ √ 9 + 10 9 2 × 3 = 3 + 1 6 (b) This is already a linear function, so we simply plug in x 1: y (1) = 5 2 = 3. (c) We have y prime = cos( x ). At x = 0 . 1, we set x = 0, so our approximation is sin(0 . 1) ≈ sin(0) + cos(0)(0 . 1 0) = 0 . 1. At x = π + 0 . 1, we set x = π : sin( π + 0 . 1) ≈ sin( π ) + cos( π )0 . 1 = . 1. 12.2 Use the method of linear approximation to find the cube root of (a) 0 . 065 (Hint: 3 √ . 064 = 0 . 4) (b) 215 (Hint: 3 √ 216 = 6) v.2005.1  September 4, 2009 1 Math 102 Problems Chapter 12 Detailed Solution: We need to approximate the values of the function f ( x ) = 3 √ x at x = 0 . 065 and x = 215. Taking the first derivative of f ( x ) gives f prime ( x ) = 1 3 x 2 3 = 1 3( 3 √ x ) 2 . Using linear approximation we have f ( x ) ≈ f ( x ) + f prime ( x )( x x ) = 3 √ x + 1 3( 3 √ x ) 2 ( x x ) . (a) x = 0 . 065 and x = 0 . 064, so 3 √ . 065 ≈ 3 √ . 064 + 1 3( 3 √ . 064) 2 (0 . 065 . 064) = 0 . 4 + 1 3 · . 16 · . 001 ≈ . 40208 (b) x = 215 and x = 216, so 3 √ 215 ≈ 3 √ 216 + 1 3( 3 √ 216) 2 (215 216) = 6 + 1 3 · 36 · ( 1) ≈ 5 . 99074 12.3 Use the data in the graph in Figure 12.1 to make the best approximation you can to f (2 . 01). (2, 1) (3, 0) y = f(x) Figure 12.1: Figure for Problem 12.3 Detailed Solution: Using linear approximation, we know that f ( x ) ≈ f ( a ) + f prime ( a )( x a ) . v.2005.1  September 4, 2009 2 Math 102 Problems Chapter 12 From Figure 12.1 we see that a = 2, f ( a ) = 1, and f prime ( a ) is the slope of the line which can be calculated from the graph: (1 0) / (2 3) = 1. Thus f ( x ) ≈ 1 1( x 2) = 1 + (2 x ) = 3 x. Substituting in x = 2 . 01 leads to f (2 . 01) ≈ 3 2 . 01 = 0 . 99 . 12.4 Using linear approximation, find the value of (a) tan44 ◦ , given tan 45 ◦ = 1, sec 45 ◦ = √ 2, and 1 ◦ ≈ . 01745 radians. (b) sin 61 ◦ , given sin 60 ◦ = √ 3 2 , cos 60 ◦ = 1 2 and 1 ◦ ≈ . 01745 radians. Detailed Solution: (a) We want to evaluate the function f ( x ) = tan x at x = 44 ◦ . The formula for linear approxi mation gives f ( x ) ≈ f ( x )+ f prime ( x )( x x ) = tan x +(sec 2 x )( x x ). Substituting x = 44 ◦ and x = 45 ◦ tan44 ◦ ≈ tan45 ◦ + (sec 2 45 ◦ )(44 ◦ 45 ◦ ) = 1 + 2( . 01745) ≈ . 98255 ....
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 Winter '09
 Allard
 Method, Quadratic equation, Continuous function, Newton’s method

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