Section2_8 - Section 2.8 Linear approximations and...

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Unformatted text preview: (3/19/08) Section 2.8 Linear approximations and differentials Overview: In this section we return to the main concept of this chapter: the approximations of graphs by tangent lines. We discuss additional examples of tangent-line approximations and show how tangent line estimates of errors can be calculated using the notation of differentials. We finish the section with a description of some of the key steps in the history of the derivative. Topics: • Tangent-line approximations • Differentials • Error estimates • A closer look at linear approximations • History of the derivative Tangent-line approximations The tangent line y = f ( a ) + f prime ( a )( x- a ) to y = f ( x ) at x = a approximates the graph near x = a and consequently, f ( x ) ≈ f ( a ) + f prime ( a )( x- a ) for x ≈ a. (1) The function y = f ( a ) + f prime ( a )( x- a ) here, whose graph is the tangent line, is called the linear approximation or local linearization of y = f ( x ) at x = a . Example 1 Find the linear approximation y = L ( x ) of f ( x ) = 15- 4 x + x 3 at x = 2. Solution By formula (1) with a = 2, the linear approximation is L ( x ) = f (2) + f prime (2)( x- 2). We calculate f (2) = 15- 4(2) + 2 3 = 15- 8 + 8 = 15. Then we use the differentiation formulas from Section 2.4 to write f prime ( x ) = d dx (5- 4 x + x 3 ) =- 4 + 3 x 2 , and calculate f prime (2) =- 4+3(- 2) 2 =- 4+12 = 8. The linear approximation is L ( x ) = 25 +8( x- 2). The graphs of y = f ( x ) and of its approximation y = L ( x ) are in Figure 1. Notice that L ( x ) approximates f ( x ) well for x near 2 but not for x far from 2. square x 1 2 3 y 10 20 y = f ( x ) y = L ( x ) FIGURE 1 157 p. 158 (3/19/08) Section 2.8, Linear approximations and differentials Example 2 One hour after leaving Toulouse, France, on a test flight, the French/British supersonic passenger jet Concord had traveled 975 miles and was flying 1520 miles per hour (twice the speed of sound). (1) Approximately how far was it from Toulouse six minutes (0 . 1 hours) later? Solution We let s = s ( t ) be the Concord’s distance from Toulouse t hours after take off. We are given that s (1) = 975 miles and s prime (1) = 1520 miles per hour, and we want to calculate the approximate value of s (1 . 1). By (1) with s ( t ) in place of f ( x ) and a = 1, the linear approximation of s = s ( t ) at t = 1 is L ( t ) = 950 + 1520( t- 1) . Therefore, s (1 . 1) ≈ L (1 . 1) = 975 + 1520(1 . 1- 1) = 1127 miles. The plane was approximately 1127 miles from Toulouse one hour and six minutes after take-off. square Differentials For a function y = f ( x ) with a derivative at x = a , the differentials dx and df represent corresponding changes in x and y on the tangent line to the graph at x = a (Figure 2)....
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Section2_8 - Section 2.8 Linear approximations and...

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