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Chapter 2 CSM

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NOT FOR SALE 2 LIMITS AND DERIVATIVES 2.1 The Tangent and Velocity Problems 1. (a) Using  (15 250), we construct the following table: slope =     5 (5 694) 694−250 5−15 = − 444 = −444 10 10 (10 444) 444−250 10−15 = − 194 = −388 5 20 (20 111) 111−250 20−15 = − 139 = −278 5 25 (25 28) 28−250 25−15 30 (30 0) 0−250 30−15 (b) Using the values of  that correspond to the points closest to  ( = 10 and  = 20), we have −388 + (−278) = −333 2 = − 222 10 = −222 = − 250 15 = −166 (c) From the graph, we can estimate the slope of the tangent line at  to be −300 9 = −333. 2. (a) Slope = 2948 − 2530 42 − 36 = 418 6 ≈ 6967 (b) Slope = 2948 − 2661 42 − 38 = 287 4 = 7175 (c) Slope = 2948 − 2806 42 − 40 = 142 2 = 71 (d) Slope = 3080 − 2948 44 − 42 = 132 2 = 66 From the data, we see that the patient’s heart rate is decreasing from 71 to 66 heartbeatsminute after 42 minutes. After being stable for a while, the patient’s heart rate is dropping. 3. (a)  = 1 ,  (2 −1) 1−  (i) 15 (ii) 19 (iii) 199 (iv) 1999 (v) 25 (vi) 21 (vii) 201 (viii) 2001 (b) The slope appears to be 1. (c) Using  = 1, an equation of the tangent line to the ( 1(1 − ))   (15 −2) 2 (199 −1010 101) 1010 101 (25? ...
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