Chap.2-3-Sol - P 15,250 t Q slope= m PQ 5 5,694 694 250 5...

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Unformatted text preview: P 15,250 ( ) t Q slope= m PQ 5 5,694 ( ) 694 ¡ 250 5 ¡ 15 = ¡ 444 10 = ¡ 44.4 10 10,444 ( ) 444 ¡ 250 10 ¡ 15 = ¡ 194 5 = ¡ 38.8 20 20,111 ( ) 111 ¡ 250 20 ¡ 15 = ¡ 139 5 = ¡ 27.8 25 25,28 ( ) 28 ¡ 250 25 ¡ 15 = ¡ 222 10 = ¡ 22.2 30 30,0 ( ) ¡ 250 30 ¡ 15 = ¡ 250 15 = ¡ 16.6 t P t =10 t =20 ¡ 38.8+ ¡ 27.8 ( ) 2 = ¡ 33.3 P ¡ 300 9 = ¡ 33.3 = 2948 ¡ 2530 42 ¡ 36 = 418 6 ¢ 69.67 = 2948 ¡ 2661 42 ¡ 38 = 287 4 =71.75 = 2948 ¡ 2806 42 ¡ 40 = 142 2 =71 = 3080 ¡ 2948 44 ¡ 42 = 132 2 =66 71 66 / 42 1. (a) Using , we construct the following table: (b) Using the values of that correspond to the points closest to ( and ), we have (c) From the graph, we can estimate the slope of the tangent line at to be . 2. (a) Slope (b) Slope (c) Slope (d) Slope From the data, we see that the patient’s heart rate is decreasing from to heartbeats minute after minutes. After being stable for a while, the patient’s heart rate is dropping. 1 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems y= x / 1+ x ( ) P 1, 1 2 x Q m PQ (i) 0.5 0.5,0.333333 ( ) 0.333333 (ii) 0.9 0.9,0.473684 ( ) 0.263158 (iii) 0.99 0.99,0.497487 ( ) 0.251256 (iv) 0.999 0.999,0.499750 ( ) 0.250125 (v) 1.1 1.5,06 ( ) 0.2 (vi) 1.5 1.1,0.523810 ( ) 0.238095 (vii) 1.01 1.01,0.502488 ( ) 0.248756 (viii) 1.001 1.001,0.500250 ( ) 0.249875 1 4 y ¡ 1 2 = 1 4 ( x ¡ 1) y = 1 4 x + 1 4 y =ln x P (2,ln2) x Q m PQ (i) 1.5 1.5,0.405465 ( ) 0.575364 (ii) 1.9 1.9,0.641854 ( ) 0.512933 (iii) 1.99 1.99,0.688135 ( ) 0.501254 (iv) 1.999 1.999,0.692647 ( ) 0.500125 (v) 2.5 2.5,0.916291 ( ) 0.446287 (vi) 2.1 2.1,0.741937 ( ) 0.487902 (vii) 2.01 2.01,0.698135 ( ) 0.498754 (viii) 2.001 2.001,0.693647 ( ) 0.499875 1 2 y ¡ ln 2= 1 2 ( x ¡ 2) y = 1 2 x ¡ 1+ln 2 3. (a) For the curve and the point (b) The slope appears to be . (c) or . 4. For the curve and the point : (a) (b) The slope appears to be . (c) or (d) 2 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems y = y ( t )=40 t ¡ 16 t 2 t =2 y =40(2) ¡ 16(2) 2 =16 2 2+ h v ave = y (2+ h ) ¡ y (2) (2+ h ) ¡ 2 = 40(2+ h ) ¡ 16(2+ h ) 2 ¡ 16 h = ¡ 24 h ¡ 16 h 2 h = ¡ 24 ¡ 16 h h ¢ [2,2.5] h =0.5 v ave = ¡ 32 / [2,2.1] h =0.1 v ave = ¡ 25.6 / [2,2.05] h =0.05 v ave = ¡ 24.8 / [2,2.01] h =0.01 v ave = ¡ 24.16 / t =2 h ¡ 24 / t t + h 58( t + h ) ¡ 0.83( t + h ) 2 ¡ 58 t ¡ 0.83 t 2 ( ) h = 58 h ¡ 1.66 th ¡ 0.83 h 2 h =58 ¡ 1.66 t ¡ 0.83 h h ¢ t =1 58 ¡ 1.66 ¡ 0.83 h =56.34 ¡ 0.83 h 1,2 h =1,55.51 / 1,1.5 h =0.5 55.925 / 1,1.1 h =0.1 56.257 / 1,1.01 h =0.01 56.3317 / 1,1.001 h =0.001 56.33917 / 1 56.34 / 1,3 h =2 v ave = 13 6 / 1,2 h =1 v ave = 7 6 / 1,1.5 h =0.5 v ave = 19 24 / 1,1.1 h =0.1 v ave = 331 600 / h 5. (a) . At , . The average velocity between times and is , if ....
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This note was uploaded on 09/26/2011 for the course GF sdf taught by Professor Sdfg during the Spring '11 term at Columbia School of Broadcasting.

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Chap.2-3-Sol - P 15,250 t Q slope= m PQ 5 5,694 694 250 5...

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