Chap.2-3-Sol

# Chap.2-3-Sol - Stewart Calculus ET 5e 0534393217;2 Limits...

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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 ( ) 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 ² 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

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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 ³ 0 [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 0 ± 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 ³ 0 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 0 5. (a) . At , . The average velocity between times and is , if . (i) : , ft s (ii) : , ft s (iii) : , ft s (iv) : , ft s (b) The instantaneous velocity when ( approaches ) is ft s. 6. The average velocity between and seconds is if . (a) Here , so the average velocity is . (i) : m s (ii) : , m s (iii) : , m s (iv) : , m s (v) : , m s (b) The instantaneous velocity after second is m s. 7. (a) (i) : , ft s (ii) : , ft s (iii) : , ft s (iv) : , ft s (b) As approaches , the velocity approaches 3 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems

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3 6 = 1 2 / t =2 t =2+ h s (2+ h ) ± s (2) h h =3 ´ v av = s (5) ± s (2) 5 ± 2 = 178 ± 32 3 = 146 3 ² 48.7 / h =2 ´ v av = s (4) ± s (2) 4 ± 2 = 119 ± 32 2 = 87 2 =43.5 / h =1 ´ v av = s (3) ± s (2) 3 ± 2 = 70 ± 32 1 =38 / 0.8,0 ( ) 5,118 ( ) t =2 118 ± 0 5 ± 0.8 ² 28 / y =sin (10 ± / x ) P 1,0 ( ) x Q m PQ 2 2,0 ( ) 0 ft s. (c) (d) 8. Average velocity between times and is given by . (a) (i) ft s (ii) ft s (iii) ft s (b) Using the points and from the approximate tangent line, the instantaneous velocity at is about ft s. 9. For the curve and the point : (a) 4 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems
1.5 1.5,0.8660 ( ) 1.7321 1.4 1.4, ± 0.4339 ( ) ± 1.0847 1.3 1.3, ± 0.8230 ( ) ± 2.7433 1.2 1.2,0.8660 ( ) 4.3301 1.1 1.1, ± 0.2817 ( ) ± 2.8173 x Q m PQ 0.5 0.5,0 ( ) 0 0.6 0.6,0.8660 ( ) ± 2.1651 0.7 0.7,0.7818 ( ) ± 2.6061 0.8 0.8,1 ( ) ± 5 0.9 0.9, ± 0.3420 ( ) 3.4202 x 1 P x ± 1 x =1.001 Q 1.001, ± 0.0314 ( ) m PQ ² ± 31.3794 x =0.999 Q 0.999,0.0314 ( ) m PQ = ± 31.4422 ± 31.4108 P ± 31.4 As approaches , the slopes do not appear to be approaching any particular value.

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