Solucionario del calculo stewart

Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems

1. (a) Using P ( 15,250 ) , we construct the following table: t 5 10 20 25 30 Q slope=m
PQ

( 5,694 ) ( 10,444 ) ( 20,111 ) ( 25,28 ) ( 30,0 )

694 250 444 = = 44.4 5 15 10 444 250 194 = = 38.8 10 15 5 111 250 139 = = 27.8 20 15 5 28 250 222 = = 22.2 25 15 10 0 250 250 = = 16.6 30 15 15(b) Using the values of t that correspond to the points closest to P ( t=10 and t=20 ), we have 38.8+ ( 27.8 ) = 33.3 2 (c) From the graph, we can estimate the slope of the tangent line at P to be 300 = 33.3 . 9

2. 2948 42 2948 (c) Slope = 42 (a) Slope = 2530 418 2948 = 69.67 (b) Slope = 36 6 42 2806 142 3080 (d) Slope = = =71 40 2 44 2661 287 = =71.75 38 4 2948 132 = =66 42 2

From thedata, 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.
1

Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems

3. (a) For the curve y=x/ ( 1+x ) and the point P 1, x (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) 0.5 0.9 0.99 0.9991.1 1.5 1.01 1.001 Q m
PQ

1 2

( 0.5,0.333333) ( 0.9,0.473684 ) ( 0.99,0.497487) ( 0.999,0.499750 ) ( 1.5,06 ) ( 1.1,0.523810 ) ( 1.01,0.502488 ) ( 1.001,0.500250 )

0.333333 0.263158 0.251256 0.250125 0.2 0.238095 0.248756 0.249875

1 . 4 1 1 1 1 (c) y = (x 1) or y= x+ . 2 4 4 4 (b) The slope appears to be 4. For the curve y=ln x and the point P(2,ln2) : (a) x (i) (ii) (iii) (iv) (v)(vi) (vii) (viii) 1.5 1.9 1.99 1.999 2.5 2.1 2.01 2.001 Q m
PQ

( 1.5,0.405465) ( 1.9,0.641854 ) ( 1.99,0.688135) ( 1.999,0.692647) ( 2.5,0.916291 ) ( 2.1,0.741937) ( 2.01,0.698135) ( 2.001,0.693647)
1 . 2

0.575364 0.512933 0.501254 0.500125 0.446287 0.487902 0.498754 0.499875

(b) The slope appears to be (c) y ln 2= (d)

1 1 (x 2) or y= x 1+ln 2 2 2

2

Stewart Calculus ET 5e0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems

5. (a) y=y(t)=40t 16t . At t=2 , y=40(2) 16(2) =16 . The average velocity between times 2 and 2+h y(2+h) y(2) 40(2+h) 16(2+h) 16 24h 16h is v = = = = 24 16h , if h 0 . ave (2+h) 2 h h (i) [2,2.5] : h=0.5 , v = 32 ft / s (ii) [2,2.1] : h=0.1 , v = 25.6 ft / s
ave ave 2 2

2

2

(iii) [2,2.05] : h=0.05 , vave= 24.8ft / s (iv) [2,2.01] : h=0.01 , vave= 24.16 ft / s (b) The instantaneous velocity when t=2 ( h approaches 0 ) is 24 ft / s. 6. The average velocity between t and t+h seconds is 58(t+h) 0.83(t+h) 58t 0.83t 58h 1.66th 0.83h = =58 1.66t 0.83h if h 0 . h h (a) Here t=1 , so the average velocity is 58 1.66 0.83h=56.34 0.83h . (i) (ii) 1,1.5 : h=0.5 , 55.925 m / s 1,2 : h=1,55.51 m / s (iii) 1,1.1 :h=0.1 , 56.257 m / s (iv) 1,1.01 : h=0.01 , 56.3317 m / s (v) 1,1.001 : h=0.001 , 56.33917 m / s (b) The instantaneous velocity after 1 second is 56.34 m / s. 7. (a) (i) (iii) 13 (ii) ft / s ave 6 19 1,1.5 : h=0.5 , v = ft / s (iv) ave 24 1,3 : h=2 , v = 7 ft / s ave 6 331 1,1.1 : h=0.1 , v = ft / s ave 600 1,2 : h=1 , v =
2

(

2

)

2

(b) As h approaches 0 , the velocity approaches
3Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems

3 1 = ft / s. 6 2

(c)

(d) 8. Average velocity between times t=2 and t=2+h is given by (a) (i) (ii) (iii) s(5) av 5 s(4) v = av 4 s(3) v = av 3 s(2) 178 32 146 = = 48.7 ft / s 2 3 3 s(2) 119 32 87 = = =43.5 ft / s 2 2 2 s(2) 70 32 = =38 ft / s 2 1 s(2+h) s(2) . h

h=3 h=2 h=1

v =(b) Using the points ( 0.8,0 ) and ( 5,118 ) from the approximate tangent line, the instantaneous 118 0 28 ft / s. velocity at t=2 is about 5 0.8

9. For the curve y=sin (10 /x) and the point P ( 1,0 ) : (a) m x Q
PQ

2

( 2,0 )

0
4

Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems

1.5 1.4 1.3 1.2 1.1 x 0.5 0.6 0.7 0.8 0.9

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