Chapter 3 Solutions Manual

Chapter 3 Solutions Manual - February 8, 2005 10:25...

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February 8, 2005 10:25 l38-ch03 Sheet number 1 Page number 41 black 41 CHAPTER 3 The Derivative EXERCISE SET 3.1 1. (a) m tan = (50 10) / (15 5) =40 / 10 = 4 m/s (b) t (s) 4 10 20 v (m/s) 3. (a) m tan = (600 0) / (20 2 . 2) = 600 / 17 . 8 33 . 71 m/s (b) m tan (820 600) / (20 16) = 220 / 4 = 55 m/s The speed is increasing with time. 5. From the fgure: t s t 0 t 1 t 2 (a) The particle is moving ±aster at time t 0 because the slope o± the tangent to the curve at t 0 is greater than that at t 2 . (b) The initial velocity is 0 because the slope o± a horizontal line is 0. (c) The particle is speeding up because the slope increases as t increases ±rom t 0 to t 1 . (d) The particle is slowing down because the slope decreases as t increases ±rom t 1 to t 2 . 7. It is a straight line with slope equal to the velocity. 9. (a) m sec = f (1) f (0) 1 0 = 2 1 =2 (b) m tan = lim x 1 0 f ( x 1 ) f (0) x 1 0 = lim x 1 0 2 x 2 1 0 x 1 0 = lim x 1 0 2 x 1 =0 (c) m tan = lim x 1 x 0 f ( x 1 ) f ( x 0 ) x 1 x 0 = lim x 1 x 0 2 x 2 1 2 x 2 0 x 1 x 0 = lim x 1 x 0 (2 x 1 +2 x 0 ) =4 x 0 (d) –2 –1 1 2 1 2 3 x y Secant Tangent 11. (a) m sec = f (3) f (2) 3 2 = 1 / 3 1 / 2 1 = 1 6 (b) m tan = lim x 1 2 f ( x 1 ) f (2) x 1 2 = lim x 1 2 1 /x 1 1 / 2 x 1 2 = lim x 1 2 2 x 1 2 x 1 ( x 1 2) = lim x 1 2 1 2 x 1 = 1 4
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February 8, 2005 10:25 l38-ch03 Sheet number 2 Page number 42 black 42 Chapter 3 (c) m tan = lim x 1 x 0 f ( x 1 ) f ( x 0 ) x 1 x 0 = lim x 1 x 0 1 /x 1 1 /x 0 x 1 x 0 = lim x 1 x 0 x 0 x 1 x 0 x 1 ( x 1 x 0 ) = lim x 1 x 0 1 x 0 x 1 = 1 x 2 0 (d) x y Secant Tangent 1 4 13. (a) m tan = lim x 1 x 0 f ( x 1 ) f ( x 0 ) x 1 x 0 = lim x 1 x 0 ( x 2 1 1) ( x 2 0 1) x 1 x 0 = lim x 1 x 0 ( x 2 1 x 2 0 ) x 1 x 0 = lim x 1 x 0 ( x 1 + x 0 )=2 x 0 (b) m tan =2( 1) = 2 15. (a) m tan = lim x 1 x 0 f ( x 1 ) f ( x 0 ) x 1 x 0 = lim x 1 x 0 x 1 x 0 x 1 x 0 = lim x 1 x 0 1 x 1 + x 0 = 1 2 x 0 (b) m tan = 1 2 1 = 1 2 17. (a) 72 F at about 4:30 P.M. (b) about (67 43) / 6=4 F/h (c) decreasing most rapidly at about 9 P.M.; rate of change of temperature is about 7 F/h (slope of estimated tangent line to curve at 9 P.M.) 19. (a) during the ±rst year after birth (b) about 6 cm/year (slope of estimated tangent line at age 5) (c) the growth rate is greatest at about age 14; about 10 cm/year (d) t (yrs) Growth rate (cm/year) 51 01 52 0 10 20 30 40 21. (a) (40) 3 / 10=20 , 238 . 6ft (b) v ave =20 , 238 . 6 / 40 = 505 . 96 ft/s (c) Solve s = t 3 / 10 = 135 ,t 7 . 53 so v ave = 135 / 7 . 53=17 . 93 ft/s. (d) v inst = lim t 1 40 t 3 1 / 10 (40) 3 / 10 t 1 40 = lim t 1 40 ( t 3 1 40 3 ) ( t 1 40) 10 = lim t 1 40 1 10 ( t 2 1 +40 t 1 + 1600) = 1517 . 89 ft/s 23. (a) v ave = 6(4) 4 6(2) 4 4 2 = 720 ft/min
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February 8, 2005 10:25 l38-ch03 Sheet number 3 Page number 43 black Exercise Set 3.2 43 (b) v inst = lim t 1 2 6 t 4 1 6(2) 4 t 1 2 = lim t 1 2 6( t 4 1 16) t 1 2 = lim t 1 2 6( t 2 1 + 4)( t 2 1 4) t 1 2 = lim t 1 2 6( t 2 1 + 4)( t 1 + 2) = 192 ft/min EXERCISE SET 3.2 1. f 0 (1)=2, f 0 (3)=0, f 0 (5) = 2, f 0 (6) = 1 3. (b) m = f 0 (2)=3 (c) the same, f 0 5.
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Chapter 3 Solutions Manual - February 8, 2005 10:25...

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