01_ISMCHAP1

01_ISMCHAP1 - INSTRUCTOR'S SOLUTIONS MANUAL SECTION...

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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 1.1 (PAGE 61) CHAPTER 1. LIMITS AND CONTINUITY Section 1.1 Examples of Velocity, Growth Rate, and Area (page 61) 1. Average velocity = ± x ± t = ( t + h ) 2 t 2 h m/s. 2. h Avg. vel. over [2 , 2 + h ] 1 5.0000 0 . 1 4.1000 0 . 01 4.0100 0 . 001 4.0010 0 . 0001 4.0001 3. Guess velocity is v = 4 m/s at t = 2s . 4. Average volocity on [2 , 2 + h ]is ( 2 + h ) 2 4 ( 2 + h ) 2 = 4 + 4 h + h 2 4 h = 4 h + h 2 h = 4 + h . As h approaches 0 this average velocity approaches 4 m/s 5. x = 3 t 2 12 t + 1 m at time t s. Average velocity over interval [1 , 2] is ( 3 × 2 2 12 × 2 + 1 ) ( 3 × 1 2 12 × 1 + 1 ) 2 1 =− 3 m/s. Average velocity over interval [2 , 3] is ( 3 × 3 2 12 × 3 + 1 ) ( 3 × 2 2 12 × 2 + 1 ) 3 2 = 3 m/s. Average velocity over interval [1 , 3] is ( 3 × 3 2 12 × 3 + 1 ) ( 3 × 1 2 12 × 1 + 1 ) 3 1 = 0 m/s. 6. Average velocity over [ t , t + h 3 ( t + h ) 2 12 ( t + h ) + 1 ( 3 t 2 12 t + 1 ) ( t + h ) t = 6 th + 3 h 2 12 h h = 6 t + 3 h 12 m/s . This average velocity approaches 6 t 12 m/s as h ap- proaches 0. At t = 1 the velocity is 6 × 1 12 6 m/s. At t = 2 the velocity is 6 × 2 12 = 0 m/s. At t = 3 the velocity is 6 × 3 12 = 6 m/s. 7. At t = 1 the velocity is v 6 < 0 so the particle is moving to the left. At t = 2 the velocity is v = 0 so the particle is station- ary. At t = 3 the velocity is v = 6 > 0 so the particle is moving to the right. 8. Average velocity over [ t k , t + k ]i s 3 ( t + k ) 2 12 ( t + k ) + 1 [3 ( t k ) 2 12 ( t k ) + 1] ( t + k ) ( t k ) = 1 2 k ± 3 t 2 + 6 tk + 3 k 2 12 t 12 k + 1 3 t 2 + 6 3 k 2 + 12 t 12 k + 1 ² = 12 24 k 2 k = 6 t 12 m/s , which is the velocity at time t from Exercise 7. 9. y 1 2 t 1234 5 y = 2 + 1 π sin t ) Fig. 1.1.9 At t = 1 the height is y = 2 ft and the weight is moving downward. 10. Average velocity over [1 , 1 + h 2 + 1 π sin π( 1 + h ) ³ 2 + 1 π sin π ´ h = sin + π h ) π h = sin π cos h ) + cos π sin h ) π h sin h ) π h . h Avg. vel. on [1 , 1 + h ] 1 . 0000 0 0 . 1000 -0.983631643 0 . 0100 -0.999835515 0 . 0010 -0.999998355 11. The velocity at t = 1 is about v 1 ft/s. The “ indicates that the weight is moving downward. 23
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SECTION 1.1 (PAGE 61) R. A. ADAMS: CALCULUS 12. We sketched a tangent line to the graph on page 55 in the text at t = 20. The line appeared to pass through the points ( 10 , 0 ) and ( 50 , 1 ) . On day 20 the biomass is growing at about ( 1 0 )/( 50 10 ) = 0 . 025 mm 2 /d. 13. The curve is steepest, and therefore the biomass is grow- ing most rapidly, at about day 45. 14. a) profit 25 50 75 100 125 150 175 year 2000 2001 2002 2003 2004 Fig. 1.1.14 b) Average rate of increase in profits between 2002 and 2004 is 174 62 2004 2002 = 112 2 = 56 (thousand$/yr). c) Drawing a tangent line to the graph in (a) at t = 2002 and measuring its slope, we find that the rate of increase of profits in 1992 is about 43 thousand$/year. Section 1.2 Limits of Functions (page 68) 1. From inspecting the graph y x 11 1 y = f ( x ) Fig. 1.2.1 we see that lim x →− 1 f ( x ) = 1 , lim x 0 f ( x ) = 0 , lim x 1 f ( x ) = 1 . 2. From inspecting the graph y x 123 1 y = g ( x ) Fig. 1.2.2 we see that lim x 1 g ( x ) does not exist (left limit is 1, right limit is 0) lim x 2 g ( x ) = 1 , lim x 3 g ( x ) = 0 .
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01_ISMCHAP1 - INSTRUCTOR'S SOLUTIONS MANUAL SECTION...

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