app-b - Appendix B Exercise B1 2. a. The general...

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Exercise B1 2. a. The general antiderivative of f ( x )=12 x 2 – 24 x +1 is F ( x 1 } 1 3 } x 3 2 – 24 1 } 1 2 } x 2 2 + x + C =4 x 3 – 12 x 2 + x + C. Since F (1) = –2, we have: F (1)=4 12+1+ C = –2. Thus, –7 + C = –2 C =5 . The specific antiderivative is F ( x )=4 x 3 – 12 x 2 + x +5. b. The general antiderivative of f ( x )=3 Ï x w – sin x is F ( x 1 } 2 3 } x } 3 2 } 2 + cos x + C =2 x } 3 2 } + cos x + C. Since F (0) = 0, we have: F (0)=0+1+ C =0 . Thus, C = –1 and the specific antiderivative is F ( x )=2 x } 3 2 } + cos x – 1. d. The general antiderivative of f ( x )= e 3 x } 2 1 x } is F ( x } 1 3 } e 3 x } 1 2 } ln x + C. Since F (1) = e 3 , we have: F (1) = } 1 3 } e 3 } 1 2 } ln 1 + C = } 1 3 } e 3 – 0 + C = e 3 . Thus, C = } 2 3 } e 3 and the specific antiderivative is F ( x } 1 3 } e 3 x } 1 2 } ln x + } 2 3 } e 3 . e. The general antiderivative of f ( x } Ï x 3 x + 2 1 w } is F ( x )= (2) 1 } 1 3 } 2 ( x 3 +1) } 1 2 } + C. = } 2 3 } ( x 3 } 1 2 } + C Since F (0) = 4, we have: F (0) = } 2 3 } (0+1) } 1 2 } + C = } 2 3 } + C . Thus, C = } 1 3 0 } and the specific antiderivative is F ( x } 2 3 } Ï x 3 w + } 1 3 0 } . f. The general antiderivative of f ( x )= cos x sin 4 x is F ( x } 1 5 } (sin x ) 5 + C. Since F (0) = –1, we have: F (0) = } 1 5 } (sin 0) 5 + C = C = –1. The specific antiderivative is F ( x )= sin 5 x – 1. 3. We wish to determine a function P ( t ) that gives the population at any time t . We are given that the rate of change of the population is 3 + 4 t } 1 3 } . Using P' ( t )=3+4 t } 1 3 } , we can determine P ( t ), the general antiderivative: P ( t t +4 1 } 3 4 } t } 4 3 } 2 + C =3 t +3 t } 4 3 } + C. In order to determine the specific population function, we use the fact that the current population is 10 000, i.e., P (0) = 10 000: P (0) = 10 000 0+0+ C = 10 000 C = 10 000. Thus, the population at any time t is given by P ( t t t } 4 3 } + 10 000. Six months from now the population will be P (6) = 3(6)+3(6) } 4 3 } + 10 000 = 10 081. 4. We first need to determine a function V ( t ) that gives the volume of water in the tank at any time t . Since the water is leaking from the tank at the rate of } 5 t 0 } L/min, we have V' ( t } 5 t 0 } . Thus, V ( t )=– } 5 1 0 } 1 } 1 2 } t 2 2 + C =– } 1 1 00 } t 2 + C. Since V = 400 L at time t , C = 400. The volume of water in the tank at any time t is V ( t } 1 1 00 } t 2 + 400. Appendix B 271 Appendix B
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To determine when the tank will be empty, we solve V ( t )=0: } 1 1 00 } t 2 + 400 = 0 t 2 = 40 000 t = 200, t 0. The tank will be empty 200 min or 3 h 20 min from the time at which there were 400 L of water in it. 5. Let the measure of the inner radius of a water pipe at any time t be r ( t ). We are given that r' ( t )= –0.02 e –0.002 t cm/year. a. Thus, r ( t 1 } –0. 1 002 } e –0.002 t 2 + C = 10 e –0.002 t + C. Since r =1 when t =0: 1= 10 e °+ C =10+ C C = –9. The inner radius of a pipe at any time t is r ( t )=10 e –0.002 t – 9. b. When t =3 years, r (3) = 10 e (–0.002)(3) – 9 = 0.94 After three years, the inner radius of a pipe will be 0.94 cm. c. The pipe will be completely blocked when r =0 . To determine when this occurs, we solve r ( t 10 e –0.002 t – 9 = 0 e –0.002 t = } 1 9 0 } –0.002 t = ln(0.9), by definition t = } ln 0 ( . 0 0 . 0 9 2 ) } = 52.68. The pipe will be completely blocked in approximately 52.7 years from the time when its inner radius is 1 cm.
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app-b - Appendix B Exercise B1 2. a. The general...

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