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PracticeProblems for Exponential Growth and Exponential Decay Exponential Growth: 1.A population of raccoons in a state park quadruples every 40 years. a. Find the function that models the growth of this population. Since you have not been given the initial population, write that as P 0 in the equation. b. If the population at time t = 20 is 200 raccoons, what was the initial population? c. What is the population at time t = 10? d. At what time does the population consist of 1200 raccoons? e. Write the population function using an appropriate base so that the exponent consists only of the variable t. f. What is the annual growth rate for this population of raccoons? State your answer in sentence form that indicated what the annual PERCENTAGE growth rate is. g. Write the population function using natural base e. h. Use the form of the equation P(t) = 100 ekt to find the population at time t = 25 i. Use the form of the equation equation P(t) = 100 ekt to find the time at which the population equals 750 raccoons. Exponential Decay: 2. A radioactive substance decays in a manner so that at time t = 50 days, 30% of the substance remains. a. Find the exponential function that models the amount of the substance that remains at time t days. b. If the amount at time t = 20 days is 150 grams , what was the initial amount? c. What is the amount at time t = 12 days? d. At what time is there 100 grams of this substance? e. Write the population function using an appropriate base so that the exponent consists only of the variable t. f. At the end of one day, what percent of the initial amount still remains? State your answer in sentence form that indicates what the remaining PERCENTAGE is. g. What is the daily decay rate for this substance? State your answer in sentence form that indicates what the daily PERCENTAGE decay rate is. h. Write the amount function using natural base e. i. Use the form of the equation A(t) = 242.8 ekt to find the amount at time t = 30 days j. Use the form of the equation A(t) = 242.8 ekt to find the time at which the 80 grams of the substance remain. k. Find the half life of this substance: There are two methods, depending on which version of the equation you choose to use. The answers to each will differ slightly due to the rounding that occurred when you calculated k for the equation that used base e. Solution to Practice Problem for Exponential Growth 1.A population of raccoons in a state park quadruples every 40 years. a. Find the function that models the growth of this population. Since you have not been given the initial population, write that as P 0 in the equation. t 40 P (t ) P0 4 b. If the population at time t = 20 is 200 raccoons, what was the initial population? t 40 P (t ) P0 4 20 200 P0 4 40 0.5 200 P0 4 200 P0 40.5 200 200 200 P0 0.5 100 raccoons 2 4 4 c. What is the population at time t = 10? t t 40 40 P(t ) P0 4 100 4 10 P(10) 100 4 40 141. 42 141 raccoons d. At what time does the population consiste of 1200 raccoons? t 40 P (t ) 100 4 t 1200 100 4 40 t 12 4 40 t 40 ln 12 ln 4 t ln 12 ln 4 40 ln 12 71.699 71.7 years t = 40 ln 4 e. Write the population function using an appropriate base so that the exponent consists only of the variable t. t 401 t P(t ) 100 4 1001. 035 3 f. What is the annual growth rate for this population of raccoons? State your answer in sentence form that indicated what the annual percentage growth rate is. 1. 035 3 1 .0353 The population is increasing at the rate of 3.53% per year. g. Write the population function using natural base e. t 1 t P(t ) 100 4 40 1001. 035 3 100ekt 1 40 4 1. 035 3 e k 1 1 k ln 4 40 ln 4 0.03 466 40 OR k ln 1.0353 0.03 469 Therefore k 0.03 47 t 1 0.03 47t t P(t ) 100 4 40 1001. 035 3 100e h. Use the form of the equation P(t) = 100 ekt 0.03 47( 25) P(25) 100e 238 raccoons i. Use the form of the equation P(t) = 100 ekt equals 750 raccoons 750 100e0.03 47t 750 100 e 0.03 47t 7.5 e0.0347t ln 7.5 0.0347t t ln 7.5 58 years 0.03 47 to find the population at time t = 25 to find the time at which the population Solution to Practice Problem for Exponential Decay A radioactive substance decays in a manner so that at time t = 50 days, 30% of the substance remains. a. Find the exponential function that models the amount of the substance that remains at time t days. t A(t ) A0 .30 50 b. If the amount at time t = 20 days is 150 grams , what was the initial amount? 50t A(t ) A0 .30 20 150 A0 .30 50 .4 150 P0 0.30 150 P0 0 . 3 0 .4 150 P0 242. 8 grams 0.30.4 c. What is the amount at time t = 12 days? t t 50 50 A(t ) A0 0.30 242. 8 0.30 12 A(12) 242. 8 0.30 50 181. 873 grams d. At what time is there 100 grams of this substance? t 50 A(t ) 242. 8 0.30 t 100 242. 8 0.30 50 t 100 242. 8 0.30 50 0.41186 0.30 t 50 t 50 ln 0.411 86 ln 0.30 t ln 0.41186 ln 0.30 50 ln 0.411 86 t= 50 ln 0.30 t 36. 84 days 36.839 e. Write the population function using an appropriate base so that the exponent consists only of the variable t. t t 1 A(t ) 242.8 0.30 50 242.8 0.30 50 242.8 0.976 2t f. At the end of one day, what percent of the initial amount still remains? State your answer in sentence form that indicates what the remaining PERCENTAGE is. b 0.976 2 At the end of one day, 97.62% of the initial amount still remains. g. What is the daily decay rate for this substance? State your answer in sentence form that indicates what the daily percentage decay rate is. 1 0.976 2 0.023 8 The substance is decaying at the rate of 2.38% per day. hg. Write the amount function using natural base e. t 1 50 A(t ) 242.8 0.30 242.8 0.976 2t 0.30 1 50 0.976 2 ek 1 1 50 k ln 0.30 ln 0.30 .02 407 9 50 OR k ln 0.976 2 .02 408 8 Therefore k 0.0241 t 1 50 A(t ) 242.8 0.30 242.8 0.976 2t 242.8e 0.0241 t i. Use the form of the equation A(t) = 242.8 ekt to find the amount at time t = 30 days A(30) 242.8e0.0241(30) 117.83 grams j. Use the form of the equation A(t) = 242.8 ekt of the substance remain. 80 242. 8e 0.0241t 80 e0.0241 t 242 .8 0.329 49 e0.0241t ln 0.329 49 0.0241t t ln 0.329 49 46. 067 days 0.0241 to find the time at which the 80 grams k. Find the half life of this substance: There are two methods, depending on which version of the equation you choose to use. The answers to each will differ slightly due to the rounding that occurred when you calculated k for the equation that used base e. Method 1 t 50 A(t ) 242.8 0.30 t 50 (0.5)( 242.8) 242.8 0.30 t 50 0.5 0.30 t 50 ln 0.5 ln 0.30 t ln 0.5 ln 0.30 50 t 50 ln 0.5 28. 786 days ln 0.30 Method 2 A(t ) 242.8e0.0241t (0.5)(242.8) 242.8e0.0241t 0.5 e0.0241 t ln 0.5 0.0241t t ln 0.5 28. 761 days 0.0241