Practice problems for Exponential Growth and Exponential Decay - DOC

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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.
             40 
               t
 P(t )  P0 4 
                
                

b. If the population at time t = 20 is 200 raccoons, what was the initial population?
              40 
                 t
 P(t )  P0  4 
                   
                   
               20 
 200  P0  4 40 
                  
                  
          
 200  P 40
               0.5


 200
       P0
 40.5
      200 200 200
 P0  0.5        100 raccoons
      4     4   2

c. What is the population at time t = 10?
             40 
                t
                         t 
 P(t )  P0  4   100 4 40 
                           
                           
                10

 P(10)  100 4 40   141. 42  141 raccoons
                
                

d. At what time does the population consiste of 1200 raccoons?
                 40 
                     t
  P(t )  100   4 
                        
                        
                     t
                         
 1200  100 4 40 
                        
                        
           t
               
 12   4 40 
              
              
               40 
                   t
 ln 12  ln   4 
                      
                      
                ln 4
            t
 ln 12 
           40
         ln 12
 t = 40           71.699  71.7 years
          ln 4
e. Write the population function using an appropriate base so that the exponent consists
only of the variable t.
                    t
             40 
               1

 P(t )  100 4   1001. 0353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.
 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
 P(t )  100 4 40   1001. 0353  100ekt
                                   t
                  
                  
   1
 4 40    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
 P(t )  100 4 40   1001.0353  100e
                                  t       0.03 47t
             
             

h. Use the form of the equation P(t) = 100 ekt       to find the population at time t = 25
                   0.03 47(25)
 P(25)  100e                     238 raccoons

i. Use the form of the equation P(t) = 100 ekt       to find the time at which the population
equals 750 raccoons
 750  100e0.03 47t
 750
 100
          e 0.03 47 t
 7.5  e 0.0347t
 ln 7.5  0.0347t
         ln 7.5
 t              58 years
        0.03 47
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?
             50  t
             .30 
 A(t )  A0          
                     
                20
                     
 150  A0  .30 50 
                    
                    
 150  P0 0.30 .4




 150
         P0
0.3 0.4
        150
P0             242. 8 grams
       0.3 0.4

c. What is the amount at time t = 12 days?
                   t
                                      t
                                          
             0.30 50   242. 8 0.30 50 
 A(t )  A0                            
                                       
                       12
                           
 A(12)  242. 8 0.30 50   181. 873 grams
                          
                          

d. At what time is there 100 grams of this substance?
                      t
                          
                0.30 50 
 A(t )  242. 8          
                         
                    t
                         
 100  242. 8 0.30 50 
                        
                        
                 t
  100
 242. 8    0.30 50
                       t

0.411 86  0.30       50


                       t
                          
                 0.30 50 
ln 0.41186  ln          
                         
ln 0.41186  ln 0.30
              t
             50
             ln 0.41186
 t=   50 ln 0.30  36.839
t  36. 84 days
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 0.3050   242.8 0.3050   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
                       
 A(t )  242.8 0.3050   242.8  0.976 2t
                      
                      
        1

 0.30   50
              0.976 2  e k
              1
                  
         0.30 50   1 ln 0.30  .02 407 9
 k  ln           50
                 
OR k  ln 0.976 2  .02 408 8
Therefore k  0.0241
                               t
                    1
                        
 A(t )  242.8 0.30 50 
                           242.8  0.976 2t  242.8e0.0241t
                       
                       


i. Use the form of the equation A(t) = 242.8 ekt to find the amount at time t = 30 days
  A(30)  242.8e0.0241(30)  117.83 grams


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.
              0.0241t
 80  242. 8e
  80  e  0.0241t
 242 .8
 0.329 49  e0.0241t
 ln 0.329 49  0.0241t
     ln 0.329 49
t                46. 067 days
        0.0241
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 
A(t )  242 .8  0.30 50 
                        
                        
                                t 
(0.5)( 242 .8)  242 .8   0.30 50 
                                  
                                  
            t
0.5  0.30 50

                   t 
ln 0.5  ln  0.30 50 
                     
                     
          t
ln 0.5  ln 0.30
         50

          ln 0.5
t  50            28. 786 days
         ln 0.30


Method 2
 A(t )  242.8e0.0241t

(0.5)(242.8)  242.8e0.0241t

0.5  e0.0241t

ln 0.5  0.0241t

      ln 0.5
t             28. 761 days
      0.0241

 t             28. 761 days
       0.0241

				
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