When interest is compounded CONTINUOUSLY:
𝐴 = 𝑃𝑒 ^{𝑟𝑡}
Problem 1 :
Madeline decided to invest her $500 Christmas money. She found a bank offering 5.5% compounded continuously. What will be her ending balance after investing for 4 years?
Solution :
Formula for compound interest,
𝐴 = 𝑃𝑒 ^{𝑟𝑡}
P = 500, r = 5.5%, t = 4
𝐴 = 500𝑒 ^{0.055(4)}
= 500𝑒 ^{0.22}
= 500 (1.2460)
= 623.02
Approximately $623.
Problem 2 :
Your parents starts a college fund when you enter kindergarten at age 5. They invest $14,000 in an account that earns 0.8% annual interest compounded continuously.
If you are 18 years old when you enter college, how much money is available in your college find at that time, to the nearest dollar?
P = 14000, r = 0.8%, t = 13
𝐴 = 14000𝑒 ^{0.08(13)}
= 14000𝑒 ^{1.04}
= 14000 (2.828)
= 39604.7
Approximately $39605.
Problem 3 :
The population P (in thousands), Nevada can be modeled
P = 134𝑒^{kt}
where t is the year with t = 0, corresponding to the year 1990. In 2000 the population was 18000.
Solution :
P = 134000𝑒^{kt }-----(1)
After 10 years, the population is 18000.
18000 = 134000𝑒^{k(10)}
18000/134000 = 𝑒^{10k}
1.343 = 𝑒^{10k}
ln (1.343) = 10k
0.2949 = 10k
k = 0.2949/10
k = 0.02949
k = 0.0295
Approximately the value of k is 0.0295.
Applying the value of k in (1), we get
P = 134000𝑒^{0.0295t }-----(1)
To figure out the population in 2010, we have to apply t = 20.
P = 134000𝑒^{0.0295(20)}
= 134000𝑒^{0.59}
= 134000(1.8038)
= 134000(1.8038)
= 241719.6
So, the approximate value is 241720.
Problem 4 :
The population (in thousands) of Las vegas, Nevada by
P = 258000𝑒^{kt}
where t is the year with t = 0 corresponding to the year 1990. In 2000 the population was 478000.
Solution :
P = 258000𝑒^{kt }-----(1)
After 10 years, the population is 478000.
478000 = 258000𝑒^{k(10)}
478000/258000 = 𝑒^{10k}
1.853 = 𝑒^{10k}
ln (1.8527) = 10k
0.6166 = 10k
k = 0.6166/10
k = 0.06166
k = 0.0617
Approximately the value of k is 0.0617
Applying the value of k in (1), we get
P = 258000𝑒^{0.0617t }-----(1)
To figure out the population in 2010, we have to apply t = 20.
P = 258000𝑒^{0.0617 }^{(20)}
= 258000𝑒^{1.234}
= 258000(3.4345)
= 886101
So, the required value is 886101.
Problem 5 :
The initial bacterium count in the culture is 500. A biologist later makes a sample count of bacteria in the culture and finds that the relative rate of growth is 40% per hour.
Solution :
A = 500, r = 40%
y = 500𝑒^{(0.40)t }
y = 500𝑒^{0.40t}
After 10 years.
y = 500𝑒^{0.40(10)}
= 500𝑒^{4}
= 500(54.5755)
= 27287.75
= 27288.
So, after 10 hours the estimated count of bacteria is 27288.
Problem 6 :
A certain breed of rabbit was introduces onto a small island about 8 years ago. The current rabbit population on the island is estimated to be 4100, with a relative growth rate of 55% per year.
Solution :
Initial value = A, k = 55% or 0.55
After 8 years, the population of rabbit is 4100.
4100 = A𝑒^{0.55(8)}
4100 = A𝑒^{4.4}
A = 4100/𝑒^{4.4}
A = 4100/81.41
A = 50.36
Population after 12 years :
y = A𝑒^{0.55(12)}
Applying the value of A, we get
y = 50.36𝑒^{6.6}
= 50.36(734.59)
= 36993.9524
Problem 7 :
The population of the world in 2000 was 6.1 billion and the estimated related growth rate was 1.4% per year. If the population continues to grow at this rate, when will it reach 122 billion ? y = A𝑒^{kt}
Solution :
Initial value = 6.1
Growth rate = 1.4% = 0.014
At t = ?, the value of A will become 122 billion.
122 = 6.1𝑒^{0.014t}
122/6.1 = 𝑒^{0.014t}
20 = 𝑒^{0.014t}
ln (20) = 0.014 t
2.9957 = 0.014 t
t = 2.9957/0.014
t = 213.97
Approximately after 214 years, the population will become 122 billion.
Problem 8 :
A culture starts with 8600 bacteria. After 1 hour the count 10000.
Find the function that models the number of bacteria after t hours. y = A𝑒^{kt}
Solution :
y = A𝑒^{kt}
Initial value = 8600
After 1 hour, the number of bacteria = 10000
10000 = 8600𝑒^{k(1)}
10000/8600 = 𝑒^{k(1)}
1.1627= 𝑒^{k(1)}
ln (1.1627) = k
k = 0.1507
y = 8600𝑒^{0.1507t}
Finding the number of bacteria after 2 hours :
y = 8600𝑒^{0.1507(2)}
= 8600𝑒^{0.3014}
= 8600(1.3517)
= 11624.62
Approximately 11625.
Number of bacteria doubles :
17200 = 8600𝑒^{0.1507t}
17200/8600 = 𝑒^{0.1507t}
2 = 𝑒^{0.1507t}
ln 2 = 0.1507t
t = ln2 / 0.1507
t = 0.6931/0.1507
t = 4.59
t = 4.6 hours
Mar 14, 24 10:44 PM
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