COMPOUND CONTINUOUSLY PROBLEMS

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.

• Find the value of k for the model. Round the answer for four decimal places.
• Use the model to predict the population in 2010.

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.

• Find the value of k for the model. Round your decimal nearest four digits.
• Use the model to predict the population in 2010.

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.

• Find a function that models the number of bacteria after t hours. y = a𝑒^kt 
• What is the estimated count after 10 hours.

Solution :

A = 500, r = 40%

y = 500𝑒^(0.40)t 

y = 500𝑒^0.40t

After 10 years.

y = 500𝑒^0.40(10)

= 500𝑒⁴

= 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.

• What was the initial size of the rabbit population, use y = A𝑒^kt
• Estimate the population 12 years from now.

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

• Find the bacteria after 2 hours.
• After how many hours the number of bacteria will become double.

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