GROWTH FACTOR AND GROWTH RATE

For example, you invest $500 in a bank account which pays 10% interest per year compounded annually.

A = P(1 + ʳ⁄n)^nt

Here,

P = 500

r = 10% or 0.1

n = 1 (interest compounded annually, 1 time in a year)

Then, we have

A = 500(1 + ⁰^.¹⁄₁)^(1)t

A = 500(1 + 0.1)^t

A = 500(1.1)^t

Growth factor = 1.1

Growth rate = 1.1 - 1 = 0.1 or 10%

Meaning :

Solved Problems

Problem 1 :

Mr. Kevin invests $100 in a bank account which pays 10% interest per year compounded semiannually. Find the growth factor and the growth rate.

Solution :

Compount Interest Formula :

A = P(1 + ʳ⁄n)^nt

Here,

P = 100

r = 10% or 0.1

n = 2

(interest compounded semiannually, 2 times in a year)

Then, we have

A = 100(1 + ⁰^.¹⁄2)^(2)t

A = 500(1 + 0.05)^2t

A = 500[(1.05)²]^t

A = 500(1.1025)^t

Growth factor = 1.1025

Growth rate = 1.1025 - 1

= 0.1025

= 10.25% per year

Problem 2 :

You invest $100 in bank which pays 10% interest per year compounded continuously. State the growth rate and the growth factor.

Solution :

Continuous Compounding Formula :

A = Pe^rt

Here,

P = 2500

r = 10% or 0.1

t = 10

e ≈ 2.71828

Then, we have

A = 100(2.71828)^0.1t

= 100[(2.71828)^0.1]^t

= 100(1.1052)^t

Growth factor = 1.1052

Growth rate = 1.1052 - 1

= 0.1052

= 10.52% per year

Problem 3 :

A population of rabbits is observed to grow by 4% each month. Write an exponential function to model the population P(t) over t years assuming continuous growth. Use P_o to denote the initial population and state the growth factor and the growth rate.

Solution :

Exponential function to model the population P(t) over t years :

P(t) = P_oe^rt

Here,

r = 4% or 0.04

e ≈ 2.71828

Since the population of rabbits grows by 4% each month, time has to be in months. To convert t years to months, multiply t by 12.

t years = 12t months

Then, we have

P(t) = P_o(2.71828)^0.04(12t)

P(t) = P_o(2.71828)^0.48t

P(t) = P_o[(2.71828)^0.48]^t

P(t) = P_o(1.616)^t

Growth factor = 1.616

Growth rate = 1.616 - 1

= 0.616

= 6.16% per year

Problem 4 :

Cafine has a half-life of 2 hours, meaning the amount of caffeine in the body decreases by 50% every 2 hours. Write a general formula, C(t), to represent the amount of caffeine (in mg) after t hours. Use a_o to represent the initial amount. State the decay factor and the decay rate.  

Solution :

Half-life Decay Formula :

A = P(½)^ᵗ⁄d

Here,

A = C(t)

P = a_o

d = 2 (half-life time)

Then, we have

C(t) = a_o(½)^ᵗ⁄2

C(t) = a_o[(½)^1⁄2]^ᵗ

C(t) = a_o(0.707)^ᵗ

Growth factor = 0.707

Growth rate = 1 - 0.707

= 0.293

= 29.3% per hour

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