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The functions in Problems \(17-20\) represent exponential growth or decay. What is the initial quantity? What is the growth rate? State if the growth rate is continuous. $$P=3.2 e^{0.03 t}$$

Short Answer

Expert verified
Initial quantity is 3.2, growth rate is 3%, and growth is continuous.

Step by step solution

01

Identify Initial Quantity

The formula given is an exponential function of the form \( P = P_0 e^{rt} \), where \( P_0 \) is the initial quantity. In our case, the initial quantity \( P_0 \) is 3.2.
02

Identify Growth Rate

In the formula \( P = P_0 e^{rt} \), the value of \( r \) represents the growth rate. Here, \( r = 0.03 \) which means the growth rate is 3% per time unit.
03

Determine Growth Type

The function uses the constant \( e \), which is the base of the natural logarithm. This indicates that the growth is continuous because \( e^x \) describes continuous growth or decay.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Initial Quantity
When dealing with exponential functions, the concept of initial quantity is crucial. It represents the starting value of whatever is being measured—often before any time has passed.
This value is depicted in the exponential equation of the form \( P = P_0 e^{rt} \), where \( P_0 \) stands for the initial quantity.

For example, in the given function \( P = 3.2 e^{0.03t} \), the initial quantity is \( 3.2 \). This means that at the very beginning of the observed period, the quantity of interest is \( 3.2 \). It's the baseline from which all growth (or decay) starts.
  • Think of \( P_0 \) as the very first value we record.
  • It guides us in projecting forward in time.
The initial quantity is foundational, as it sets the stage for understanding how the system evolves with time.
Exploring Growth Rate
The growth rate in an exponential function is the rate at which the initial quantity increases or decreases over time.
In the equation \( P = P_0 e^{rt} \), the term \( r \) serves as the growth rate.

In our function \( P = 3.2 e^{0.03t} \), the value of \( r \) is \( 0.03 \), which signifies that the quantity is growing at a rate of 3% per time unit.

It's important to note:
  • The growth rate tells us how quickly the quantity is changing.
  • A positive \( r \) implies growth, while a negative \( r \) indicates decay.
This percentage rate is handy because it normalizes the growth, allowing us to compare different systems easily based solely on their rates.
What is Continuous Growth?
Continuous growth is a type of growth characterized by the use of the mathematical constant \( e \), approximately equal to 2.718. This constant forms the base of natural logarithms.
Its presence in an exponential equation like \( P = P_0 e^{rt} \) indicates continuous growth.

With \( e \) as the base, our given function \( P = 3.2 e^{0.03t} \) displays a continuous growth pattern.
  • Continuous growth means the rate of growth is consistent at any point in time, without interruptions.
  • It models natural processes accurately, such as population growth or compound interest.
This seamless and perpetually ongoing process makes it a powerful concept in mathematical modeling and real-life applications.

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Most popular questions from this chapter

Kleiber's Law states that the metabolic needs (such as calorie requirements) of a mammal are proportional to its body weight raised to the 0.75 power. \(^{86}\) Surprisingly, the daily diets of mammals conform to this relation well. Assuming Kleiber's Law holds: (a) Write a formula for \(C,\) daily calorie consumption, as a function of body weight, \(W\) (b) Sketch a graph of this function. (You do not need scales on the axes.) (c) If a human weighing 150 pounds needs to consume 1800 calories a day, estimate the daily calorie requirement of a horse weighing 700 lbs and of a rabbit weighing 9 lbs. (d) On a per-pound basis, which animal requires more calories: a mouse or an elephant?

For Problems \(1-16,\) solve for \(t\) using natural logarithms. $$2 P=P e^{0.3 t}$$

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A person is to be paid 2000 for work done over a year. Three payment options are being considered. Option 1 is to pay the 2000 in full now. Option 2 is to pay \(\$ 1000\) now and \(\$ 1000\) in a year. Option 3 is to pay the full 2000 in a year. Assume an annual interest rate of \(5 \%\) a year, compounded continuously. (a) Without doing any calculations, which option is the best option financially for the worker? Explain. (b) Find the future value, in one year's time, of all three options. (c) Find the present value of all three options.

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