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Chapter 6: Problem 1

Describing Values Describe what the values of \(C\) and \(k\) represent in the exponential growth and decay model \(y=C e^{k t} .\)

Short Answer

Expert verified
In the exponential growth and decay model \(y=C e^{k t}\), \(C\) represents the initial value of the quantity under observation when \(t=0\), and \(k\) signifies the rate of growth or decay. If \(k>0\), it's an instance of growth, but if \(k<0\), it represents decay.

Step by step solution

01

Understanding the Exponential Model

The exponential growth and decay model is in the form \(y=C e^{k t}\) where \(y\) represents the quantity that's growing or decaying over time. The variable \(t\) represents the time period. \(C\) and \(k\) are constants.
02

Interpreting the constant C

In the exponential growth and decay model, \(C\) represents the initial value of the quantity at time \(t=0\). It is the starting point for the process of growth or decay.
03

Interpreting the constant k

The constant \(k\) represents the rate of growth or decay. If \(k>0\) the process is one of exponential growth where the quantity increases over time. However, if \(k<0\), the process is one of exponential decay, where the quantity decreases over time.

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Key Concepts

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

Initial Value
In the context of exponential growth and decay, the term \(C\) in the equation \(y=C e^{k t}\) serves a crucial role. The initial value, \(C\), is the starting point of the quantity you are observing. This is the value of \(y\) when \(t = 0\). In other words:
  • If you have a population of bacteria, \(C\) represents the initial number of bacteria present before any growth occurs.
  • In financial terms, \(C\) could be the initial amount of money in a savings account before any interest is added.
Understanding \(C\) helps you know where your exponential journey begins. It's like the "launching pad" for the growth or decay process. Without an initial value, it would be difficult to track changes over time.
Growth Rate
The growth rate is described by the constant \(k\) in the equation \(y=C e^{k t}\). Whenever \(k>0\), the situation describes exponential growth. This means the quantity is increasing as time progresses. Here鈥檚 why the growth rate matters:
  • It determines how quickly the initial value increases.
  • A larger \(k\) indicates faster growth.
  • This could reflect how quickly a population of animals increases if there are ample resources.
Think of \(k\) as a speedometer for your growth journey. The greater \(k\) is, the faster your quantity will increase. It is crucial in understanding how dynamic and rapid the changes can be in your scenario. Whether you're examining growth in nature, finance, or any area of interest, knowing the growth rate helps to predict future values accurately.
Decay Rate
The decay rate is also represented by the constant \(k\) but reflects a decrease in quantity when \(k < 0\). In other words, it鈥檚 the reverse of exponential growth, indicating that:
  • The quantity decreases over time.
  • A more negative \(k\) suggests faster decay.
  • This could illustrate the decay of radioactive substances, where atoms diminish over time.
By understanding the decay rate, you gain insight into how quickly something is diminishing. It's essential in areas such as environmental science, physics, and even economics, where resources or values might decrease over time. With \(k\) being negative, it becomes a compass guiding expectations about the reduction trends.

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Most popular questions from this chapter

In Exercises \(17-24,\) find the particular solution of the first- order linear differential equation for \(x>0\) that satisfies the initial condition. $$\begin{array}{ll}{\text { Differential Equation }} & {\text { Initial Condition }} \\ {y^{\prime}+(2 x-1) y=0} & {y(1)=2}\end{array}$$

Using a Logistic Equation In Exercises 53 and 54 , the logistic equation models the growth of a population. Use the equation to (a) find the value of \(k,(\) b) find the carrying capacity, (c) find the initial population, (d) determine when the population will reach 50% of its carrying capacity, and (e) write a logistic differential equation that has the solution \(P(t).\) $$P(t)=\frac{5000}{1+39 e^{-0.2 t}}$$

Integrating Factor Explain why you can omit the constant of integration when finding an integrating factor.

Compound Interest In Exercises 49 and 50 , find the time necessary for \(\$ 1000\) to double when it is invested at rate \(r\) and compounded (a) annually, (b) monthly, (c) daily, and (d) continuously. $$r=7 \%$$

A 200 -gallon tank is half full of distilled water. Starting at time \(t=0,\) a solution containing 0.5 pound of concentrate per gallon is admitted to the tank at a rate of 5 gallons per minute, and the well-stirred mixture is withdrawn at a rate of 3 gallons per minute. (a) At what time will the tank be full? (b) At the time the tank is full, how many pounds of concentrate will it contain? (c) Repeat parts (a) and (b), assuming that the solution entering the tank contains 1 pound of concentrate per gallon.
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