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Chapter 8: Problem 37

(a) Make a conjecture about the effect on the graphs of \(y=y_{0} e^{k t}\) and \(y=y_{0} e^{-k t}\) of varying \(k\) and keeping \(y_{0}\) fixed. Confirm your conjecture with a graphing utility. (b) Make a conjecture about the effect on the graphs of \(y=y_{0} e^{k t}\) and \(y=y_{0} e^{-k t}\) of varying \(y_{0}\) and keeping \(k\) fixed. Confirm your conjecture with a graphing utility.

Short Answer

Expert verified
Varying \( k \) changes growth/decay rates, while varying \( y_0 \) shifts the graph vertically.

Step by step solution

01

Understanding the Exponential Functions

The given functions are exponential: \( y = y_0 e^{kt} \) and \( y = y_0 e^{-kt} \). Here, \( y_0 \) is the initial value, \( k \) is the growth or decay rate, and \( t \) is the time variable. The behavior of each function depends on the sign and magnitude of \( k \), and the value of \( y_0 \).
02

Effect of Varying \( k \)

Consider the exponential growth \( y = y_0 e^{kt} \). When \( k > 0 \), the function shows growth, and when \( k < 0 \), it depicts decay. Increasing \( k \) results in a steeper increase in growth or decay. The larger \( |k| \), the faster the growth or decay rate. For \( y = y_0 e^{-kt} \), the pattern reverses: \( k > 0 \) leads to decay, and \( k < 0 \) results in growth.
03

Conjecture for Varying \( k \)

The primary effect of increasing \( k \) is that the graph of \( y = y_0 e^{kt} \) becomes steeper and grows faster, while the graph of \( y = y_0 e^{-kt} \) decays faster. Decreasing \( k \) results in slower growth or decay rates.
04

Using a Graphing Utility for Part (a)

Plot both functions for various values of \( k \) while keeping \( y_0 \) fixed. Observe how increasing \( k \) causes the graph of \( y = y_0 e^{kt} \) to rise more sharply, whereas for \( y = y_0 e^{-kt} \), it descends more steeply. Confirm that these observations match the conjecture.
05

Effect of Varying \( y_0 \)

When \( y_0 \) varies and \( k \) is fixed, the function's initial value adjusts, shifting the entire graph up or down. A larger \( y_0 \) results in reaching higher values, while a smaller \( y_0 \) limits the graph to lower values. \( y_0 \) controls the starting point of the graph on the y-axis but doesn't affect the growth or decay rate.
06

Conjecture for Varying \( y_0 \)

Changing \( y_0 \) affects only the vertical position of the graphs. Increasing \( y_0 \) shifts graphs upward, and decreasing \( y_0 \) shifts them downward, without altering their shape or growth/decay characteristics.
07

Using a Graphing Utility for Part (b)

Graph the functions for various values of \( y_0 \) while \( k \) remains constant. Notice how the graphs move up with higher \( y_0 \) and down with lower \( y_0 \). The shape remains unchanged, confirming the conjecture that \( y_0 \) only affects vertical shifts.

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Key Concepts

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

Growth and Decay Rate
Exponential functions like \( y = y_0 e^{kt} \) or \( y = y_0 e^{-kt} \) show unique growth or decay patterns based on the parameter \( k \). This parameter, \( k \), dictates how fast the function grows or decays over time \( t \). Here's the breakdown:

  • When \( k > 0 \), the function \( y = y_0 e^{kt} \) exhibits growth. As \( k \) increases, the rate at which it rises becomes steeper, indicating a quicker growth pace.
  • Conversely, \( y = y_0 e^{-kt} \) decays if \( k > 0 \). The higher the value of \( k \), the sharper the decline in the graph. It signifies a rapid decay rate.
  • When \( k < 0 \), these behaviors flip. \( y = y_0 e^{kt} \) now decays, whereas \( y = y_0 e^{-kt} \) grows, albeit at rates defined by the magnitude of \(|k|\).
Understanding these effects helps predict how the graph will behave simply by changing the value of \( k \). Larger values intensify the growth or decay rate, resulting in graphs that either spike up quickly or drop down sharply.
Initial Value
The initial value, denoted as \( y_0 \) in exponential functions, is a crucial determinant of the starting point of the graph on the y-axis. Changing \( y_0 \) influences where the function begins, but not how it grows or decays.

  • When you increase \( y_0 \), the graph shifts upward. This is because \( y_0 \) sets the scale for the entire function's initial condition, meaning it starts higher on the y-axis.
  • Conversely, decreasing \( y_0 \) moves the graph downward, reflecting a lower starting value.
  • The shape and rate of growth or decay remains consistent regardless of \( y_0 \). Hence, \( y_0 \) acts as a vertical translation factor, moving the graph up or down without altering its trajectory.
Students should note that although \( y_0 \) defines the initial state, the overall behavior in terms of growth or decay is solely controlled by the parameter \( k \). Understanding this separation helps differentiate the role of initial conditions versus growth rates in exponential functions.
Graphing Utility
A graphing utility is a powerful tool for visualizing how exponential functions behave as parameters change. It allows students to concretely see the effect of varying \( k \) and \( y_0 \). Here's how you can utilize it effectively:

  • Start by entering the function \( y = y_0 e^{kt} \) or \( y = y_0 e^{-kt} \) into the utility. Choose different values for \( k \) while keeping \( y_0 \) constant. Observe the sharpness of the graph as it grows or decays.
  • Next, change \( y_0 \) while maintaining a constant \( k \). Notice how the graph shifts vertically without altering its growth or decay pattern.
  • Using these visualizations confirms theoretical predictions about growth and decay rates as well as the vertical influence of the initial value.
This interactive approach provides practical insights into mathematical concepts. By experimenting with a graphing utility, students can solidify their understanding of exponential functions, aligning visual observations with theoretical conjectures.

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Most popular questions from this chapter

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of x. $$ e^{-y} \sin x-y^{\prime} \cos ^{2} x=0 $$

(a) Suppose that a particle moves along an \(s\) -axis in such a way that its velocity \(v(t)\) is always half of \(s(t)\) Find a differential equation whose solution is \(s(t)\) (b) Suppose that an object moves along an \(s\) -axis in such a way that its acceleration \(a(t)\) is always twice the velocity. Find a differential equation whose solution is \(s(t) .\)

(a) Sketch some typical integral curves of the differential equation \(y^{\prime}=-x / y .\) (b) Find an equation for the integral curve that passes through the point \((3,4) .\)

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of x. $$ y^{\prime}-(1+x)\left(1+y^{2}\right)=0 $$

Find a solution to the initial-value problem. \(x^{2} y^{\prime}+2 x y=0, y(1)=2 \quad[\text {Hint}:\) Interpret the left-hand side of the equation as the derivative of a product of two functions. \(]\)
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