Problem 11 - Differential Equations and Linear Algebra

Question

Make a sketch of the given function on \([0, \infty)\) and express it in terms of the unit step function. $$f(t)=\left\\{\begin{array}{cc}2, & 0 \leq t<1 \\\2 e^{(t-1)}, & t>1\end{array}\right.$$

Step-by-Step Solution

Answer

The given function can be sketched as a horizontal line at \(y = 2\) for \(0 \leq t < 1\) and an exponential curve starting at \(t=1\) and \(f(t)=2\), growing exponentially as t increases. The function in terms of the unit step function is: $$ f(t) = 2u(t) - 2u(t-1) + 2e^{(t-1)}u(t-1) $$

Step 1: Sketch the function over \([0, \infty)\)

To sketch the given function, we will consider the two intervals separately: 1) For \(0 \leq t < 1\), the function is a constant, \(f(t) = 2\). It will be a horizontal line at \(y=2\) over the interval [0, 1). 2) For \(t>1\), the function is given by \(f(t) = 2e^{(t-1)}\). This is an exponential function, where the graph starts at \(t=1\) and \(f(t)=2\) and grows exponentially as t increases. Combining these two pieces of information, we can sketch the graph of the function.

Step 2: Express the function in terms of the unit step function

To express the given function in terms of the unit step function, we need to manipulate the unit step function to match the intervals and expressions given for the function. For the interval \(0 \leq t < 1\), the function takes the constant value 2. We can use the unit step function to represent this as follows: $$ 2u(t) - 2u(t-1) $$ For the interval \(t > 1\), the function takes the form \(2e^{(t-1)}\). We need the unit step function to help us shift the interval and represent the function properly: $$ 2e^{(t-1)}(u(t-1)) $$ Now we can combine them to represent the entire function in terms of the unit step function: $$ f(t) = 2u(t) - 2u(t-1) + 2e^{(t-1)}u(t-1) $$

Key Concepts

Piecewise Functions

Piecewise functions, as the name suggests, are defined by different expressions for different intervals of the input variable. They are like patchwork quilts, where each patch is a different function, operating in its own range:
- Each "piece" corresponds to a specific part of the domain. In our case, two pieces exist: a constant function and an exponential function.
- The first piece \(f(t) = 2\) governs the function over \(0 \leq t < 1\). This means for any value of \(t\) in this interval, \(f(t)\) will be 2, creating a horizontal line when graphed.- The second piece \(f(t) = 2e^{(t-1)}\) is in effect for \(t > 1\). Here, the function rapidly increases as \(t\) grows, characteristic of exponential growth.
To better manipulate these intervals in problems, we often turn to the unit step function, which helps "turn on" or "turn off" a part of the piecewise function at specific points.

Exponential Functions

Exponential functions are a key concept in mathematics, particularly noticeable for their role in growth and decay processes.
- In our piecewise function, the segment \(f(t) = 2e^{(t-1)}\) for \(t > 1\) is an exponential function.
- The base, \(e\), is a constant approximately equal to 2.718, often referred to as Euler's number, which provides a natural basis for exponential growth.- Exponentials demonstrate rapid increase or decrease depending on the function's structure. In our function, as \(t\) increases past 1, the output \(f(t)\) will grow faster and faster due to the exponential term \(e^{(t-1)}\).
This characteristic of exponential functions makes them important in modeling real-life phenomena such as population growth, radioactive decay, and interest calculations.

Graph Sketching

Graph sketching is a powerful tool to visually interpret mathematical functions and understand their behaviors over different intervals.
- Begin by identifying key points and transitions in the function. In our piecewise function, take note of where the function changes its expression, specifically at \(t = 1\) here.
- For \(0 \leq t < 1\), you draw a straight horizontal line at \(y = 2\). This indicates the constant value of the function in this range.
- After \(t > 1\), the exponential growth part \(2e^{(t-1)}\) starts. Beginning at this pivot point, sketch the curve that rises steeply as \(t\) increases beyond 1.Utilizing graphs not only provides a visual representation of the function but also helps uncover critical points and potential discontinuities, making complex functions much easier to comprehend.