Question

As in Example 2, use the definition to find the Laplace transform for \(f(t)\), if it exists. In each exercise, the given function \(f(t)\) is defined on the interval \(0 \leq t<\infty\). If the Laplace transform exists, give the domain of \(F(s)\). In Exercises 9-12, also sketch the graph of \(f(t)\). $$ f(t)=e^{3 t} $$

Short answer

Question: Find the Laplace transform of the given function \(f(t) = e^{3t}\), its domain, and sketch the graph of \(f(t)\). Answer: The Laplace transform of \(f(t) = e^{3t}\) is \(F(s) = \frac{1}{3-s}\) for \(s<3\). The domain of the Laplace transform is \(s<3\). The graph of the given function represents an exponential growth.

Step-by-step solution

Use the Definition of Laplace Transform

We'll start by using the definition of Laplace transform, integrate the product of given function \(f(t)\) and \(e^{-st}\). So, $$F(s) = \int_0^\infty e^{3t}e^{-st} dt$$ To simplify the integrand, we can combine the exponents: $$F(s) = \int_0^\infty e^{(3-s)t} dt$$ Now, we can integrate with respect to \(t\): $$F(s) = \left[-\frac{1}{3-s}e^{(3-s)t}\right]_0^\infty$$ The next step is to evaluate this expression at the limits of integration.

Evaluate the Limits of Integration

Evaluate the expression at t = 0 and t = ∞: $$F(s) = \lim_{T\to\infty} \left[-\frac{1}{3-s}e^{(3-s)T}+\frac{1}{3-s}\right]$$ In order for the Laplace transform to exist, the first term in the bracket should be zero as T approaches infinity: $$\lim_{T\to\infty} -\frac{1}{3-s}e^{(3-s)T} = 0$$ In order for this expression to be zero, \((3-s)\) must be greater than 0: $$(3-s) > 0$$ $$s < 3$$ Therefore, the Laplace transform exists only for \(s < 3\). So, the Laplace transform of the given function is $$F(s) = \frac{1}{3-s},\quad s<3$$ and its domain is \(s<3\).

Sketch the Graph of \(f(t)\)

The given function is \(f(t) = e^{3t}\), which represents an exponential growth as \(t\) increases. For \(t=0\), \(f(0)=e^0=1\). As the value of \(t\) increases, the value of \(f(t)\) also increases rapidly. To sketch the graph: 1. Make sure the y-axis intercept is at 1. 2. The function is always positive, so it never intersects the x-axis. 3. The graph increases rapidly as \(t\) increases. To summarize: 1. The Laplace transform of \(f(t) = e^{3t}\) is \(F(s) = \frac{1}{3-s}\) for \(s<3\). 2. The domain of the Laplace transform is \(s<3\). 3. The graph of the given function represents an exponential growth.

Key concepts

Exponential Growth

Exponential growth refers to a process where a quantity increases rapidly over time. In mathematical terms, an exponential growth function can be expressed as \( f(t) = e^{kt} \), where \( k \) is a positive constant. This constant determines the rate of growth. The larger the value of \( k \), the faster the quantity grows.

In our example, the function \( f(t) = e^{3t} \) is given. Here, \( k = 3 \), indicating the rapid rate of growth for \( f(t) \). At \( t = 0 \), the value of \( f(t) \) is \( e^0 = 1 \). As \( t \) increases, \( f(t) \) grows exponentially larger.

Exponential functions like this are crucial to model real-world phenomena such as population growth, radioactive decay, and certain financial investments.

• Always positive: The function \( e^{3t} \) is always greater than zero.
• Rapid increase: The value of \( f(t) \) quickly becomes very large as \( t \) increases.

Integrals and Limits

Integrals and limits are fundamental building blocks in calculus. They help us understand the behavior and properties of functions, particularly over specific intervals.

In the context of the Laplace transform, we use an integral to find \( F(s) \), the transformed function. We take the integral of the product of the input function \( f(t) \) and the exponential term \( e^{-st} \). This becomes \( \int_0^\infty e^{(3-s)t} dt \).

Once we set up the integral, evaluating limits helps determine the conditions under which our transformed function exists. Specifically, we want the evaluated integral to not "blow up" as it approaches infinity. For our example, we found that the integral converges only if \( s < 3 \). Otherwise, the term \( e^{(3-s)T} \) would become unbounded as \( T \) goes to infinity.

• Integration: By calculating definite integrals, we find accumulated values over an interval, helping us define the Laplace transform function.
• Limit evaluation: Limits tell us where the function's expression remains valid and bounded.

Function Domain

Understanding the domain of a function is essential because it identifies all possible inputs for which the function is defined. In the case of the Laplace transform, we determined the domain of \( F(s) \) from conditions on the limits in the integral evaluation.

For our specific problem, the domain of the Laplace transform function \( F(s) = \frac{1}{3-s} \) is limited to values where \( s < 3 \). This is because, for \( s \geq 3 \), the integral would not converge as \( e^{(3-s)T} \) grows without bounds, making the Laplace transform undefined.

Understanding domains helps ensure that any calculations or applications involving the function will produce real and meaningful results. For any function, its domain is the set of "allowable" input values.

• Non-convergent integrals indicate domain boundaries where functions become undefined.
• Domains ensure functions are usable within specific contexts, offering stability in calculations.