Question

Find \(\mathcal{L}\\{f(t)\\}\) for each of the functions \(f\) defined. $$ f(t)=\left\\{\begin{array}{ll} t, & 03 \end{array}\right. $$

Short answer

The Laplace transform of the given piecewise function is: \(\mathcal{L}\\{f(t)\\} = \frac{1}{s^2}(1-e^{-3s})\).

Step-by-step solution

Identify the given function

We have a piecewise function \(f(t)\) defined as: $$ f(t)=\left\{\begin{array}{ll} t, & 03 \end{array}\right. $$

Find the Laplace transform for the first case

For the first case, \(f(t) = t\) for \(0<t<3\). We will use the Laplace transform definition to compute this part. The Laplace transform of \(f(t)\) is given by: $$ \mathcal{L}\\{f(t)\\} = \int_0^\infty f(t)e^{-st}dt $$ Since this is a piecewise-defined function, in the first case we will compute the integral from 0 to 3: $$ \mathcal{L}\\{f(t)\\}_{(0<t<3)} = \int_0^3 te^{-st}dt $$ We will use integration by parts to compute this integral. Let \(u=t\) and \(dv=e^{-st}dt\). Then, \(du=dt\) and \(v = -\frac{1}{s}e^{-st}\). Applying the integration by parts formula, we get: $$ \mathcal{L}\\{f(t)\\}_{(0<t<3)} = \left[-\frac{1}{s}te^{-st}\right]_0^3 +\frac{1}{s} \int_0^3 e^{-st} dt $$ Now, integrate the remaining term: $$ \mathcal{L}\\{f(t)\\}_{(0<t<3)} = \left[-\frac{1}{s}te^{-st}\right]_0^3 -\frac{1}{s^2}\left[e^{-st}\right]_0^3 $$ Applying the limits: $$ \mathcal{L}\\{f(t)\\}_{(0<t<3)}= -\frac{3}{s}e^{-3s}+\frac{1}{s^2}(1-e^{-3s}) $$

Find Laplace transform for the second case

For the second case, \(f(t) = 3\) for \(t>3\). Now, we compute the Laplace transform: $$ \mathcal{L}\\{f(t)\\}_{(t>3)} = \int_3^\infty 3e^{-st}dt $$ Integrate with respect to \(t\): $$ \mathcal{L}\\{f(t)\\}_{(t>3)} = -\frac{3}{s}\left[e^{-st}\right]_3^\infty $$ Applying the limits: $$ \mathcal{L}\\{f(t)\\}_{(t>3)} = \frac{3}{s}e^{-3s} $$

Combine the results

Now, we will combine the results from Steps 2 and 3 to find the Laplace transform of the entire piecewise function. Function \(f(t)\) can be represented by a combination of two cases: $$ \mathcal{L}\\{f(t)\\} = \mathcal{L}\\{f(t)\\}_{(03)} $$ Substituting the values found in Steps 2 and 3, we get: $$ \mathcal{L}\\{f(t)\\} = -\frac{3}{s}e^{-3s}+\frac{1}{s^2}(1-e^{-3s}) + \frac{3}{s}e^{-3s} $$ Simplifying the expression: $$ \mathcal{L}\\{f(t)\\} = \frac{1}{s^2}(1-e^{-3s}) $$ So, the Laplace transform of the given piecewise function is: $$ \mathcal{L}\\{f(t)\\} = \frac{1}{s^2}(1-e^{-3s}) $$

Key concepts

Piecewise Function

A piecewise function is a type of function that is defined by different expressions based on the input or the domain it's evaluated over. In simpler terms, a piecewise function consists of "pieces", and each piece can have its own formula depending on the range of values it considers. This allows for flexibility and precision in defining complex behaviors over specific intervals.

• Think of it like a tailor-made suit, where each piece of fabric is sewn together to create the complete outfit.
• Each segment of the piecewise function is valid for a specific interval. In our case, we have two segments: one where the function is equal to \(t\) for \(0 < t < 3\), and another where it is equal to \(3\) for \(t > 3\).

When calculating mathematical properties, like the Laplace Transform, each segment must be handled separately before combining the results to form a complete analysis of the function.

Integration by Parts

Integration by Parts is a powerful technique used in calculus to compute the integral of products of functions. It's particularly useful when direct integration is complex or impossible. This method stems from the product rule for differentiation and is expressed by the formula:\[\int u \, dv = uv - \int v \, du\]

• First, identify parts of the integral as \(u\) and \(dv\).
• Differential \(du\) is then found by differentiating \(u\), and \(v\) is calculated by integrating \(dv\).
• Finally, substitute into the integration by parts formula.

The choice of \(u\) and \(dv\) can influence the ease of solving the integral, so sometimes trial and error is necessary. In our example, for the integral of \(te^{-st}\), setting \(u=t\) and \(dv=e^{-st}dt\), we simplify the process by reducing the power of \(t\) through differentiation.

Transform of a Piecewise Function

Finding the Laplace Transform of a piecewise function involves calculating the transform for each piece separately, given its domain constraints, and then combining these results. This ensures each section of the piecewise function's behavior is accurately represented in the transform output.

• The Laplace Transform is particularly handy when working with piecewise functions, as it seamlessly adapts to different intervals.
• It's crucial to respect the boundaries of each piece when performing integration; extend beyond them, and accuracy is compromised.

In the exercise, each part of the function — from \(t=0\) to \(t=3\) and from \(t>3\) — required its own transform computation, which were then combined. This methodology not only ensures precision but also aids in visualizing the overall contribution of each interval to the final transform.

Integration Limits

Integration limits dictate the bounds of integration — from where to where the integral calculation should occur. This is crucial when dealing with piecewise functions. Each piece of the function must be integrated within its own set of boundaries.

• In mathematical analysis, setting correct integration limits ensures the computed values are valid and relevant to that function's segment.
• A mistake in integration limits can lead to inaccurate results, as each piece is defined and behaves distinctly within its intervals.

For this exercise, we integrated different expressions: one from \(0\) to \(3\), and another from \(3\) to \(\infty\). Notice how the integral calculation is confined within these specific boundaries to maintain its relevance and accuracy. Properly setting integration limits is an integral part of solving problems involving discontinuous or segment-defined functions.