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Chapter 5: Problem 8

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 \(f(t)\). $$ f(t)=(t-2)^{2} $$

Short Answer

Expert verified
Answer: The Laplace transform of \(f(t)=(t-2)^2\) is \(F(s) = \frac{4}{s} + \frac{4}{s^2} + \frac{2}{s^4}\) with a domain of \(s > 0\). The graph of \(f(t)\) is a parabola starting at the vertex (2,0) and expanding to the right.

Step by step solution

01

Recall the definition of the Laplace transform

The Laplace transform of a function \(f(t)\) is defined as: $$ F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) dt $$ We are given that \(f(t)=(t-2)^2\). So, we will find the Laplace transform using this definition.
02

Evaluate the integral

Using the given function \(f(t) = (t-2)^2\), we will now find the Laplace transform, \(F(s)\) by evaluating the integral: $$ F(s) = \int_0^\infty e^{-st} (t-2)^2 dt $$ To solve this integral, we can use integration by parts. Let \(u = (t-2)^2\) and \(dv = e^{-st} dt\). Then, \(du = 2(t-2) dt\) and \(v = \frac{-1}{s} e^{-st}\). Using integration by parts twice, we get: $$ F(s) = \frac{-1}{s} e^{-st} (t-2)^2 \Bigr|_0^\infty + \frac{2}{s} \int_0^\infty e^{-st} (t-2) dt $$ Performing integration by parts again on the remaining integral, let \(u = (t-2)\) and \(dv = e^{-st} dt\). Then, \(du = dt\) and \(v = \frac{-1}{s} e^{-st}\). $$ F(s) = \frac{-1}{s} e^{-st} (t-2)^2 \Bigr|_0^\infty + \frac{2}{s} \left( \frac{-1}{s} e^{-st} (t-2) \Bigr|_0^\infty - \frac{1}{s} \int_0^\infty e^{-st} dt \right) $$ Now evaluate the limits, noting that terms with \(e^{-st}\) as \(t \to \infty\) will go to zero if \(s > 0\). Thus, we obtain: $$ F(s) = \frac{4}{s} + \frac{4}{s^2} - \frac{2}{s^3} \int_0^\infty e^{-st} dt $$ The remaining integral can be easily evaluated, resulting in: $$ F(s) = \frac{4}{s} + \frac{4}{s^2} - \frac{2}{s^3} \left( \frac{-1}{s} \right) = \frac{4}{s} + \frac{4}{s^2} + \frac{2}{s^4} $$
03

Find the domain of \(F(s)\)

The domain of \(F(s)\) will be the range of \(s\) for which the Laplace transform exists. We've already seen that for \(s > 0\), we can evaluate the integral. Therefore, the domain of \(F(s)\) is \(s > 0\).
04

Graph \(f(t)\)

In order to sketch the graph of \(f(t) = (t-2)^2\), note that this is a quadratic function with a vertex at \((2, 0)\) and opens upwards. The graph will look like a parabola, but it is only defined for \(t \geq 0\). Thus, it is a parabola starting at the vertex \((2,0)\) and expanding to the right. In conclusion, the Laplace transform of \(f(t)= (t-2)^2\) is \(F(s) = \frac{4}{s} + \frac{4}{s^2} + \frac{2}{s^4}\) with a domain of \(s > 0\), and the graph of \(f(t)\) is a parabola starting at the vertex \((2,0)\) and expanding to the right.

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Key Concepts

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

Integration by Parts
Integration by parts is a technique used to simplify the integration of products of functions. It is particularly useful when the integral involves polynomials and exponential functions. The formula for integration by parts is given by:\[ \int u \, dv = uv - \int v \, du \]Here, you choose parts of your integral to be \(u\) and \(dv\), and then find their derivatives or antiderivatives \(du\) and \(v\), respectively. For the function \(f(t) = (t-2)^2\), we performed integration by parts twice. By setting \(u = (t-2)^2\) and \(dv = e^{-st} \, dt\), we found:
  • \(du = 2(t-2) \, dt\)
  • \(v = \frac{-1}{s} e^{-st}\)
The repeated application helped us simplify the integral further, allowing us to handle polynomials easily together with the exponential decay term \(e^{-st}\). Remember, the goal of using integration by parts is to break down a complex integral into simpler parts that are easy to evaluate.
Quadratic Function
A quadratic function is a type of polynomial function that is represented as \(f(t) = at^2 + bt + c\). In this exercise, the function you are dealing with is \(f(t) = (t-2)^2\).This is a quadratic function opening upwards, as it can be expanded to \(f(t) = t^2 - 4t + 4\). Here are key characteristics:
  • Vertex: The lowest point of the parabola occurs at the vertex, located at \((2, 0)\). This vertex-form representation highlights where the graph changes direction.
  • Symmetry: Quadratic functions have an axis of symmetry. For this function, the axis is the vertical line at \(t = 2\).
  • Zeroes: Setting \((t-2)^2 = 0\), we solve for the root \(t = 2\), confirming the vertex is also a point where the function equals zero.
Understanding the structure of quadratic functions assists in sketching their graphs and analyzing their properties.
Domain of Transform
When determining the domain of a Laplace transform, you are essentially identifying the values of \(s\) for which the transform is defined. For the function \(f(t) = (t-2)^2\), the integral defining the Laplace transform is:\[F(s) = \int_0^\infty e^{-st} (t-2)^2 \, dt\]The principle is that the exponential term \(e^{-st}\) must dominate over the polynomial term \((t-2)^2\) to ensure convergence. The domain of \(F(s)\) is crucial, as it reveals where this balance holds.
  • For the integral to converge, \(s\) must be > 0.
  • This ensures \(e^{-st}\) diminishes sufficiently rapidly as \(t\) approaches infinity to balance the quadratic growth of \((t-2)^2\).
Therefore, the domain of \(F(s)\) is \(s > 0\), confirming for these values, the Laplace transform exists and converges.
Graph Sketching
Graph sketching involves visually representing the function \(f(t) = (t-2)^2\). This function is a classic example of a quadratic, and its graph will be a parabola.
  • Shape: Being a quadratic, the graph is a U-shaped parabola, facing upwards.
  • Starting Point: The graph begins at the vertex \((2, 0)\), positioned on the horizontal axis since \(f(2) = 0\).
  • Direction: It extends to the right because the function is only defined for \(t \geq 0\).
  • Axis of Symmetry: There's a vertical line of symmetry through \(t = 2\), as the graph is mirror-symmetrical about this line.
Graph sketching, in this context, helps visualize the behavior of the function in the domain \(0 \leq t < \infty\), ensuring a clear understanding of the function's properties.

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Most popular questions from this chapter

Assume a body of mass \(m\) moves along a horizontal surface in a straight line with velocity \(v(t)\). The body is subject to a frictional force proportional to velocity and is propelled forward with a periodic propulsive force \(f(t)\). Applying Newton's second law, we obtain the following initial value problem: $$m v^{\prime}+k v=f(t), \quad t \geq 0, \quad v(0)=v_{0} .$$ Assume that \(m=1 \mathrm{~kg}, k=1 \mathrm{~kg} / \mathrm{s}\), and \(v_{0}=1 \mathrm{~m} / \mathrm{s}\). (a) Use Laplace transform methods to determine \(v(t)\) for the propulsive force \(f(t)\), where \(f(t)\) is given in newtons. (b) Plot \(v(t)\) for \(0 \leq t \leq 10\) [this time interval spans the first five periods of \(f(t)\) ]. In Exercise 17, explain why \(v(t)\) is constant on the interval \(0 \leq t \leq 1\). $$ f(t)=t / 2, \quad 0 \leq t<2, \quad f(t+2)=f(t) $$

Use the Laplace transform to solve the initial value problem. $$y^{\prime \prime}+4 y=\sin 2 t, \quad y(0)=1, \quad y^{\prime}(0)=0$$

Solve the given initial value problem, in which inputs of large amplitude and short duration have been idealized as delta functions. Graph the solution that you obtain on the indicated interval. \(\frac{d}{d t}\left[\begin{array}{l}y_{1} \\\ y_{2}\end{array}\right]=\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{l}y_{1} \\\ y_{2}\end{array}\right]+\left[\begin{array}{l}0 \\\ 1\end{array}\right]-\delta(t-1)\left[\begin{array}{l}1 \\\ 0\end{array}\right], \quad\left[\begin{array}{l}y_{1}(0) \\\ y_{2}(0)\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right], \quad 0 \leq t \leq 2\)

Find the inverse Laplace transform. $$F(s)=\frac{4 s^{2}+s+1}{s^{3}+s}$$

Give the form of the partial fraction expansion for the given rational function \(F(s)\). You need not evaluate the constants in the expansion. However, if the denominator of \(F(s)\) contains irreducible quadratic factors of the form \(s^{2}+2 \alpha s+\beta^{2}, \beta^{2}>\alpha^{2}\), complete the square and rewrite this factor in the form \((s+\alpha)^{2}+\omega^{2}\). $$F(s)=\frac{s^{4}+5 s^{2}+2 s-9}{\left(s^{2}+8 s+17\right)^{2}(s-2)^{2}}$$
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