/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 $$\mathscr{L}\left\\{2 t^{4}\rig... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 19

$$\mathscr{L}\left\\{2 t^{4}\right\\}=2 \frac{4 !}{s^{5}}$$

Short Answer

Expert verified
\(\mathscr{L}\{2t^4\} = \frac{48}{s^5}\)

Step by step solution

01

Identify the Laplace Transform Formula

The Laplace Transform of a function \(t^n\) is given by \(\mathscr{L}\{t^n\} = \frac{n!}{s^{n+1}}\). For any constant \(c\), the transform is \(\mathscr{L}\{c \cdot t^n\} = c \cdot \frac{n!}{s^{n+1}}\).
02

Recognize Given Function and Constant

The given function is \(2t^4\), which means \(c = 2\) and \(n = 4\). We need the transform of \(2 \cdot t^4\).
03

Apply the Formula

Substitute \(c = 2\) and \(n = 4\) into the Laplace Transform formula to get: \[\mathscr{L}\{2t^4\} = 2 \cdot \frac{4!}{s^{4+1}} = 2 \cdot \frac{4!}{s^5}\].
04

Simplification of the Result

Calculate \(4!\), which is 4 factorial: \(4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24\). Therefore, \[\mathscr{L}\{2t^4\} = 2 \cdot \frac{24}{s^5} = \frac{48}{s^5}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Factorial Function
The factorial function, denoted as "!", is a basic but powerful concept in mathematics. It's a way to find the product of all positive integers up to a specified number. For example, the factorial of 4, written as 4!, is calculated as follows:
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \] This tool becomes handy especially in calculus and algebra when solving problems that involve repeated multiplication.
  • Factorials are used in permutations and combinations to calculate possible arrangements.
  • They are also crucial in various calculus applications, such as in the coefficients of the Taylor series.
  • In engineering, understanding how to manipulate factorials helps in determining system behaviors formally.
Factorials grow very rapidly with larger numbers, so they are useful when managing exponential functions, such as those involved in the Laplace transform.
Transform of Monomials
In the context of Laplace transforms, understanding how to handle monomials is vital. A monomial is simply a product of a constant and a power of a single variable, for instance, \( t^n \). For this, the Laplace transform formula is a key identifier:
\[ \mathscr{L}\{t^n\} = \frac{n!}{s^{n+1}} \] Here,
  • \(t^n\) represents the basic monomial form.
  • "\(s\)" is a complex frequency parameter from the transform's domain.
When you have a constant affecting the monomial, the Laplace transform maintains linearity. So, for a constant \(c\) multiplied by \(t^n\), it adapts like this:

  • \( \mathscr{L}\{c \cdot t^n \} = c \cdot \frac{n!}{s^{n+1}} \)
Practically, this allows us to tackle ordinary differential equations in engineering where such terms naturally occur.
Calculus in Engineering
Calculus is a cornerstone of engineering, providing the tools to model and solve real-world phenomena. Within engineering contexts, Laplace transforms serve as a method to switch from the time domain into the complex frequency domain. This shift is essential as it simplifies the process of analyzing systems, particularly those defined by differential equations.
  • System Behavior: Engineers use calculus to understand the behavior and stability of systems, such as circuits or mechanical dynamics.
  • Signal Processing: Calculus underpins signal processing techniques, such as filtering, which are crucial in telecommunications.
  • Control Systems: In designing control systems, engineers rely on calculus concepts to ensure compatibility and efficiency between different components.
By leveraging the Laplace transform, engineers can convert complex problems into easier-to-handle algebraic equations. This not only speeds up analysis but also delivers more precise and manageable solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Assuming that (c) of Theorem 4.1 is applicable with a complex exponent, we have $$\mathscr{L}\left\\{e^{(a+i b) t}\right\\}=\frac{1}{s-(a+i b)}=\frac{1}{(s-a)-i b} \frac{(s-a)+i b}{(s-a)+i b}=\frac{s-a+i b}{(s-a)^{2}+b^{2}}.$$ By Euler's formula, \(e^{i \theta}=\cos \theta+i \sin \theta,\) so $$\begin{aligned} \mathscr{L}\left\\{e^{(a+i b) t}\right\\} &=\mathscr{L}\left\\{e^{a t} e^{i b t}\right\\}=\mathscr{L}\left\\{e^{a t}(\cos b t+i \sin b t)\right\\} \\ &=\mathscr{L}\left\\{e^{a t} \cos b t\right\\}+i \mathscr{L}\left\\{e^{a t} \sin b t\right\\} \\ &=\frac{s-a}{(s-a)^{2}+b^{2}}+i \frac{b}{(s-a)^{2}+b^{2}}. \end{aligned}$$ Equating real and imaginary parts we get $$\mathscr{L}\left\\{e^{a t} \cos b t\right\\}=\frac{s-a}{(s-a)^{2}+b^{2}} \quad \text { and } \quad \mathscr{L}\left\\{e^{a t} \sin b t\right\\}=\frac{b}{(s-a)^{2}+b^{2}}.$$

Taking the Laplace transform of the system gives \\[ \begin{aligned} (s-4) \mathscr{L}\\{x\\}+s^{3} \mathscr{L}\\{y\\} &=\frac{6}{s^{2}+1} \\ (s+2) \mathscr{L}\\{x\\}-2 s^{3} \mathscr{L}\\{y\\} &=0 \end{aligned} \\] so that \\[ \mathscr{L}\\{x\\}=\frac{4}{(s-2)\left(s^{2}+1\right)}=\frac{4}{5} \frac{1}{s-2}-\frac{4}{5} \frac{s}{s^{2}+1}-\frac{8}{5} \frac{1}{s^{2}+1} \\] and \\[ \mathscr{L}\\{y\\}=\frac{2 s+4}{s^{3}(s-2)\left(s^{2}+1\right)}=\frac{1}{s}-\frac{2}{s^{2}}-2 \frac{2}{s^{3}}+\frac{1}{5} \frac{1}{s-2}-\frac{6}{5} \frac{s}{s^{2}+1}+\frac{8}{5} \frac{1}{s^{2}+1} \\] Then \\[ x=\frac{4}{5} e^{2 t}-\frac{4}{5} \cos t-\frac{8}{5} \sin t \\] and \\[ y=1-2 t-2 t^{2}+\frac{1}{5} e^{2 t}-\frac{6}{5} \cos t+\frac{8}{5} \sin t \\]

$$\mathscr{L}^{-1}\left\\{\frac{1}{s^{2}+3 s}\right\\}=\mathscr{L}^{-1}\left\\{\frac{1}{3} \cdot \frac{1}{s}-\frac{1}{3} \cdot \frac{1}{s+3}\right\\}=\frac{1}{3}-\frac{1}{3} e^{-3 t}$$

(a) By Kirchhoff's first law we have \(i_{1}=i_{2}+i_{3} .\) By Kirchhoff's second law, on each loop we have \(E(t)=R i_{1}+L_{1} i_{2}^{\prime}\) and \(E(t)=R i_{1}+L_{2} i_{3}^{\prime}\) or \(L_{1} i_{2}^{\prime}+R i_{2}+R i_{3}=E(t)\) and \(L_{2} i_{3}^{\prime}+R i_{2}+R i_{3}=E(t)\) (b) Taking the Laplace transform of the system \\[ \begin{aligned} 0.01 i_{2}^{\prime}+5 i_{2}+5 i_{3} &=100 \\ 0.0125 i_{3}^{\prime}+5 i_{2}+5 i_{3} &=100 \end{aligned} \\] gives \\[ (s+500) \mathscr{L}\left\\{i_{2}\right\\}+500 \mathscr{L}\left\\{i_{3}\right\\}=\frac{10,000}{s} \\] \\[ 400 \mathscr{L}\left\\{i_{2}\right\\}+(s+400) \mathscr{L}\left\\{i_{3}\right\\}=\frac{8,000}{s} \\] so that \\[ \mathscr{L}\left\\{i_{3}\right\\}=\frac{8,000}{s^{2}+900 s}=\frac{80}{9} \frac{1}{s}-\frac{80}{9} \frac{1}{s+900} \\] Then \\[ i_{3}=\frac{80}{9}-\frac{80}{9} e^{-900 t} \quad \text { and } \quad i_{2}=20-0.0025 i_{3}^{\prime}-i_{3}=\frac{100}{9}-\frac{100}{9} e^{-900 t} \\] (c) \(i_{1}=i_{2}+i_{3}=20-20 e^{-900 t}\)

$$\mathscr{L}\left\\{e^{2 t}-2+e^{-2 t}\right\\}=\frac{1}{s-2}-\frac{2}{s}+\frac{1}{s+2}$$
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Waves Physics

Read Explanation

Rotational Dynamics

Read Explanation

Physics of Motion

Read Explanation

Quantum Physics

Read Explanation

Electricity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.