Chapter 12: Problem 18
a) Find \(\mathscr{L}\left\\{\frac{d}{d t} \sin \omega t\right\\}\) b) Find \(\mathscr{L}\left\\{\frac{d}{d t} \cos \omega t\right\\}\) c) Find \(\mathscr{L}\left\\{\frac{d^{3}}{d t^{3}} t^{2}\right\\}\) d) Check the results of parts \((a),(b),\) and \((c)\) by first differentiating and then transforming.
Short Answer
Step by step solution
Apply the Laplace Transform Property for Derivatives
Part (a) - Find \(\mathscr{L}\{\frac{d}{dt} \sin \omega t\}\)
Part (b) - Find \(\mathscr{L}\{\frac{d}{dt} \cos \omega t\}\)
Simplify the Result of Part (b)
Part (c) - Find \(\mathscr{L}\{\frac{d^{3}}{d t^{3}} t^{2}\}\)
Verification of Results for Part (a)
Verification of Results for Part (b)
Verification of Results for Part (c)
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 Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative of Functions
The Laplace Transform comes into play by transforming these derivatives into algebraic forms. The transformation is represented as:\[ \mathscr{L}\left\{ \frac{d}{dt}f(t) \right\} = s \cdot \mathscr{L}\{f(t)\} - f(0) \]
This formula shows that the Laplace Transform of a derivative depends on the transform of the original function \(f(t)\), the variable \(s\) which is a complex number, and the initial value \(f(0)\). The Laplace Transform simplifies the process of dealing with derivatives, especially in systems involving differential equations.
Sinusoidal Functions
The Laplace Transform of a sinusoidal function, such as \(\sin \omega t\) or \(\cos \omega t\), is especially useful in controlling systems and electrical circuits. The transforms are given by:
- \(\mathscr{L}\{\sin \omega t\} = \frac{\omega}{s^2 + \omega^2}\)
- \(\mathscr{L}\{\cos \omega t\} = \frac{s}{s^2 + \omega^2}\)
By applying the Laplace Transform to the derivative of these sinusoidal functions, we convert them to algebraic forms facilitating easy manipulation and solution.
Polynomial Functions
Finding the Laplace Transform of polynomial functions involves the use of formulaic transformations. For example, for the function \(t^2\), the Laplace Transform is \(\mathscr{L}\{t^2\} = \frac{2}{s^3}\).
When dealing with higher-order derivatives, like \(\frac{d^3}{dt^3} t^2\), the Laplace Transform can be applied sequentially according to the derivative property. Each step further simplifies the function until a constant or zero is reached. This process highlights the power of Laplace Transforms in reducing the complexity involved in solving differential equations with polynomial terms.
Differentiation Verification
After applying the Laplace Transform to derivatives and obtaining results, differentiation verification involves transforming back to ensure consistency. This means technically reversing the process by manually differentiating the original function and then applying the Laplace Transform:
- For \(\sin \omega t\), differentiating gives \(\omega \cos \omega t\). Transforming \(\omega \cos \omega t\) confirms \(\frac{s \omega}{s^2 + \omega^2}\).
- For \(\cos \omega t\), the differentiation yields \(-\omega \sin \omega t\), leading to \(-\frac{\omega^2}{s^2 + \omega^2}\), confirming the transformed result.