Question

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

The Laplace transforms are: a) \(\frac{s \omega}{s^2 + \omega^2}\), b) \(-\frac{\omega^2}{s^2 + \omega^2}\), c) 2. The steps confirm these results.

Step-by-step solution

Apply the Laplace Transform Property for Derivatives

The Laplace Transform of the derivative of a function, \[ \mathscr{L}\{ \frac{d}{dt}f(t) \} = s \cdot \mathscr{L}\{f(t)\} - f(0)\]is used in each part. For sinusoidal and polynomial functions, calculate the initial values.

Part (a) - Find \(\mathscr{L}\{\frac{d}{dt} \sin \omega t\}\)

We know \(\mathscr{L}\{\sin \omega t\} = \frac{\omega}{s^2 + \omega^2}\). Use the Laplace Transform property:\[ \mathscr{L}\{\frac{d}{dt} \sin \omega t\} = s \cdot \frac{\omega}{s^2 + \omega^2} - \sin 0 = \frac{s \omega}{s^2 + \omega^2}\]since \(\sin(0) = 0\).

Part (b) - Find \(\mathscr{L}\{\frac{d}{dt} \cos \omega t\}\)

We have \(\mathscr{L}\{\cos \omega t\} = \frac{s}{s^2 + \omega^2}\). Thus,\[ \mathscr{L}\{\frac{d}{dt} \cos \omega t\} = s \cdot \frac{s}{s^2 + \omega^2} - \cos 0 = \frac{s^2}{s^2 + \omega^2} - 1\]since \(\cos(0) = 1\).

Simplify the Result of Part (b)

The expression from the previous step can be simplified:\[ \frac{s^2}{s^2 + \omega^2} - 1 = \frac{s^2 - (s^2 + \omega^2)}{s^2 + \omega^2} = -\frac{\omega^2}{s^2 + \omega^2}\]

Part (c) - Find \(\mathscr{L}\{\frac{d^{3}}{d t^{3}} t^{2}\}\)

First find \(\mathscr{L}\{t^2\} = \frac{2}{s^3}\). Use the derivative property three times. We find:\[ \mathscr{L}\{\frac{d}{dt} t^2\} = s \cdot \frac{2}{s^3} - t^2(0) = \frac{2}{s^2} \]\[ \mathscr{L}\{\frac{d^2}{dt^2} t^2\} = s \cdot \frac{2}{s^2} - 0 = \frac{2}{s} \]\[ \mathscr{L}\{\frac{d^3}{dt^3} t^2\} = s \cdot \frac{2}{s} - 0 = 2 \]

Verification of Results for Part (a)

Differentiate \(\sin \omega t\) to get \(\omega \cos \omega t\), then find its Laplace Transform:\[ \mathscr{L}\{\omega \cos \omega t\} = \omega \cdot \frac{s}{s^2 + \omega^2} = \frac{s \omega}{s^2 + \omega^2}\]This confirms the result of part (a).

Verification of Results for Part (b)

Differentiate \(\cos \omega t\) to get \(-\omega \sin \omega t\), then find its Laplace Transform:\[ \mathscr{L}\{-\omega \sin \omega t\} = -\omega \cdot \frac{\omega}{s^2 + \omega^2} = -\frac{\omega^2}{s^2 + \omega^2}\]This confirms the result of part (b).

Verification of Results for Part (c)

Differentiate \(t^2\) three times: first is \(2t\), second is \(2\), third is \(0\). The Laplace Transform of \(0\) is \(0\), confirming the result of part (c).

Key concepts

Derivative of Functions

Understanding the derivative of a function is pivotal in calculus and mathematical analysis. The derivative measures how a function changes as its input changes. It's the rate at which something happens and is represented as \(\frac{d}{dt}f(t)\) for a function \(f(t)\). The derivative gives us insight into how the function behaves over time or space.

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

Sinusoidal functions like sine and cosine are fundamental in describing oscillatory phenomena such as waves and vibrations. These functions are pivotal in many fields like engineering, physics, and even signal processing.

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:

• \(\mathscr{L}\{\sin \omega t\} = \frac{\omega}{s^2 + \omega^2}\)
• \(\mathscr{L}\{\cos \omega t\} = \frac{s}{s^2 + \omega^2}\)

These expressions convert time-domain oscillations into the frequency domain, where analysis and manipulation can be more accessible.

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

Polynomial functions, such as \(t^2\), are another critical element in mathematical modeling. These functions appear in scenarios ranging from simple parabolic motion in physics to more complex financial models.

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

Verification of differentiation results is vital to ensure accuracy and validity in mathematical solutions. This step involves confirming the correctness of results obtained by comparing them against expected outcomes.

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.

The verification ensures that solutions for derivatives through Laplace Transforms remain reliable and robust.