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Magazine Chapter 15: Problem 1
Use the expression for the transform of a second derivative to obtain the transform of \(\cos k t\).
Short Answer
Expert verified
The Laplace transform of \( \cos(kt) \) is \( \frac{s}{s^2 + k^2} \).
Step by step solution
01
Recall the Laplace Transform Definition
The Laplace transform of a function \( f(t) \) is given by \( \ \mathcal{L} \{ f(t) \} = \int_{0}^{\infty} e^{-st} f(t) dt \). For functions involving derivatives, there are special formulas that can be used to simplify the process.
02
Use the Transform of a Second Derivative
The Laplace transform of a second derivative \( f''(t) \) is given by \( \ \mathcal{L} \{ f''(t) \} = s^2 \mathcal{L} \{ f(t) \} - s f(0) - f'(0) \). This will be crucial for finding the Laplace transform of \( \ \cos(kt) \) since the differential equation for cosine involves its second derivative.
03
Consider the Differential Equation for Cosine
The function \( \cos(kt) \) satisfies the differential equation \( \ f''(t) = -k^2 f(t) \). Here, \( f(t) = \cos(kt) \). Using this equation will allow us to apply the formula from Step 2 to find its Laplace transform.
04
Substitute into the Transform Equation
Substitute \( \ f''(t) = -k^2 f(t) \) into the transform equation: \( \ \mathcal{L} \{ -k^2 f(t) \} = s^2 \mathcal{L} \{ f(t) \} - sf(0) - f'(0) \). Solve for \( \mathcal{L} \{ f(t) \} \) to find \( \ s^2 \mathcal{L} \{ \cos(kt) \} = -k^2 \mathcal{L} \{ \cos(kt) \} + s \cdot 1 + 0 \).
05
Solve for the Laplace Transform of Cosine
After substituting and rearranging, we have \( \ s^2 \mathcal{L} \{ \cos(kt) \} + k^2 \mathcal{L} \{ \cos(kt) \} = s \). Factoring leads to \( \ \mathcal{L} \{ \cos(kt) \} (s^2 + k^2) = s \), so \( \ \mathcal{L} \{ \cos(kt) \} = \frac{s}{s^2 + k^2} \).
06
Final Step: Verify the Initial Conditions
For initial conditions of cosine, \( f(0) = 1 \) and \( f'(0) = 0 \) complies with what we used. Thus, the Laplace transform \( \ \mathcal{L} \{ \cos(kt) \} = \frac{s}{s^2 + k^2} \) is correct and verified.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second Derivative
In calculus, the second derivative of a function gives insights into the function's concavity, or how it curves. It is the derivative of the derivative of a function, essentially measuring how the rate of change of a function itself changes.
The Laplace transform of the second derivative is a useful tool in solving differential equations, as it transforms complex functions into a more straightforward algebraic form.
The Laplace transform of the second derivative is a useful tool in solving differential equations, as it transforms complex functions into a more straightforward algebraic form.
- The rule for the Laplace transform of the second derivative of a function \( f(t) \) is \( \mathcal{L} \{ f''(t) \} = s^2 \mathcal{L} \{ f(t) \} - s f(0) - f'(0) \).
- This equation shows how the behavior of the function at the initial time (\( t = 0 \)) impacts the transform.
Differential Equation
A differential equation is any equation which involves the derivatives of a function. These equations describe the relationship between a function and its derivatives, capturing phenomena such as motion, growth, and decay.
In the case of \( \cos(kt) \):
In the case of \( \cos(kt) \):
- It satisfies the differential equation \( f''(t) = -k^2 f(t) \).
- This equation implies that the second derivative of the cosine function is proportional to the function itself, with a negative sign, reflecting its oscillatory nature.
Cosine Function
The cosine function \( \cos(kt) \) is a fundamental trigonometric function that describes periodic oscillations, like those of waves or circular motion.
Some interesting properties include:
This result is essential because it simplifies analyzing systems that exhibit periodic behavior by converting the time domain function into the s-domain. Consequently, it allows for easier manipulation and solution of differential equations.
Some interesting properties include:
- It is continuous and differentiable for all real numbers.
- The period of \( \cos(kt) \) is \( \frac{2\pi}{k} \), meaning it repeats its values in regular intervals.
This result is essential because it simplifies analyzing systems that exhibit periodic behavior by converting the time domain function into the s-domain. Consequently, it allows for easier manipulation and solution of differential equations.
Initial Conditions
Initial conditions specify the properties of a function at the start of the analysis, usually when \( t = 0 \). They are crucial for solving differential equations as they help pin down the specific solution among many possibilities.
For the cosine function:
Not only do correct initial conditions justify the solution process, but they confirm that the transformed function \( \mathcal{L} \{ \cos(kt) \} = \frac{s}{s^2 + k^2} \) satisfies the setup of the problem accurately.
For the cosine function:
- At \( t = 0 \), \( f(0) = \cos(0) = 1 \).
- The derivative at \( t = 0 \) is \( f'(0) = -k\sin(0) = 0 \).
Not only do correct initial conditions justify the solution process, but they confirm that the transformed function \( \mathcal{L} \{ \cos(kt) \} = \frac{s}{s^2 + k^2} \) satisfies the setup of the problem accurately.
