Sketch the graph of the function below and determine its Laplace transform. (t- 3)2u(t- 3)
Sketch the graph of the function below and determine its Laplace transform. (t- 3)2u(t- 3)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.2: Exponential Functions
Problem 25E
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Question
![**Problem Statement:**
Sketch the graph of the function below and determine its Laplace transform.
\[
(t - 3)^2 u(t - 3)
\]
**Explanation:**
1. **Function Description:**
- The function \((t - 3)^2\) is a quadratic function shifted to the right by 3 units.
- The unit step function \(u(t - 3)\) indicates that the function is zero for \(t < 3\) and active for \(t \geq 3\).
2. **Graph Explanation:**
- Before \(t = 3\), the function is zero due to the unit step function.
- At \(t = 3\), the function begins as a quadratic, opening upwards from zero.
3. **Laplace Transform:**
- Apply the Laplace transform to the function: First, consider the Laplace transform of \( (t-3)^2 \), then use the property of the step function in Laplace transforms.
- The Laplace transform of \((t-a)^n u(t-a)\) is \(\frac{e^{-as} n!}{s^{n+1}}\).
Given the function is \((t-3)^2 u(t-3)\), the Laplace transform will be:
\[
\frac{e^{-3s} 2!}{s^3} = \frac{2e^{-3s}}{s^3}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F894299d3-9d11-45b0-ba35-3f62b5e6a781%2F905fe18c-ab52-4e02-b614-309f5902e20f%2Fegktqsr_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Sketch the graph of the function below and determine its Laplace transform.
\[
(t - 3)^2 u(t - 3)
\]
**Explanation:**
1. **Function Description:**
- The function \((t - 3)^2\) is a quadratic function shifted to the right by 3 units.
- The unit step function \(u(t - 3)\) indicates that the function is zero for \(t < 3\) and active for \(t \geq 3\).
2. **Graph Explanation:**
- Before \(t = 3\), the function is zero due to the unit step function.
- At \(t = 3\), the function begins as a quadratic, opening upwards from zero.
3. **Laplace Transform:**
- Apply the Laplace transform to the function: First, consider the Laplace transform of \( (t-3)^2 \), then use the property of the step function in Laplace transforms.
- The Laplace transform of \((t-a)^n u(t-a)\) is \(\frac{e^{-as} n!}{s^{n+1}}\).
Given the function is \((t-3)^2 u(t-3)\), the Laplace transform will be:
\[
\frac{e^{-3s} 2!}{s^3} = \frac{2e^{-3s}}{s^3}
\]
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