Math Problem Statement
laplace :f\left(t\right)=\left{t^2:0\le :t<:2,:1-t-3t^2:t\ge :2\right}
Solution
You are asking to compute the Laplace transform of a piecewise function. Let's break the function down and find its Laplace transform step by step.
Given:
\begin{cases} t^2 & \text{for } 0 \leq t < 2 \\ 1 - t - 3t^2 & \text{for } t \geq 2 \end{cases}$$ To compute the Laplace transform of this piecewise function, we can use the **second shifting theorem**, which handles functions defined in intervals (like this one). Here's the general strategy: ### 1. Break $$f(t)$$ into two separate pieces and find the Laplace transform of each part. - For $$0 \leq t < 2$$, the function is $$f_1(t) = t^2$$. - For $$t \geq 2$$, the function is $$f_2(t) = 1 - t - 3t^2$$. ### 2. Laplace Transform of $$f_1(t) = t^2$$ (for $$0 \leq t < 2$$): The Laplace transform of $$t^n$$ is $$\frac{n!}{s^{n+1}}$$, so the Laplace transform of $$t^2$$ is: $$\mathcal{L}(t^2) = \frac{2!}{s^3} = \frac{2}{s^3}$$ ### 3. Laplace Transform of $$f_2(t) = 1 - t - 3t^2$$ (for $$t \geq 2$$): We will apply the **second shifting theorem** to deal with the interval $$t \geq 2$$. The second shifting theorem says that if a function $$g(t)$$ is shifted by $$t_0$$, then the Laplace transform is given by: $$\mathcal{L}(g(t - t_0)) = e^{-t_0 s} \mathcal{L}(g(t))$$ So, we must shift $$f_2(t)$$ by 2 units. We can write: $$f_2(t) = 1 - t - 3t^2 = g(t - 2) \quad \text{where} \quad g(t) = 1 - t - 3t^2$$ Now, find the Laplace transform of $$g(t) = 1 - t - 3t^2$$: - The Laplace transform of 1 is $$\frac{1}{s}$$ - The Laplace transform of $$t$$ is $$\frac{1}{s^2}$$ - The Laplace transform of $$t^2$$ is $$\frac{2}{s^3}$$ So: $$\mathcal{L}(g(t)) = \frac{1}{s} - \frac{1}{s^2} - \frac{6}{s^3}$$ ### 4. Applying the second shifting theorem: Now we apply the second shifting theorem for $$t_0 = 2$$: $$\mathcal{L}(f_2(t)) = e^{-2s} \left( \frac{1}{s} - \frac{1}{s^2} - \frac{6}{s^3} \right)$$ ### 5. Combine the results: The total Laplace transform of the piecewise function is the sum of the transforms for $$0 \leq t < 2$$ and $$t \geq 2$$: $$\mathcal{L}(f(t)) = \mathcal{L}(f_1(t)) + \mathcal{L}(f_2(t))$$ $$\mathcal{L}(f(t)) = \frac{2}{s^3} + e^{-2s} \left( \frac{1}{s} - \frac{1}{s^2} - \frac{6}{s^3} \right)$$ ### Final Result: $$\mathcal{L}(f(t)) = \frac{2}{s^3} + e^{-2s} \left( \frac{1}{s} - \frac{1}{s^2} - \frac{6}{s^3} \right)$$ Would you like further details on any step of the solution? Or have any other questions? ### Related Questions: 1. How do you compute the Laplace transform of a piecewise function with a discontinuity at $$t = 0$$? 2. What is the second shifting theorem and how is it applied in Laplace transforms? 3. How do you handle a piecewise function with different expressions for $$t \geq 0$$ and $$t < 0$$? 4. Can the Laplace transform be applied to functions with piecewise continuous but not smooth behavior? 5. How do you handle exponential functions in piecewise Laplace transforms? ### Tip: When working with piecewise functions, always break down the function into intervals, and apply the second shifting theorem to handle shifts in the function properly.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Laplace Transform
Piecewise Functions
Shifting Theorem
Formulas
Laplace Transform of t^n = n!/s^(n+1)
Second Shifting Theorem: L(g(t - t0)) = e^(-t0 s) * L(g(t))
Theorems
Second Shifting Theorem
Suitable Grade Level
Grades 11-12, College Level
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