Math Problem Statement
Solution
To address this problem, let’s break it down step-by-step.
Part (a): Representing using the Heaviside step function
The given function is defined as:
\begin{cases} t^2, & 0 \leq t < 2 \\ 6, & t \geq 2 \end{cases}$$ We want to express $$f(t)$$ using the Heaviside step function $$h(t - c)$$, where $$h(t - c)$$ is defined as 0 for $$t < c$$ and 1 for $$t \geq c$$. 1. For $$0 \leq t < 2$$, $$f(t) = t^2$$. 2. For $$t \geq 2$$, $$f(t)$$ switches to the constant value 6. We can rewrite $$f(t)$$ in terms of the Heaviside function as follows: $$f(t) = t^2 + (6 - t^2) h(t - 2)$$ Here’s the reasoning: - For $$t < 2$$: $$h(t - 2) = 0$$, so $$f(t) = t^2$$. - For $$t \geq 2$$: $$h(t - 2) = 1$$, so $$f(t) = t^2 + (6 - t^2) = 6$$. ### Part (b): Finding the Laplace Transform $$F(s) = \mathcal{L}\{f(t)\}$$ The Laplace transform of $$f(t) = t^2 + (6 - t^2) h(t - 2)$$ involves two parts: 1. The Laplace transform of $$t^2$$, which is straightforward. 2. The Laplace transform of $$(6 - t^2) h(t - 2)$$, which uses the shifting property of Laplace transforms for functions multiplied by a Heaviside function. 1. **Laplace Transform of $$t^2$$:** $$\mathcal{L}\{t^2\} = \frac{2}{s^3}$$ 2. **Laplace Transform of $$(6 - t^2) h(t - 2)$$:** - Rewrite $$(6 - t^2) h(t - 2)$$ as a shifted function. Define a shifted variable $$u = t - 2$$, so $$t = u + 2$$. - $$f(t) = 6 - (u + 2)^2 = 6 - (u^2 + 4u + 4) = -u^2 - 4u + 2$$. Using the shifting property $$\mathcal{L}\{g(t - a) h(t - a)\} = e^{-as} G(s)$$, we find the Laplace transform for thisAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Heaviside Step Function
Laplace Transform
Formulas
Heaviside step function h(t - c)
Laplace transform of a piecewise function
Shifting property of the Laplace transform
Theorems
Shifting Theorem for Laplace Transform
Suitable Grade Level
Undergraduate (Calculus, Differential Equations)
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