Laplace Even Conservative 768042

1. **Problem Statement:** (a) Find the inverse Laplace transform of $$L(f) = \frac{2(e^{-s} - e^{-3s})}{s^2 - 4}$$. (b) (i) Obtain the even periodic extension of the given function $$f(t)$$ defined on $$[0,2]$$. (ii) Sketch the periodic extension for $$0 \leq t \leq 5$$. (c) Given $$\mathbf{F}(x,y,z) = z\mathbf{i} + y\mathbf{j} + x\mathbf{k}$$, (i) Show that $$\mathbf{F}$$ is conservative. (ii) Find the work done by $$\mathbf{F}$$ along the curve $$\mathbf{r}(t) = 3t\mathbf{i} - 5t\mathbf{j} + t^2\mathbf{k}, -1 \leq t \leq 1$$. --- 2. **Part (a) Inverse Laplace Transform:** - Recall the Laplace transform shift theorem: $$\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)$$ where $$u(t)$$ is the unit step function. - The denominator $$s^2 - 4 = (s-2)(s+2)$$ corresponds to the Laplace transform of $$\sinh(2t)$$ or $$\cosh(2t)$$. - We rewrite: $$L(f) = 2 \left( \frac{e^{-s}}{s^2 - 4} - \frac{e^{-3s}}{s^2 - 4} \right)$$ - Using linearity and shift theorem: $$f(t) = 2\left[ u(t-1)g(t-1) - u(t-3)g(t-3) \right]$$ where $$g(t) = \mathcal{L}^{-1}\left\{ \frac{1}{s^2 - 4} \right\}$$. - Find $$g(t)$$: $$\mathcal{L}\{\sinh(2t)\} = \frac{2}{s^2 - 4} \implies \mathcal{L}^{-1}\left\{ \frac{1}{s^2 - 4} \right\} = \frac{\sinh(2t)}{2}$$. - Substitute back: $$f(t) = 2\left[ u(t-1) \frac{\sinh(2(t-1))}{2} - u(t-3) \frac{\sinh(2(t-3))}{2} \right] = u(t-1)\sinh(2(t-1)) - u(t-3)\sinh(2(t-3))$$. --- 3. **Part (b)(i) Even Periodic Extension:** - The original function $$f(t)$$ is defined on $$[0,2]$$. - The even extension $$f_e(t)$$ is defined for $$[-2,2]$$ by: $$f_e(t) = \begin{cases} f(t), & 0 \leq t \leq 2 \\ f(-t), & -2 \leq t < 0 \end{cases}$$ - The periodic extension repeats $$f_e(t)$$ with period $$T=4$$. - This means for all $$t$$: $$f_{per}(t) = f_e(t - 4k) \quad \text{where } k \in \mathbb{Z} \text{ such that } t-4k \in [-2,2]$$. --- 4. **Part (b)(ii) Sketch for $$0 \leq t \leq 5$$:** - Sketch the original $$f(t)$$ on $$[0,2]$$. - Reflect it about $$t=0$$ to get the even extension on $$[-2,0]$$. - Repeat this pattern every 4 units to cover $$[0,5]$$. - The graph will show a wave-like pattern symmetric about $$t=0, 4$$. --- 5. **Part (c)(i) Show $$\mathbf{F}$$ is conservative:** - A vector field $$\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$$ is conservative if its curl is zero: $$\nabla \times \mathbf{F} = \mathbf{0}$$. - Compute curl: $$\nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right)\mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\mathbf{k}$$ - Given $$P = z, Q = y, R = x$$: $$\frac{\partial R}{\partial y} = \frac{\partial x}{\partial y} = 0$$ $$\frac{\partial Q}{\partial z} = \frac{\partial y}{\partial z} = 0$$ $$\frac{\partial R}{\partial x} = \frac{\partial x}{\partial x} = 1$$ $$\frac{\partial P}{\partial z} = \frac{\partial z}{\partial z} = 1$$ $$\frac{\partial Q}{\partial x} = \frac{\partial y}{\partial x} = 0$$ $$\frac{\partial P}{\partial y} = \frac{\partial z}{\partial y} = 0$$ - Substitute: $$\nabla \times \mathbf{F} = (0 - 0)\mathbf{i} - (1 - 1)\mathbf{j} + (0 - 0)\mathbf{k} = \mathbf{0}$$ - Hence, $$\mathbf{F}$$ is conservative. --- 6. **Part (c)(ii) Work done by $$\mathbf{F}$$ along $$\mathbf{r}(t)$$:** - Work done $$W = \int_C \mathbf{F} \cdot d\mathbf{r}$$. - Since $$\mathbf{F}$$ is conservative, work depends only on endpoints. - Find potential function $$\phi$$ such that $$\nabla \phi = \mathbf{F}$$. - From $$\mathbf{F} = (z, y, x)$$: $$\frac{\partial \phi}{\partial x} = z$$ $$\frac{\partial \phi}{\partial y} = y$$ $$\frac{\partial \phi}{\partial z} = x$$ - Integrate $$\frac{\partial \phi}{\partial x} = z$$ w.r.t. $$x$$: $$\phi = xz + g(y,z)$$ - Differentiate w.r.t. $$y$$: $$\frac{\partial \phi}{\partial y} = \frac{\partial g}{\partial y} = y$$ - Integrate w.r.t. $$y$$: $$g(y,z) = \frac{y^2}{2} + h(z)$$ - Now $$\phi = xz + \frac{y^2}{2} + h(z)$$. - Differentiate w.r.t. $$z$$: $$\frac{\partial \phi}{\partial z} = x + h'(z) = x$$ - So $$h'(z) = 0 \implies h(z) = C$$ constant. - Potential function: $$\phi(x,y,z) = xz + \frac{y^2}{2} + C$$ - Evaluate work: $$W = \phi(\mathbf{r}(1)) - \phi(\mathbf{r}(-1))$$ - Compute $$\mathbf{r}(1) = (3(1), -5(1), 1^2) = (3, -5, 1)$$ - Compute $$\mathbf{r}(-1) = (3(-1), -5(-1), (-1)^2) = (-3, 5, 1)$$ - Calculate: $$\phi(3,-5,1) = 3 \times 1 + \frac{(-5)^2}{2} = 3 + \frac{25}{2} = 3 + 12.5 = 15.5$$ $$\phi(-3,5,1) = (-3) \times 1 + \frac{5^2}{2} = -3 + \frac{25}{2} = -3 + 12.5 = 9.5$$ - Work done: $$W = 15.5 - 9.5 = 6$$ --- **Final answers:** (a) $$f(t) = u(t-1)\sinh(2(t-1)) - u(t-3)\sinh(2(t-3))$$ (b)(i) Even periodic extension with period $$4$$ defined by reflecting $$f(t)$$ about $$t=0$$ and repeating. (b)(ii) Sketch shows repeated even extension over $$[0,5]$$. (c)(i) $$\mathbf{F}$$ is conservative since $$\nabla \times \mathbf{F} = \mathbf{0}$$. (c)(ii) Work done $$W = 6$$.