Applied Math III Assignment

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1. Laplace Transforms (a) Laplace Transform of $f(t) = \cos^2 t + t^2u(t-2)$ Use identity: $\cos^2 t = \frac{1 + \cos(2t)}{2}$ $\mathcal{L}\{\cos^2 t\} = \mathcal{L}\left\{\frac{1}{2} + \frac{1}{2}\cos(2t)\right\} = \frac{1}{2s} + \frac{s}{2(s^2+4)}$ For $t^2u(t-2)$, use time-shifting property: $\mathcal{L}\{g(t-a)u(t-a)\} = e^{-as}\mathcal{L}\{g(t)\}$ Here $a=2$. Let $g(t-2) = t^2$. Then $g(t) = (t+2)^2 = t^2 + 4t + 4$. $\mathcal{L}\{t^2\} = \frac{2}{s^3}$, $\mathcal{L}\{4t\} = \frac{4}{s^2}$, $\mathcal{L}\{4\} = \frac{4}{s}$ $\mathcal{L}\{t^2u(t-2)\} = e^{-2s}\mathcal{L}\{(t+2)^2\} = e^{-2s}\left(\frac{2}{s^3} + \frac{4}{s^2} + \frac{4}{s}\right)$ Total Laplace Transform: $F(s) = \frac{1}{2s} + \frac{s}{2(s^2+4)} + e^{-2s}\left(\frac{2}{s^3} + \frac{4}{s^2} + \frac{4}{s}\right)$ (b) Solve System of Differential Equations using Laplace Transform Given system: $x' - y = 2\sin t$ $y' + x = 0$ with $x(0)=1, y(0)=0$. Let $\mathcal{L}\{x(t)\} = X(s)$ and $\mathcal{L}\{y(t)\} = Y(s)$. Apply Laplace Transform: $sX(s) - x(0) - Y(s) = \frac{2}{s^2+1} \Rightarrow sX(s) - Y(s) = \frac{2}{s^2+1} + 1 = \frac{s^2+3}{s^2+1}$ (Eq. 1) $sY(s) - y(0) + X(s) = 0 \Rightarrow X(s) + sY(s) = 0$ (Eq. 2) From (Eq. 2), $X(s) = -sY(s)$. Substitute into (Eq. 1): $s(-sY(s)) - Y(s) = \frac{s^2+3}{s^2+1}$ $-(s^2+1)Y(s) = \frac{s^2+3}{s^2+1} \Rightarrow Y(s) = -\frac{s^2+3}{(s^2+1)^2}$ Use partial fractions/table for $Y(s)$: $\frac{s^2+3}{(s^2+1)^2} = \frac{A_1s+B_1}{s^2+1} + \frac{A_2s+B_2}{(s^2+1)^2}$ $s^2+3 = (A_1s+B_1)(s^2+1) + A_2s+B_2 = A_1s^3+B_1s^2+(A_1+A_2)s+(B_1+B_2)$ Comparing coefficients: $A_1=0, B_1=1, A_1+A_2=0 \Rightarrow A_2=0, B_1+B_2=3 \Rightarrow B_2=2$ So, $Y(s) = -\left(\frac{1}{s^2+1} + \frac{2}{(s^2+1)^2}\right)$ Using $\mathcal{L}^{-1}\left\{\frac{1}{s^2+a^2}\right\} = \frac{1}{a}\sin(at)$ and $\mathcal{L}^{-1}\left\{\frac{1}{(s^2+a^2)^2}\right\} = \frac{1}{2a^3}(\sin(at)-at\cos(at))$ for $a=1$: $y(t) = -\left(\sin t + 2 \cdot \frac{1}{2}(\sin t - t\cos t)\right) = -\sin t - (\sin t - t\cos t) = t\cos t - 2\sin t$ Find $x(t)$ using $x(t) = -y'(t)$: $y'(t) = \frac{d}{dt}(t\cos t - 2\sin t) = (\cos t - t\sin t) - 2\cos t = -t\sin t - \cos t$ $x(t) = -(-t\sin t - \cos t) = t\sin t + \cos t$ Solution: $x(t) = t\sin t + \cos t$, $y(t) = t\cos t - 2\sin t$ 2. Conservative Vector Fields Vector field: $\mathbf{F} = (3x^2+3y^2z \sin xz)\mathbf{i} + (ay \cos xz+bz)\mathbf{j} + (3xy^2 \sin xz + 5y)\mathbf{k}$ A vector field $\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$ is conservative if $\nabla \times \mathbf{F} = \mathbf{0}$, which means the following partial derivatives must be equal: $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ $\frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}$ $\frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y}$ Calculate partial derivatives: $P = 3x^2+3y^2z \sin xz$ $Q = ay \cos xz+bz$ $R = 3xy^2 \sin xz + 5y$ $\frac{\partial P}{\partial y} = 6yz \sin xz$ $\frac{\partial Q}{\partial x} = -ayz \sin xz$ Equating them: $6yz \sin xz = -ayz \sin xz \Rightarrow a = -6$ (if $yz \sin xz \neq 0$) $\frac{\partial P}{\partial z} = 3y^2 \sin xz + 3y^2z(x \cos xz)$ $\frac{\partial R}{\partial x} = 3y^2 \sin xz + 3xy^2(z \cos xz) = 3y^2 \sin xz + 3xy^2z \cos xz$ These are already equal, so no new information about $a$ or $b$. $\frac{\partial Q}{\partial z} = -axy \sin xz + b$ $\frac{\partial R}{\partial y} = 6xy \sin xz + 5$ Equating them: $-axy \sin xz + b = 6xy \sin xz + 5$ Substitute $a=-6$: $-(-6)xy \sin xz + b = 6xy \sin xz + 5$ $6xy \sin xz + b = 6xy \sin xz + 5 \Rightarrow b = 5$ Therefore, the vector field is conservative when $a=-6$ and $b=5$. 3. Evaluate the Flux Integral Evaluate $\iint_S \mathbf{F} \cdot \mathbf{n} \,dS$, where $\mathbf{F}(x, y, z) = (x+1, y+1, 2z)$, and $S$ is the part of the paraboloid $z = 4-x^2-y^2$ that lies above the triangle $0 \le x \le 1; 0 \le y \le 1-x$, oriented with a negative z-component for its unit normal. Use the divergence theorem for a closed surface, but $S$ is open. We can use the formula for flux through a surface $z = g(x,y)$: $\iint_S \mathbf{F} \cdot \mathbf{n} \,dS = \iint_D (-P \frac{\partial g}{\partial x} - Q \frac{\partial g}{\partial y} + R) \,dA$ for upward normal. Here $g(x,y) = 4-x^2-y^2$. $\frac{\partial g}{\partial x} = -2x$, $\frac{\partial g}{\partial y} = -2y$. $\mathbf{F} = (P, Q, R) = (x+1, y+1, 2z)$. Since $z=g(x,y)$, $R = 2(4-x^2-y^2) = 8-2x^2-2y^2$. The normal has a negative z-component, so we use the downward normal: $\iint_S \mathbf{F} \cdot \mathbf{n} \,dS = \iint_D (P \frac{\partial g}{\partial x} + Q \frac{\partial g}{\partial y} - R) \,dA$ The region $D$ is the triangle in the $xy$-plane: $0 \le x \le 1$, $0 \le y \le 1-x$. Substitute into the integral formula: $\iint_D ((x+1)(-2x) + (y+1)(-2y) - (8-2x^2-2y^2)) \,dA$ $= \iint_D (-2x^2-2x -2y^2-2y -8+2x^2+2y^2) \,dA$ $= \iint_D (-2x - 2y - 8) \,dA$ Set up the double integral over the triangular region: $\int_0^1 \int_0^{1-x} (-2x - 2y - 8) \,dy \,dx$ Inner integral: $\int_0^{1-x} (-2x - 2y - 8) \,dy = [-2xy - y^2 - 8y]_0^{1-x}$ $= -2x(1-x) - (1-x)^2 - 8(1-x)$ $= -2x+2x^2 - (1-2x+x^2) - 8+8x$ $= -2x+2x^2 - 1+2x-x^2 - 8+8x$ $= x^2 + 8x - 9$ Outer integral: $\int_0^1 (x^2 + 8x - 9) \,dx = \left[\frac{x^3}{3} + 4x^2 - 9x\right]_0^1$ $= \frac{1}{3} + 4 - 9 = \frac{1}{3} - 5 = \frac{1-15}{3} = -\frac{14}{3}$ The flux integral is $-\frac{14}{3}$. 4. Stokes' Theorem Evaluate $\oint_C z^2e^{x^2} \,dx + xy^2 \,dy + \tan^{-1} y \,dz$, where $C$ is the circle $x^2+y^2=9$, oriented counterclockwise as viewed from above. Stokes' Theorem: $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \,dS$ Here $\mathbf{F} = (z^2e^{x^2}, xy^2, \tan^{-1} y)$. The curve $C$ is the circle $x^2+y^2=9$ in the $xy$-plane (implicitly $z=0$). Choose $S$ to be the disk $x^2+y^2 \le 9$ in the $xy$-plane ($z=0$). For this surface, the upward normal vector is $\mathbf{n} = \mathbf{k} = (0,0,1)$. The orientation of $C$ (counterclockwise as viewed from above) matches this normal. Calculate the curl of $\mathbf{F}$: $\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2e^{x^2} & xy^2 & \tan^{-1} y \end{vmatrix}$ $\mathbf{i}\left(\frac{\partial}{\partial y}(\tan^{-1} y) - \frac{\partial}{\partial z}(xy^2)\right) = \mathbf{i}\left(\frac{1}{1+y^2} - 0\right) = \frac{1}{1+y^2}\mathbf{i}$ $-\mathbf{j}\left(\frac{\partial}{\partial x}(\tan^{-1} y) - \frac{\partial}{\partial z}(z^2e^{x^2})\right) = -\mathbf{j}(0 - 2ze^{x^2}) = 2ze^{x^2}\mathbf{j}$ $+\mathbf{k}\left(\frac{\partial}{\partial x}(xy^2) - \frac{\partial}{\partial y}(z^2e^{x^2})\right) = \mathbf{k}(y^2 - 0) = y^2\mathbf{k}$ So, $\nabla \times \mathbf{F} = \left(\frac{1}{1+y^2}\right)\mathbf{i} + (2ze^{x^2})\mathbf{j} + (y^2)\mathbf{k}$. Now calculate $(\nabla \times \mathbf{F}) \cdot \mathbf{n}$: Since $\mathbf{n} = \mathbf{k}$, $(\nabla \times \mathbf{F}) \cdot \mathbf{k} = y^2$. The integral becomes $\iint_S y^2 \,dS$. The surface $S$ is the disk $x^2+y^2 \le 9$ in the $xy$-plane, so $dS = dA$. $\iint_{x^2+y^2 \le 9} y^2 \,dA$. Use polar coordinates: $x = r\cos\theta$, $y = r\sin\theta$, $dA = r \,dr \,d\theta$. The region is $0 \le r \le 3$, $0 \le \theta \le 2\pi$. $\int_0^{2\pi} \int_0^3 (r\sin\theta)^2 r \,dr \,d\theta = \int_0^{2\pi} \int_0^3 r^3\sin^2\theta \,dr \,d\theta$ Inner integral: $\int_0^3 r^3\sin^2\theta \,dr = \sin^2\theta \left[\frac{r^4}{4}\right]_0^3 = \frac{81}{4}\sin^2\theta$ Outer integral: $\int_0^{2\pi} \frac{81}{4}\sin^2\theta \,d\theta$ Use identity $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$: $\frac{81}{4} \int_0^{2\pi} \frac{1-\cos(2\theta)}{2} \,d\theta = \frac{81}{8} \left[\theta - \frac{1}{2}\sin(2\theta)\right]_0^{2\pi}$ $= \frac{81}{8} ((2\pi - 0) - (0 - 0)) = \frac{81}{8} (2\pi) = \frac{81\pi}{4}$ The line integral is $\frac{81\pi}{4}$.