\thepage } %} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \subsection*{Ma 227 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Exam III B \ Solutions\hfill 4/26/04} \subsection*{{\protect\normalsize Name: \protect\underline{\hspace{2.5in}} \hfill ID: \protect\underline{\hspace{2.0in}}}} \subsection*{{\protect\normalsize Lecture Section: \protect\underline{% \hspace{2.5in}} \hfill Lecturer: \protect\underline{\hspace{2in}}}} \hfill {\small \textit{I pledge my honor that I have abided by the Stevens Honor System.}\hspace{2pt} \ \ \ \ \ \ \ \ \ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_% \_\_\_\_} \hfill \paragraph{You may not use a calculator, cell phone, or computer while taking this exam. All work must be shown to obtain full credit. Credit will not be given for work not reasonably supported. When you finish, be sure to sign the pledge.} \vspace{1pt} Score on Problem \ \#1 \_\_\_\_\_\_\_\_\vspace{0.2cm} \qquad \qquad \qquad \qquad\ \ \ \#2 \_\_\_\_\_\_\_\_\vspace{0.2cm} \qquad \qquad \qquad \qquad\ \ \ \#3 \_\_\_\_\_\_\_\_\vspace{0.2cm} \qquad \qquad \qquad \qquad \qquad \qquad \qquad\ \ \qquad \qquad \qquad \qquad\ \ \qquad \qquad \qquad \qquad\ \ \qquad \qquad \qquad \qquad\ \ Total Score \ \ \ \ \ \ \ \ \ \ \ \ \ \ \_\_\_\_\_\_\_\_\bigskip \pagebreak \begin{description} \item[{1a [15 pts.]}] Find the work done by the force field \begin{equation*} \vec{F}\left( x,y\right) =2y\vec{i}+3x\vec{j} \end{equation*} \end{description} along the plane path that is the graph of the parabola $y=2x^{2}$ from $% A=\left( 1,2\right) $ to $B=\left( 2,8\right) $. Solution: The path may be parametrized as $C:x=t,y=2t^{2}$ \ $1\leq t\leq 2.$ Then $\vec{r}\left( t\right) =t\vec{i}+2t^{2}\vec{j},$ $\vec{r}^{\prime }\left( t\right) =\vec{i}+4t\vec{j},$ \ $\vec{F}\left( t\right) =2(2t^{2})% \vec{i}+3t\vec{j}$ \begin{equation*} \int_{C}\vec{F}\cdot d\vec{r}=\int_{1}^{2}\vec{F}\left( t\right) \cdot \vec{r% }^{\prime }\left( t\right) dt=\int_{1}^{2}\left( 4t^{2}+12t^{2}\right) dt=\int_{1}^{2}16t^{2}dt=\left. 16\frac{t^{3}}{3}\right] _{1}^{2}=16\left( \frac{7}{3}\right) =\frac{112}{3} \end{equation*} \begin{description} \item[{1b [15 pts.]}] If the function $f\left( x,y,z\right) $ has continuous second-order partial derivatives, show that \begin{equation*} curl\left( grad\text{ }f\right) =0 \end{equation*} \end{description} \vspace{1pt}Solution: \ $\nabla f=grad$ $f=\frac{\partial f}{\partial x}\vec{% i}+\frac{\partial f}{\partial y}\vec{j}+\frac{\partial f}{\partial z}\vec{k}$ so% \begin{eqnarray*} \nabla \times \nabla f &=&\left\vert \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{% \partial f}{\partial z}% \end{array}% \right\vert =\vec{i}\left\vert \begin{array}{cc} \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}% \end{array}% \right\vert -\vec{j}\left\vert \begin{array}{cc} \frac{\partial }{\partial x} & \frac{\partial }{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial xz}% \end{array}% \right\vert +\vec{k}\left\vert \begin{array}{cc} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}% \end{array}% \right\vert \\ &=&\left( \frac{\partial ^{2}f}{\partial y\partial z}-\frac{\partial ^{2}f}{% \partial z\partial y}\right) \vec{i}+\left( \frac{\partial ^{2}f}{\partial z\partial x}-\frac{\partial ^{2}f}{\partial x\partial z}\right) \vec{j}% +\left( \frac{\partial ^{2}f}{\partial x\partial y}-\frac{\partial ^{2}f}{% \partial x\partial z}\right) \vec{k}=\vec{0} \end{eqnarray*} \begin{description} \item[{2a $\left[ 20\text{ pts.}\right] $}] Find a function $\Phi \left( x,y,z\right) $ such that $\nabla \Phi =\vec{F},$ where% \begin{equation*} \vec{F}\left( x,y,z\right) =\left( 2xy+z^{2}\right) \vec{i}+\left( x^{2}+2yz+2z\right) \vec{j}+\left( y^{2}+2xz+2y-3\right) \vec{k} \end{equation*} \end{description} Solution: We first note the \begin{equation*} \nabla \times \left( 2xy+z^{2},x^{2}+2yz+2z,y^{2}+2xz+2y-3\right) =\left( 0,0,0\right) \end{equation*} so such a $\Phi $ exists. This is not required in order to get full credit for the solution. We have that \begin{equation*} \Phi _{x}=2xy+z^{2}\text{ \ \ \ }\Phi _{y}=x^{2}+2yz+2z\text{ \ \ \ }\Phi _{z}=y^{2}+2xz+2y-3 \end{equation*}% Starting with $\Phi _{x}$ and integrating with respect to $x$ we get% \begin{equation*} \Phi =x^{2}y+xz^{2}+h\left( y,z\right) \end{equation*}% Then% \begin{equation*} \Phi _{y}=x^{2}+\frac{\partial h}{\partial y}=x^{2}+2yz+2z \end{equation*}% Thus% \begin{equation*} \frac{\partial h}{\partial y}=2yz+2z\Rightarrow h\left( y,z\right) =y^{2}z+2yz+g\left( z\right) \end{equation*}% and therefore% \begin{equation*} \Phi =x^{2}y+xz^{2}+y^{2}z+2yz+g\left( z\right) \end{equation*} Hence% \begin{equation*} \Phi _{z}=2xz+y^{2}+2y+g^{\prime }\left( z\right) =y^{2}+2xz+2y-3 \end{equation*} Then% \begin{equation*} g^{\prime }\left( z\right) =-3\Rightarrow g\left( z\right) =-3z+k \end{equation*}% Finally we have that \begin{equation*} \Phi =x^{2}y+xz^{2}+y^{2}z+2yz+-3z+k \end{equation*} Check: SNB $\nabla \left( x^{2}y+xz^{2}+y^{2}z+2yz+-3z+k\right) =\allowbreak \left( 2xy+z^{2},x^{2}+2yz+2z,2xz+y^{2}+2y-3\right) .$ \begin{description} \item[{2b [20 pts.]}] Verify that Green's Theorem is true for the line integral% \begin{equation*} \oint_{C}\left( 2x-y^{2}\right) dx+\left( 2y-x^{2}\right) dy \end{equation*} \end{description} where $C$ is the triangle with vertices $\left( -1,1\right) ,\left( -1,-1\right) $ and $\left( 1,-1\right) .$ Solution: The triangle has vertices at $\left( -1,1\right) ,\left( -1,-1\right) $ and $\left( 1,-1\right) .$ It is shown below. $\left( -1,1,-1,-1,1,-1,-1,1\right) $\FRAME{dtbpFX}{3.0113in}{3.0113in}{0pt}{% }{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 3.0113in;height 3.0113in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute TRUE;lastEngine "MuPAD";xmin "-5";xmax "5";xviewmin "-1.002";xviewmax "1.002";yviewmin "-1.002";yviewmax "1.002";plottype 4;numpoints 49;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function \TEXUX{$\left( -1,1,-1,-1,1,-1,-1,1\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";}} We must go around the boundary in a counter clockwise direction. \ I start on the horizontal leg. \ The second leg (the hypotenuse) is parametrized using $\ x=t$ and $y=-t$ with t starting at 1 and ending at -1! \begin{eqnarray*} \oint_{C}\left( 2x-y^{2}\right) dx+\left( 2y-x^{2}\right) dy &=&\left( \int_{\left( -1,-1\right) }^{\left( 1,-1\right) }+\int_{\left( 1,-1\right) }^{\left( -1,1\right) }+\int_{\left( -1,1\right) }^{\left( -1,-1\right) }\right) \left[ \left( 2x-y^{2}\right) dx+\left( 2y-x^{2}\right) dy\right] \\ &=&\int_{-1}^{1}\left( 2x-\left( -1\right) ^{2}\right) dx+\int_{1}^{-1}\left[ \left( 2t-\left( -t\right) ^{2}\right) -\left( 2\left( -t\right) -\left( t\right) ^{2}\right) \right] dt+\int_{1}^{-1}\left( 2y-(-1)^{2}\right) dy=0 \end{eqnarray*} Also since $P=2x-y^{2}$ and $Q=2y-x^{2}$ \begin{equation*} \diint\limits_{R}\left( Q_{x}-P_{y}\right) dA=\int_{-1}^{1}\int_{-1}^{-x}\left( -2x+2y\right) dydx=0 \end{equation*} \vspace{1pt} \begin{description} \item[{3 a [10 pts.]}] Find an upward normal to the triangular surface $S$ cut off from the plane \begin{equation*} x+2y+3z=6 \end{equation*} \end{description} by the coordinate planes. Sketch the surface $S.$ Solution: \vspace{1pt}$6-3x-2y$\FRAME{dtbpFX}{3.0113in}{4.0145in}{0pt}{}{}{Plot}{% \special{language "Scientific Word";type "MAPLEPLOT";width 3.0113in;height 4.0145in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "Maple";xmin "0";xmax "2";ymin "0";ymax "3";xviewmin "0";xviewmax "5.204";yviewmin "0";yviewmax "5.204";zviewmin "0";zviewmax "10";viewset"XYZ";rangeset"XYZ";phi 45;theta 45;plottype 5;constrained TRUE;num-x-gridlines 25;num-y-gridlines 25;plotstyle "patch";axesstyle "normal";plotshading "XYZ";xis \TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function \TEXUX{$6-3x-2y$};linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";var1range "0,2";var2range "0,3";surfaceColor "[linear:XYZ:RGB:0x00ff0000:0x000000ff]";surfaceStyle "Color Patch";num-x-gridlines 25;num-y-gridlines 25;surfaceMesh "Mesh";rangeset"XY";}} Let $x=u,y=v,z=6-3u-2v.$ Then% \begin{equation*} \vec{r}\left( u,v\right) =u\vec{i}+v\vec{j}+\left( 6-3u-2v\right) \vec{k} \end{equation*} \begin{equation*} \vec{r}_{u}=\vec{i}-3\vec{k}\text{ \ \ \ }\vec{r}_{v}=\vec{j}-2\vec{k} \end{equation*} \begin{equation*} \vec{N}=\vec{r}_{u}\times \vec{r}_{v}=\left\vert \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -3 \\ 0 & 1 & -2% \end{array}% \right\vert =3\vec{i}+2\vec{j}+\vec{k} \end{equation*} This is clearly outward since it points from the origin to the point $\left( 3,2,1\right) .$ \ Alternatively, one recalls that the coefficients of the variables in the equation of a plane are the components of a normal vector. \begin{description} \item[{3 b [15 pts.]}] Give an expression for \begin{equation*} \iint_{S}\vec{F}\cdot d\vec{S}=\iint_{S}\vec{F}\cdot \vec{N}ds \end{equation*} \end{description} where% \begin{equation*} \vec{F}=z\vec{i}-xy\vec{j}+\left( x-y\right) \vec{k} \end{equation*}% and $S$ is the surface in part 3a. Do \emph{not }evaluate your expression. \vspace{1pt}Solution: \begin{equation*} \vec{F}\left( u,v\right) =\left( 6-3u-2v\right) \vec{i}-uv\vec{j}+\left( u-v\right) \vec{k} \end{equation*} so \begin{equation*} \vec{F}\cdot \vec{N}=\left( \left( 6-3u-2v\right) \vec{i}-uv\vec{j}+\left( u-v\right) \vec{k}\right) \cdot \left( 3\vec{i}+2\vec{j}+\vec{k}\right) =18-9u-6v-2uv+u-v=18-8u-7v-2uv \end{equation*} The region of integration in the $x,y-$plane is bounded by the axes and the line $3x+2y=6$ and is shown below. $3-\frac{3}{2}x$\FRAME{dtbpFX}{3.0113in}{3.0113in}{0pt}{}{}{Plot}{\special% {language "Scientific Word";type "MAPLEPLOT";width 3.0113in;height 3.0113in;depth 0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "Maple";xmin "0";xmax "2";xviewmin "-0.04";xviewmax "4";yviewmin "-0.06";yviewmax "4";viewset"XY";rangeset"X";plottype 4;constrained TRUE;numpoints 49;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};function \TEXUX{$3-\frac{3}{2}x$};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range "0,2";num-x-gridlines 49;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";function \TEXUX{$(0,3,0,0,2,0)$};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";curveColor "[flat::RGB:0000000000]";curveStyle "Line";}} Thus since $u=x$ and $y=v$% \begin{equation*} \iint_{S}\vec{F}\cdot d\vec{S}=\iint_{S}\vec{F}\cdot \vec{N}% ds=\int_{0}^{2}\int_{0}^{3-\frac{3x}{2}}\left( 18-8x-7y-2xy\right) dydx=\int_{0}^{3}\int_{0}^{2-\frac{2y}{3}}\left( 18-8x-7y-2xy\right) dxdy \end{equation*} \vspace{1pt} \end{document} %%%%%%%%%%%%% End /document/ma227_exam_3b_04s_v1_sol1.tex %%%%%%%%%%%%%