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# Untitled

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1. documentclass{memoir}
2.
3. % __________ AMS ________________________
4.   usepackage{amsmath}
5. % __________ PGF TikZ ___________________
6.   usepackage{pgfplots}
7. % usepackage{tikz}
8.   usepackage{tikz-3dplot}
9.
10. % __________ Differentials _____________________________________________________
11. % Single
12.     newcommand{diff}{d}           % If you want an upright `d', change it here
13.     newcommand{p}[1]{partial#1}
14. % ___________________ Derivatives ______________________________________________
15. % 1st derivative:
16.     newcommand{dod}[2]{dfrac{diff{#1}}{diff{#2}}}  % 'differential over differential'
17.     newcommand{pop}[2]{dfrac{p#1}{p#2}}            % 'partial over partial'
18.     newcommand{lpop}[2]{p#1/p#2}            % A 'layed down' version
19.
20. setlength{parindent}{0cm}
21. begin{document}
22.
23. textbf{Problem 5.7}par
24. centering
25. textbf{The Chain Rule and Stationary Points}\[3mm]
26. flushleft
27.   fbox{parbox{4.25in}{
28.     The function \$G(t)\$ is defined by
29.     \$\$ G(t) = F(x,y) = x^2 + y^2 + 3xy \$\$
30.     where \$x(t) = at^2\$ and \$y(t) = 2at\$. Use the chain rule to find the values of \$(x,y)\$ at which \$G(t)\$ has stationary values as a function of \$t\$. Do any of them correspond to the stationary points of \$F(x,y)\$ as a function of \$x\$ and \$y\$?
31.     }}
32. flushleft
33.
34.   vspace{3mm}
35.   noindentemph{Solution:}\[1mm]
36.   (To be terse, the derivation has been omitted.)\[2mm]
37.   The Stationary points occur at \$lpop Ft = 0\$, in which case
38.   [ 2a^2t(2t + 1)(t + 4) = 0 ]
39.   So,
40.   [ tin { -4, -1/2, 0 } ]
41.   This corresponds to the stationary points
43.   To answer the second part of the question, we differentiate \$F(x,y)\$ with respect to \$x\$, and \$y\$, and set these to zero;
44.   [ pop Fx = 2x + 3y = 0 ]
45.   [ pop Fy = 3x + 2y = 0 ]
46.   The only solution is \$(0, 0)\$.
47.
48. centering
49. begin{figure}
50.   tdplotsetmaincoords{0}{0}
51.   begin{tikzpicture}[tdplot_main_coords,scale=1.5,rotate=0]
52.     begin{axis}[domain=-6:6,y domain=-6:6]
53.       addplot3[surf] {x^2 + y^2 + 3*x*y};
54.     end{axis}
55.   end{tikzpicture}
56.   caption{\$z = x^2 + y^2 + 3xy\$}
57. end{figure}
58. flushleft
59.
60. The other two solutions for \$t\$, are the stationary points for the parabolic sheet (not shown) \$(at^2, 2at, t)\$ that intersects with the surface shown. There is also the two-dimensional version realized by looking down from \$z\$ (i.e. the projection onto the \$z\$-plane). The blue to orange colors could then be interpreted, for example, as a scalar field for temperature.
61.
62. end{document}
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