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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
  42.   [ (16a, -8a) ,quad (a/4, -a) ,quad (0, 0) ]
  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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