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authorKarl Berry <karl@freefriends.org>2006-01-09 00:49:07 +0000
committerKarl Berry <karl@freefriends.org>2006-01-09 00:49:07 +0000
commit007f67a693e4d031fd3d792df8e4d5f43e2cb2e7 (patch)
tree90d17e00e572ecb1e24764b6f29c80e098b08d29 /Master/texmf-dist/doc/latex/probsoln/samples
parent950209b26f70aa87ed07c54f82a95b6f03b7c3a0 (diff)
doc/latex
git-svn-id: svn://tug.org/texlive/trunk@84 c570f23f-e606-0410-a88d-b1316a301751
Diffstat (limited to 'Master/texmf-dist/doc/latex/probsoln/samples')
-rw-r--r--Master/texmf-dist/doc/latex/probsoln/samples/1stprncp.tex61
-rw-r--r--Master/texmf-dist/doc/latex/probsoln/samples/args.tex45
-rw-r--r--Master/texmf-dist/doc/latex/probsoln/samples/easy.tex84
-rw-r--r--Master/texmf-dist/doc/latex/probsoln/samples/implicit.tex41
4 files changed, 231 insertions, 0 deletions
diff --git a/Master/texmf-dist/doc/latex/probsoln/samples/1stprncp.tex b/Master/texmf-dist/doc/latex/probsoln/samples/1stprncp.tex
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--- /dev/null
+++ b/Master/texmf-dist/doc/latex/probsoln/samples/1stprncp.tex
@@ -0,0 +1,61 @@
+% These all involve differentiating from 1st principles
+
+\newproblem{dfp:xcube}{%
+Differentiate $f(x) = x^3$ with respect to $x$ by first principles.}{%
+\begin{eqnarray*}
+\frac{dy}{dx} & = & \lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{(x+\Delta x)^3-x^3}{\Delta x}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{(x+\Delta x)(x^2+2x\Delta x+(\Delta x)^2)-x^3}{\Delta x}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3-x^3}{\Delta x}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3}{\Delta x}\\
+ & = & \lim_{\Delta x\rightarrow 0}3x^2+3x\Delta x + (\Delta x)^2\\
+ & = & 3x^2
+\end{eqnarray*}}
+
+\newproblem{dfp:Ioverxsq}{%
+Differentiate $\displaystyle f(x) = \frac{1}{x^2}$ with respect to $x$ by first principles.}{%
+\begin{eqnarray*}
+\frac{df}{dx} & = & \lim_{\Delta x\rightarrow 0}\frac{\frac{1}{(x+\Delta x)^2}-\frac{1}{x^2}}{\Delta x}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{\frac{x^2-(x+\Delta x)^2}{x^2(x+\Delta x)^2}}{\Delta x}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{x^2-(x^2+2x\Delta x+(\Delta x)^2)}{x^2\Delta x(x+\Delta x)^2}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{-2x\Delta x-(\Delta x)^2}{x^2\Delta x(x+\Delta x)^2}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{-2x-\Delta x}{x^2(x+\Delta x)^2}\\
+ & = & \frac{-2x}{x^2x^2}\\
+ & = & -\frac{2}{x^3}
+\end{eqnarray*}}
+
+\newproblem{dfp:sqrtx}{%
+Differentiate from first principles $f(x) = \surd x$}{%
+\begin{eqnarray*}
+\frac{df}{dx} & = & \lim_{\Delta x\rightarrow 0}\frac{\sqrt{x+\Delta x}-\surd x}{\Delta x}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{(\sqrt{x+\Delta x}-\surd x)(\sqrt{x+\delta x}+\surd x)}{\Delta x(\sqrt{x+\Delta x}+\surd x)}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{x+\Delta x - x}{\Delta x(\sqrt{x+\Delta x}+\surd x)}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{\Delta x}{\Delta x(\sqrt{x+\Delta x}+\Delta x)}\\
+ & = & \lim_{\Delta x\rightarrow 0}\frac{1}{\sqrt{x+\Delta x}+\surd x}\\
+ & = & \frac{1}{2\surd x}
+\end{eqnarray*}}
+
+\newproblem{dfp:cons}{%
+Differentiate from first principles $f(x) = c$ where $c$ is a constant.}{%
+\begin{eqnarray*}
+\frac{df}{dx} & = & \lim_{\Delta x\rightarrow 0}\frac{c-c}{\Delta x}\\
+ & = & \lim_{\Delta x\rightarrow 0}0\\
+ & = & 0
+\end{eqnarray*}}
+
+\newproblem{dfp:cosx}{%
+Given
+\begin{eqnarray*}
+\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} & = & 0\\
+\lim_{x \rightarrow 0} \frac{\sin x}{x} & = & 1
+\end{eqnarray*}
+differentiate from first principles $f(x) = \cos x$.}{%
+\begin{eqnarray*}
+\frac{df}{dx} & = & \lim_{\Delta x \rightarrow 0}\frac{f(x + \Delta x) - f(x)}{\Delta x}\\
+ & = & \lim_{\Delta x \rightarrow 0} \frac{\cos(x + \Delta x) - \cos(x)}{\Delta x}\\
+ & = & \lim_{\Delta x \rightarrow 0} \frac{\cos x\cos\Delta x - \sin x\sin\Delta x - \cos x}{\Delta x}\\
+ & = & \lim_{\Delta x \rightarrow 0} \frac{\cos x(\cos\Delta x - 1) - \sin x\sin\Delta x}{\Delta x}\\
+ & = & \cos x\lim_{\Delta x \rightarrow 0}\frac{\cos\Delta x - 1}{\Delta x}
+ - \sin x\lim_{\Delta x \rightarrow 0}\frac{\sin\Delta x}{\Delta x}\\
+ & = & -1 \qquad\mbox{(using given results)}
+\end{eqnarray*}}
diff --git a/Master/texmf-dist/doc/latex/probsoln/samples/args.tex b/Master/texmf-dist/doc/latex/probsoln/samples/args.tex
new file mode 100644
index 00000000000..e3bd7de75ec
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/probsoln/samples/args.tex
@@ -0,0 +1,45 @@
+% These problems require arguments
+
+% Arguments: #1->a_2, #2->a_1 and #3->a_0
+% (Arguments must be integers)
+\newcount\ctr
+\newproblem[3]{diff:quad}{%
+\(f(x) =
+\ifnum#1=0
+\else
+\ifnum#1=1\else#1\fi x^2
+\fi
+\ifnum#2=0
+\else
+\ifnum#2>0 \ifnum#1=0 \else + \fi \fi
+\ifnum#2=1\else#2\fi x
+\fi
+\ifnum#3=0
+\else
+\ifnum#3>0 \ifnum#2=0 \ifnum#1=0 \else + \fi \else + \fi\fi
+#3
+\fi\)
+}{%
+\(f'(x) =
+\ifnum#1=0
+\else
+\ctr=2
+\multiply\ctr by #1
+\the\ctr x
+\fi
+\ifnum#2=0
+\else
+\ifnum#2>0 \ifnum#1=0 \else + \fi \fi
+#2
+\fi
+% print 0 if both #1 and #2 are 0
+\ifnum#1=0 \ifnum#2=0 0 \fi\fi
+\)
+}
+
+\newproblem[1]{diff:sin}{%
+\(f(x) = \sin(#1x)\)
+}{%
+\(f'(x) = #1\cos(#1x)\)
+}
+
diff --git a/Master/texmf-dist/doc/latex/probsoln/samples/easy.tex b/Master/texmf-dist/doc/latex/probsoln/samples/easy.tex
new file mode 100644
index 00000000000..5aa99ed04ca
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/probsoln/samples/easy.tex
@@ -0,0 +1,84 @@
+% These are all easy differentiation problems
+
+\newproblem{diffeasy:gpowh}{%
+\(f(x) = g(x)^{h(x)}.\)}{%
+\begin{eqnarray*}
+f(x) & = & e^{\ln g(x)^{h(x)}}\\
+ & = & e^{h(x)\ln g(x)}\\
+f'(x) & = & e^{h(x)\ln g(x)}(h'(x)\ln g(x) + h(x)\frac{g'(x)}{g(x)})\\
+ & = & g(x)^{h(x)}(h'(x)\ln g(x) + \frac{h(x)g'(x)}{g(x)})
+\end{eqnarray*}}
+
+\newproblem{diffeasy:arcsin}{%
+\(y = \arcsin(x)\)}{%
+\[\sin(y) = x\]
+diff. w.r.t. $x$:
+\begin{eqnarray*}
+\cos y \frac{dy}{dx} & = & 1\\
+\frac{dy}{dx} & = & \frac{1}{\cos y}\\
+ & = & \frac{1}{\sqrt{1 - \sin^2y}}\\
+ & = & \frac{1}{\sqrt{1-x^2}}.
+\end{eqnarray*}}
+
+\newproblem{diffeasy:arccos}{%
+$y = \arccos x$.}{%
+\(\cos y = x\)
+diff. w.r.t. $x$:
+\begin{eqnarray*}
+-\sin y \frac{dy}{dx} & = & 1\\
+\frac{dy}{dx} & = & \frac{-1}{\sin y}\\
+ & = & \frac{-1}{\sqrt{1-\cos^2y}}\\
+ & = & \frac{-1}{\sqrt{1-x^2}}
+\end{eqnarray*}}
+
+\newproblem{diffeasy:tan}{%
+\(y = \tan x\)}{%
+\begin{eqnarray*}
+y & = & \tan x\\
+ & = & \frac{\sin x}{\cos x}\\
+\frac{dy}{dx} & = & \frac{\cos x}{\cos x} + \sin x\times\frac{-1}{\cos^2x}\times -\sin x\\
+ & = & 1 + \tan^2x\\
+ & = & \sec^2x.
+\end{eqnarray*}}
+
+\newproblem{diffeasy:arctan}{%
+\(y = \arctan x = \tan^{-1}x\)}{%
+\[\tan y = x\]
+diff w.r.t. $x$:
+\begin{eqnarray*}
+\sec^2y\frac{dy}{dx} & = & 1\\
+\frac{dy}{dx} & = & \frac{1}{\sec^2y}\\
+ & = & \frac{1}{1+\tan^2y}\\
+ & = & \frac{1}{1+x^2}
+\end{eqnarray*}}
+
+\newproblem{diffeasy:cot}{%
+\(y = (\tan x)^{-1} = \cot x\)}{%
+\begin{eqnarray*}
+\frac{dy}{dx} & = & -(\tan x)^{-2}\sec^2x\\
+ & = & -\frac{\cos^2x}{\sin^2x}\cdot\frac{1}{\cos^2x}\\
+ & = & \frac{-1}{\sin^2x}\\
+ & = & -\csc^2x.
+\end{eqnarray*}}
+
+\newproblem{diffeasy:cosxsqsinx}{%
+$y = \cos(x^2)\sin x$.}{%
+\[\frac{dy}{dx} = -\sin(x^2)2x\sin x + \cos(x^2)\cos x\]}
+
+\newproblem{diffeasy:xlnx}{%
+$y = (x+1)\ln(x+1)$.}{%
+\begin{eqnarray*}
+\frac{dy}{dx} & = & \ln(x+1) + \frac{x+1}{x+1}\\
+ & = & 1 + \ln(x+1).
+\end{eqnarray*}}
+
+\newproblem{diffeasy:glng}{%
+$f(x) = g(x)\ln(g(x))$.}{%
+\begin{eqnarray*}
+f'(x) & = & g'(x)\ln(g(x)) + \frac{g(x)}{g(x)}g'(x)\\
+ & = & g'(x)(1+\ln(g(x))).
+\end{eqnarray*}}
+
+\newproblem{diffeasy:sinx/x}{%
+$y = \frac{\sin x}{x}$.}{%
+\[\frac{dy}{dx} = \frac{\cos x}{x} - \frac{\sin x}{x^2}\]}
diff --git a/Master/texmf-dist/doc/latex/probsoln/samples/implicit.tex b/Master/texmf-dist/doc/latex/probsoln/samples/implicit.tex
new file mode 100644
index 00000000000..00453df3045
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/probsoln/samples/implicit.tex
@@ -0,0 +1,41 @@
+% These are all implicit differentiation problems
+
+\newproblem{imd:circ}{%
+Find the gradient of the unit circle ($x^2 + y^2 = 1$).}{%
+Differentiating with respect to $x$ gives:
+\begin{eqnarray*}
+2x + 2y\frac{dy}{dx} & = & 0\\
+\frac{dy}{dx} & = & \frac{-2x}{2y}\\
+ & = & \frac{-x}{\sqrt{1-x^2}}.
+\end{eqnarray*}}
+
+\newproblem{imd:ysq:xcuov2mx}{%
+Find $\frac{dy}{dx}$, given
+\begin{displaymath}
+y^2 = \frac{x^3}{2-x}
+\end{displaymath}}{%
+Differentiating both sides w.r.t.\ $x$:
+\begin{eqnarray*}
+2y\frac{dy}{dx} & = & \frac{(2-x)3x^2 - x^3(-1)}{(2-x)^2}\\
+ & = & \frac{3x^2(2-x) + x^3}{(2-x)^2}\\
+ & = & \frac{6x^2 - 3x^3 + x^3}{(2-x)^2}\\
+ & = & \frac{6x^2-2x^3}{(2-x)^2}\\
+ & = & 2x^2\frac{3-x}{(2-x)^2}
+\end{eqnarray*}
+Therefore
+\begin{displaymath}
+y\frac{dy}{dx} = x^2\frac{3-x}{(2-x)^2}
+\end{displaymath}}
+
+\newproblem{imd:exy:IIxay}{%
+Differentiate w.r.t.\ $x$:
+\begin{displaymath}
+e^{xy} = 2x + y
+\end{displaymath}}{%
+Differentiating both sides w.r.t.\ $x$:
+\begin{eqnarray*}
+e^{xy}(1y + x\frac{dy}{dx}) & = & 2 + \frac{dy}{dx}\\
+xe^{xy}\frac{dy}{dx} - \frac{dy}{dx} & = & 2 - ye^{xy} \\
+\frac{dy}{dx}(xe^{xy}-1) & = & 2 - ye^{xy}\\
+\frac{dy}{dx} & = & \frac{2-ye^{xy}}{xe^{xy}-1}
+\end{eqnarray*}}