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author | Karl Berry <karl@freefriends.org> | 2006-01-09 00:49:07 +0000 |
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committer | Karl Berry <karl@freefriends.org> | 2006-01-09 00:49:07 +0000 |
commit | 007f67a693e4d031fd3d792df8e4d5f43e2cb2e7 (patch) | |
tree | 90d17e00e572ecb1e24764b6f29c80e098b08d29 /Master/texmf-dist/doc/latex/probsoln/samples | |
parent | 950209b26f70aa87ed07c54f82a95b6f03b7c3a0 (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')
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 new file mode 100644 index 00000000000..f857392a6ef --- /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*}} |