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author | Karl Berry <karl@freefriends.org> | 2017-07-13 23:06:59 +0000 |
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committer | Karl Berry <karl@freefriends.org> | 2017-07-13 23:06:59 +0000 |
commit | 03542c01781307bd8cb29571d8a39197a4de83e2 (patch) | |
tree | b11e221ce28f24eee3995a8efb51d2e76993472c /Master/texmf-dist/doc/latex/probsoln/samples/prob-implicit.tex | |
parent | 04f93eb1202717fcb407f1271d28f4153aaa0ca0 (diff) |
probsoln (10jul17)
git-svn-id: svn://tug.org/texlive/trunk@44783 c570f23f-e606-0410-a88d-b1316a301751
Diffstat (limited to 'Master/texmf-dist/doc/latex/probsoln/samples/prob-implicit.tex')
-rw-r--r-- | Master/texmf-dist/doc/latex/probsoln/samples/prob-implicit.tex | 52 |
1 files changed, 52 insertions, 0 deletions
diff --git a/Master/texmf-dist/doc/latex/probsoln/samples/prob-implicit.tex b/Master/texmf-dist/doc/latex/probsoln/samples/prob-implicit.tex new file mode 100644 index 00000000000..dd847d96e30 --- /dev/null +++ b/Master/texmf-dist/doc/latex/probsoln/samples/prob-implicit.tex @@ -0,0 +1,52 @@ + % This file is public domain + % + % 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:ellipse}{% +Find the gradient of the ellipse given by $4x^2 + 3y^2 = 25$.}{% +Differentiating with respect to $x$ gives: +\begin{eqnarray*} +8x + 6y\frac{dy}{dx} & = & 0\\ +\frac{dy}{dx} & = & \frac{-8x}{6y}\\ + & = & \frac{-4x}{\sqrt{25-4x^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*}} |