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MATH 180\hfill FALL 1995

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%10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#27}

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4.8, \#1.
Differentiating $x^2+xy-y^3=xy^2$ with respect to $x$ gives
$$2x+y+x\,{dy\over dx}-3y^2\,{dy\over dx}=y^2+2xy\,{dy\over dx}.$$
Thus $\displaystyle {dy\over dx}={2x+y-y^2\over 3y^2+2xy-x}$.

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4.8, \#2.
Differentiating $x^2+y^2=7$ with respect to $x$ gives
$\displaystyle 2x+2y\,{dy\over dx}=0$ and
$\displaystyle {dy\over dx}=-{x\over y}$.

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4.8, \#3.
Differentiating $x^{1/2}+y^{1/2}=25$ with respect to $x$ gives
$${1\over 2\sqrt x}+{1\over 2\sqrt y}\,{dy\over dx}=0$$
or $\displaystyle {dy\over dx}=-{\sqrt y\over \sqrt x}$.

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4.8, \#9.
Differentiating $xy^2=1$ with respect to $x$ gives
$\displaystyle y^2+2xy\; {dy\over dx}=0$.  At the point
$P=(1.-1)$ on the graph of $xy^2=1$, we have
$\displaystyle 1-2\;{dy\over dx}=0$ or $dy/dx=1/2$.
Thus the tangent line is $y=-1+{1\over 2}(x-1)=
{1\over2}x-{3\over2}$.

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4.8, \#15.
Differentiating $x^2/25+y^2/9=1$ with respect to $x$ gives
$\displaystyle dy/dx=-{9x\over 25y}$.  This is the slope
of the tangent line to the ellipse at the point $(x,y)$.
The slope is undefined when $y=0$, that is,
at the points $(5,0)$ and $(-5,0)$ of the ellipse where
the tangent line is vertical.

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4.8, \#18.
Differentiating $xe^{5y}=3y$ with respect to $x$ gives
$\displaystyle e^{5y}+xe^{5y}\cdot 5\,{dy\over dx}=3\cdot{dy\over dx}$.
At the point $(0,0)$ on the graph, this relation implies
$dy/dx=1/3$.  Thus the tangent line at $(0,0)$ is $y={1\over3}x$.
In particular, when $x=0.1$, $y$ is approximately
${1\over 3}\cdot(0.1)=0.03333$.


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