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MATH 180\hfill FALL 1995

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%10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#26}

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4.7, \#2.  $\displaystyle (\ln (t^2+1))'={1\over t^2+1}\cdot 2t$.

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4.7, \#4.  $\displaystyle (\ln(\cos \theta))'=
{1\over\cos\theta}(-\sin\theta)=-\tan\theta$.

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4.7, \#5.
$\displaystyle (\ln(1-e^{-x}))'=
{1\over 1-e^{-x}}(-e^{-x})(-1)={e^{-x}\over 1-e^{-x}}$.

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4.7, \#7.
$\displaystyle (\ln(e^x+1))'={1^x\over e^x+1}\cdot e^x=
   {e^x\over e^x+1}$.

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4.7, \#14.
$\displaystyle (\arctan (3t-4))'={1\over (3t-4)^2+1}\cdot 3
={3\over 9t^2-24t+17}$.

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4.7, \#15.
$\displaystyle (w\arcsin w)'=\arcsin w+w\cdot{1\over\sqrt{1-w^2}}$.

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4. Review, \#34.
The volume of a sphere of radius $r$ is $V={4\over3}\pi r^3$.
In the problem
$V$ and $r$ are functions of time $t$, so differentiating
both sides of this equation with respect to $t$ gives
$${dV\over dt}={4\over 3}\cdot 3\pi r^2 {dr\over dt}=
   4\pi r^2 {dr\over dt}.$$
We want $dr/dt$ given that $r=10$ and
$dV/dt=400\; \mu m^3/{\rm day}$.  Thus
$${dr\over dt}\Big |_{r=10}={dV\over dt}{1\over 4\pi r^2}\Big|_{r=10}
    ={400\over 4\pi\cdot 10^2}=.3183 \;\mu m/{\rm day}.$$

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Comment:  Seeing that the chain rule can be used on
$V={4\over3}\pi r^3$ posed difficulties.
The relevance of the chain rule may be more evident if we write
the relation as $V(t)={4\over3}\pi (r(t))^3$.

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