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MATH 180\hfill FALL 1995

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%10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#33}

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5.5, \#7(a).  The reason for the rise in the graph:  flying at higher
velocities requires correspondingly more energy.
The reason for the dip in the graph:
taking off from the ground requires more energy than staying airborn
and flying at low velocities.

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(b)  A dimension analysis shows that $a(v)=f(v)/v$, since the
units of $a(v)$, $f(v)$, $v$ are respectively
joules/meter, joules/second, meters/second

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(c)  $a(v)$ is related to the graph of $f(v)$ as follows:
Draw the line $\ell_v$ from the origin to the point $(v,f(v))$ on the
graph.  The slope of $\ell_v$ is $a(v)=f(v)/v$.  This has a local
extremum when $a'(v)=(vf'(v)-f(v))/v^2=0$, that is,
when $f'(v)=f(v)/v=a(v)$.  At that point $\ell_v$ is the tangent
line to the graph at $(v,f(v))$.

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(d)  The bird should minimize $a(v)$, since conserving energy with
respect to distance rather velocity is more important.

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5.5, \#11.
$f(x)=e^{-x^2}$ is an even function, that is, $f(-x)=f(x)$.  So the
best possible bounds on $f(x)$ on $[-0.3,0.3]$ are the same as the
best possible bounds on $[0,0.3]$.  As $x$ increases from 0 to
$0.3$, $x^2$ increases from 0 to $0.09$, while $f(x)$ decreases
from $e^0=1$ to $e^{-0.09}=.914$.   Thus $0\le  f(x)\le .914$.

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5.5, \#14.
$f(x)=x^3-4x^2+4x$, $f'(x)=3x^2-8x+4$, $f''(x)=6x-8$.
The roots of $f'(x)$ are $x=1/3$ and $x=2$.
Since $f''(1/3)<0$ and $f''(2)>0$, it follows by the
second-derivative test that $x=1/3$ is a local maximum, $x=2$ is a local
minimum.   The domain in the actual problem is restricted to
$0\le x\le 4$.  Both critical points are in this domain.  We calculate
$f$ at the critical points and at the end-points of the domain:
$$f(0)=0,\quad f(1/3)=.926,\quad f(2)=0,\quad f(4)=16.$$
So the smallest and the largest value of $f(x)$ on $[0,4]$
are 0 and 16, and thus $0\le f(x)\le 16$.


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