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MATH 180\hfill FALL 1995

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10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#8}

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2.1, \#1, 2, 3..
For \#1 the graph is a line segment of
positive slope starting at the origin, a line since the
``car is driven at a constant speed''.
For \#2 the graph is an increasing concave-up curve starting
at the origin, concave-up since the
``car is driven at an increasing speed''.
For \#3 the graph is an increasing concave-down curve starting
at the origin, concave-down since the
``its speed then decreases slowly''.

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2.1, \#5.
The average velocity over the interval $[0,0.2]$ is
$${s(0.2)-s(0)\over 0.2-0}={0.5\over0.2}=2.5 \hbox{ft/sec}$$
The average velocity over the interval $[0.2,0.4]$ is
$${s(0.4)-s(0.2)\over 0.4-0.2}={1.3\over0.2}=6.5 \hbox{ft/sec}$$
Since there are two intervals of time with $t=0.2$ as
an endpoint,
we take the average of these two  average velocities, namely
$4.5$ ft/sec.,
as an estimate for the velocity at $t=0.2$.

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2.1, \#12. The following values of
$\displaystyle{7^h-1\over h}$ for small $h$
$$\vbox{
\halign{\quad#\hfil & \quad#\hfil & \quad#\hfil & \quad#\hfil &
   \quad#\hfil & \quad#\hfil & \quad#\hfil & \quad#\hfil &
   \quad#\hfil\cr
  $h$ & $-.1$ & $-.01$ & $-.001$ &
    $-.0001$ & .0001 & .001 & .01 & .1 \cr
  Value & 1.768 & 1.927 & 1.944 & 1.946 & 1.946 & 1.948
    & 1.965 & 2.148\cr}}$$
suggest that the limit as $h\rightarrow 0$ is $1.9$ to one decimal
place.  $2.0$ is not the correct answer, since the data indicates
the limit is less than 1.95.  A number of students gave this as the
answer because of careless rounding-offs.


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