MATH 180 FALL 1995 10:00 SECTION FONG

ASSIGNMENT #8

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 concaveup 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”.

2.1, #5. The average velocity over the interval [0, 0.2] is

\frac{s(0.2) − s(0)}{0.2 − 0} = \frac{0.5}{0.2} = 2.5ft/sec

The average velocity over the interval [0.2, 0.4] is

\frac{s(0.4) − s(0.2)}{0.4 − 0.2} = \frac{1.3}{0.2} = 6.5ft/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.

2.1, #12. The following values of \frac{7^{h} − 1}{h} for small h

\begin{matrix}h & −.1 & −.01 & −.001 & −.0001 & .0001 & .001 & .01 & .1 \\ Value & 1.768 & 1.927 & 1.944 & 1.946 & 1.946 & 1.948 & 1.965 & 2.148 \end{matrix}

suggest that the limit as h → 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.