Week #2 - Limits, Continuity, and the Derivative
Section 2.1
From “Calculus, Single Variable” by Hughes-Hallett, Gleason, McCallum et. al.
Copyright 2005 by John Wiley & Sons, Inc.
This material is used by permission of John Wiley & Sons, Inc.
SUGGESTED PROBLEMS
1. The distance, s, a car has traveled on a trip is shown in the table as a function of the
time, t, since the trip started. Find the average velocity between t = 2 and t = 5
0
0
1
45
2
135
3
220
4
300
5
400
For t between t = 2 and t = 5, we have
Average Velocity =
400 − 135
265
∆s
=
=
km/hr
∆t
5−2
3
11. A car is driven at a constant speed. Sketch a graph of the distance the car has traveled
as a function of time.
Graph below:
distance (km)
Dist. vs Time for Constant Velocity
t (hours)
12. A car is driven at an increasing speed. Sketch a graph of the distance the car has traveled
as a function of time.
Graph below:
distance (km)
Dist. vs Time for Increasing Velocity
t (hours)
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13. A car starts at a high speed, and its speed then decreases slowly. Sketch a graph of the
distance the car has traveled as a function of time.
Graph below:
distance (km)
Dist. vs Time for Decreasing Velocity
t (hours)
QUIZ PREPARATION PROBLEMS
2. In a time of t seconds, a particle moves a distance of s meters from its starting point,
where s = 3t2 .
(a) Find the average velocity between
(i)
(ii)
(iii)
and
if:
,
,
.
(b) Use your answers to part (a) to estimate the instantaneous velocity of the particle
at time t = 1.
(a)
(i)
∆s
∆t
s(1 + h) − s(1)
=
(1 + h) − 1
s(1 + 0.1) − s(1)
=
(1 + .1) − 1
3(1.1)2 − 3(1)
=
0.1
.63
=
0.1
= 6.3
avg vel =
2
(ii)
∆s
∆t
3(1.01)2 − 3(1)
=
0.01
.603
=
0.01
= 6.03
avg vel =
(iii)
∆s
∆t
3(1.001)2 − 3(1)
=
0.001
.6003
=
0.001
= 6.003
avg vel =
(b) From the pattern in the previous three calculations, it seems that the instantaneous
velocity at t = 1 will be v(1) = 6.0 m/s.
15. The graph of f (t) in Figure 2.9 gives the position of a particle at time t. List the following
quantities in order, smallest to largest.
A, average velocity between t = 1 and t = 5.
B, average velocity between t = 5 and t = 6.
C, instantaneous velocity at t = 1.
D, instantaneous velocity at t = 3.
E, instantaneous velocity at t = 5.
F , instantaneous velocity at t = 6.
Figure 2.9
Note: we will consider all negative quantities to be “below” or “smaller than” any positive
quantities. The instantaneous slopes are all ’eyeballed’ based on the graph, by drawing
a tangent line and estimating the slope. You may arrive at slightly different numerical
values, though the final ranked order should still be the same.
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∆f
3.5 − 2
≃
= 0.3
∆t
5−1
3 − 3.5
∆f
≃
= −0.5
B=
∆t
6−5
C = slope at (t = 1) ≃ 1
A=
D = slope at (t = 3) ≃ 0.5
E = slope at (t = 5) ≃ −0.4hard to estimate
F = slope at (t = 6) ≃ −.6hard to estimate
In order, from largest negative to largest positive values, this would be
F, B, E, D, A, C
Note that the average velocity must be between the instantaneous velocities at the ends
of the same interval (e.g. D < A < C, and F < B < E) since the average slope will be
less extreme than the slopes at the either end of the interval on this graph.
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