James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 2.7 Derivatives and Rates of Change 143
s
0
Q{a+h, f(a+h)}
P{a, f(a)}
h
a a+h t
f(a+h)-f(a)
m PQ =
h
average velocity
The average velocity over this time interval is
average velocity − displacement
time
−
f sa 1 hd 2 f sad
h
which is the same as the slope of the secant line PQ in Figure 6.
Now suppose we compute the average velocities over shorter and shorter time intervals
fa, a 1 hg. In other words, we let h approach 0. As in the example of the falling ball,
we define the velocity (or instantaneous velocity) vsad at time t − a to be the limit of
these average velocities:
FIGURE 6
3
vsad − lim
h l 0
f sa 1 hd 2 f sad
h
This means that the velocity at time t − a is equal to the slope of the tangent line at P
(compare Equations 2 and 3).
Now that we know how to compute limits, let’s reconsider the problem of the falling
ball.
Recall from Section 2.1: The dis tance
(in meters) fallen after t seconds is
4.9t 2 .
ExamplE 3 Suppose that a ball is dropped from the upper observation deck of the
CN Tower, 450 m above the ground.
(a) What is the velocity of the ball after 5 seconds?
(b) How fast is the ball traveling when it hits the ground?
SOLUtion We will need to find the velocity both when t − 5 and when the ball hits
the ground, so it’s efficient to start by finding the velocity at a general time t. Using the
equation of motion s − f std − 4.9t 2 , we have
vstd − lim
h l 0
f st 1 hd 2 f std
h
4.9st 1 hd 2 2 4.9t 2
− lim
h l 0 h
4.9st 2 1 2th 1 h 2 2 t 2 d 4.9s2th 1 h 2 d
− lim
− lim
h l 0
h
h l 0 h
− lim
hl0
4.9hs2t 1 hd
h
− lim 4.9s2t 1 hd − 9.8t
h l 0
(a) The velocity after 5 seconds is vs5d − s9.8ds5d − 49 mys.
(b) Since the observation deck is 450 m above the ground, the ball will hit the ground
at the time t when sstd − 450, that is,
This gives
t 2 − 450
4.9
4.9t 2 − 450
and t −Î 450
4.9 < 9.6 s
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