11DIFFERENTIATION - Department of Mathematics

700 11 DIFFERENTIATION
SOLUTION ✔
FIGURE 11.3
The rocket’s altitude t seconds into flight is
given by f(t).
Feet
150,000
100,000
50,000
s
s = f(t)
20 40 60 80 100
Seconds
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t
a. The rocket’s velocity at any time t is given by
v f(t) 3t2 192t 195
b. The rocket’s velocity when t 0, 30, 50, 65, and 70 is given by
f(0) 3(0) 2 192(0) 195 195
f(30) 3(30) 2 192(30) 195 3255
f(50) 3(50) 2 192(50) 195 2295
f(65) 3(65) 2 192(65) 195 0
f(70) 3(70) 2 192(70) 195 1065
or 195, 3255, 2295, 0, and 1065 feet per second (ft/sec).
Thus, the rocket has an initial velocity of 195 ft/sec at t 0 and accelerates
to a velocity of 3255 ft/sec at t 30. Fifty seconds into the flight, the rocket’s
velocity is 2295 ft/sec, which is less than the velocity at t 30. This means
that the rocket begins to decelerate after an initial period of acceleration.
(Later on we will learn how to determine the rocket’s maximum velocity.)
The deceleration continues: The velocity is 0 ft/sec at t 65 and 1065
ft/sec when t 70. This number tells us that 70 seconds into flight the rocket
is heading back to Earth with a speed of 1065 ft/sec.
c. The results of part (b) show that the rocket’s velocity is zero when t 65.
At this instant, the rocket’s maximum altitude is
s f(65) (65) 3 96(65) 2 195(65) 5
143,655 feet
Asketch of the graph of f appears in Figure 11.3.
Refer to Example 8.
1. Use a graphing utility to plot the graph of the velocity function
v f(t) 3t2 192t 195
using the viewing rectangle [0, 120] [5000, 5000]. Then, using ZOOM and TRACE or the root-finding
capability of your graphing utility, verify that f(65) 0.
2. Plot the graph of the position function of the rocket
s f(t) t3 96t 2 195t 5
using the viewing rectangle [0, 120] [0, 150,000]. Then, using ZOOM and TRACE repeatedly, verify that
the maximum altitude of the rocket is 143,655 feet.
3. Use ZOOM and TRACE or the root-finding capability of your graphing utility to find when the rocket
returns to Earth.