A communications satellite is in geosynchronous orbit. This means that its orbit coincides with the rotation period of the Earth so that it remains above a fixed point on the surface of the Earth at all times. The altitude of the satellite is 35 , 786 km . The sender and receiver of a signal must both be within the line of sight of the satellite. What is the maximum distance along the surface of the Earth for which the sender and receiver can communicate? Take the radius of the Earth to be 6357 km and round to the nearest kilometer.
A communications satellite is in geosynchronous orbit. This means that its orbit coincides with the rotation period of the Earth so that it remains above a fixed point on the surface of the Earth at all times. The altitude of the satellite is 35 , 786 km . The sender and receiver of a signal must both be within the line of sight of the satellite. What is the maximum distance along the surface of the Earth for which the sender and receiver can communicate? Take the radius of the Earth to be 6357 km and round to the nearest kilometer.
Solution Summary: The author calculates the maximum distance along the surface of the Earth for which the sender and receiver of both the signals can communicate.
A communications satellite is in geosynchronous orbit. This means that its orbit coincides with the rotation period of the Earth so that it remains above a fixed point on the surface of the Earth at all times. The altitude of the satellite is
35
,
786
km
. The sender and receiver of a signal must both be within the line of sight of the satellite. What is the maximum distance along the surface of the Earth for which the sender and receiver can communicate? Take the radius of the Earth to be
6357
km
and round to the nearest kilometer.
Given lim x-4 f (x) = 1,limx-49 (x) = 10, and lim→-4 h (x) = -7 use the limit properties
to find lim→-4
1
[2h (x) — h(x) + 7 f(x)] :
-
h(x)+7f(x)
3
O DNE
17. Suppose we know that the graph below is the graph of a solution to dy/dt = f(t).
(a) How much of the slope field can
you sketch from this information?
[Hint: Note that the differential
equation depends only on t.]
(b) What can you say about the solu-
tion with y(0) = 2? (For example,
can you sketch the graph of this so-
lution?)
y(0) = 1
y
AN
(b) Find the (instantaneous) rate of change of y at x = 5.
In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of
change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the
following limit.
lim
h→0
-
f(x + h) − f(x)
h
The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule
defining f.
f(x + h) = (x + h)² - 5(x+ h)
=
2xh+h2_
x² + 2xh + h² 5✔
-
5
)x - 5h
Step 4
-
The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x).
-
f(x + h) f(x) =
= (x²
x² + 2xh + h² -
])-
=
2x
+ h² - 5h
])x-5h) - (x² - 5x)
=
]) (2x + h - 5)
Macbook Pro
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