A railroad bridge over New Scotland Road in Slingerlands, New York, has a low clearance for trucks. An engineer standing 20 ft away measures a 15.4 ° angle of elevation from her eye level of 5.5 ft to the bottom of the bridge. If the road is flat between the engineer and the bridge, how high over the roadway is the bottom of the bridge? Round to the nearest inch.
A railroad bridge over New Scotland Road in Slingerlands, New York, has a low clearance for trucks. An engineer standing 20 ft away measures a 15.4 ° angle of elevation from her eye level of 5.5 ft to the bottom of the bridge. If the road is flat between the engineer and the bridge, how high over the roadway is the bottom of the bridge? Round to the nearest inch.
Solution Summary: The author calculates the height of a roadway from the bottom of the bridge. An engineer measures an angle of elevation from her eye level, which is 5.5ft.
A railroad bridge over New Scotland Road in Slingerlands, New York, has a low clearance for trucks. An engineer standing
20
ft
away measures a
15.4
°
angle of elevation from her eye level of
5.5
ft
to the bottom of the bridge. If the road is flat between the engineer and the bridge, how high over the roadway is the bottom of the bridge? Round to the nearest inch.
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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