The temperature T in ° F for Kansas City, Missouri, over a several day period in April can be approximated by T t = − 5.9 cos 0.262 t − 1.245 + 48.2 , where t is the number of hours since midnight on day 1 . a. What is the period of the function? Round to the nearest hour. b. What is the significance of the term 48.2 in this model? c. What is the significance of the factor 5.9 in this model? d. What was the minimum temperature for the day? When did it occur? e. What was the maximum temperature for the day? When did it occur?
The temperature T in ° F for Kansas City, Missouri, over a several day period in April can be approximated by T t = − 5.9 cos 0.262 t − 1.245 + 48.2 , where t is the number of hours since midnight on day 1 . a. What is the period of the function? Round to the nearest hour. b. What is the significance of the term 48.2 in this model? c. What is the significance of the factor 5.9 in this model? d. What was the minimum temperature for the day? When did it occur? e. What was the maximum temperature for the day? When did it occur?
The temperature
T
in
°
F
for Kansas City, Missouri, over a several day period in April can be approximated by
T
t
=
−
5.9
cos
0.262
t
−
1.245
+
48.2
, where
t
is the number of hours since midnight on day
1
.
a. What is the period of the function? Round to the nearest hour.
b. What is the significance of the term
48.2
in this model?
c. What is the significance of the factor
5.9
in this model?
d. What was the minimum temperature for the day? When did it occur?
e. What was the maximum temperature for the day? When did it occur?
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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