Write the intervals (within the practical domain) when the first derivative is increasing and decreasing. What does this mean in terms of daylight at various times of the year?
Write the intervals (within the practical domain) when the first derivative is increasing and decreasing. What does this mean in terms of daylight at various times of the year?
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.3: Trigonometric Functions Of Real Numbers
Problem 65E
Related questions
Question
Question 6
![9/12
255
5:35
6:13
1263
758
739
The following table gives the sunrise and sunset times for various dates of the year in Lincroft, NJ.
In order to create a function from this, we need to quantify the date into "day of year",
with January 1, which we will call "Day 1" (we can assume 1 corresponds to 11:59 PM on January
1, so we'll think of the New Year beginning at x- 0). This becomes our independent variable. We
also turn the amount of time between sunrise and sunset into "hours of daylight", which becomes
our dependent variable.
12.32
12.02
11.7
11.4
11.08
9/19
9/26
103
262
269
276
5:42
5:48
5:55
6:02
6:10
6:17
6:01
5:49
5:37
5:26
5:15
5:05
, beginning
721
702
684
283
10'10
10/17
1024
290
297
665
648
108
304
311
318
631
614
10/31
11/7
11/14
11/22
11/29
126
6:25
6:34
6:42
4:56
Day of
Year
10.52
10.23
10
Date
Sunrise
Sunset
Minutes of
Hours of
4:48
Daylight
563
569
580
Daylight
4:42
600
1/3
1/10
/17
7:20
7:20
7:17
4:43
4:49
4:57
9.38
9.48
326
333
340
6:50
6:58
7:05
4:34
588
573
565
560
558
558
9.8
10
17
4:31
4:30
9.55
9.42
9.67
1/24
1/31
2/7
2/14
2/21
24
31
38
45
52
7:13
7:08
7:01
6:52
6:43
6:33
5:05
5:14
5:22
531
5:39
5:47
592
606
621
639
656
674
9.87
10.1
1035
10.65
10.93
11.23
11.55
11.85
12.17
12.48
12.78
13.1
1338
13.68
13.95
14.2
14.43
12/13
12/20
12/27
347
354
361
7:12
7:16
7:19
4:32
4:34
4:37
9.33
9.3
9.3
1. The sinusoidal function that can be used to model this data is
2/28
59
3/7
3/14
3/21
3/28
4/4
f(x) = 2.836sin (0.0172(x-80)) +12.164 , where x is the day of the year, and f(x) gives the
number of hours of daylight on day x.
a. What are the amplitude, period, and midline of the function? Round to the nearest
thousandth.
6:22
5:55
6:02
66
693
711
730
749
767
786
73
6:11
6:00
5:48
5:37
5:25
80
6:10
6:17
6.24
6:31
87
Amplitude: 2.836
94
Period
T 365.301
4/11
0.0172
Midline: y-12.164
101
4/18
4/25
5/2
5/9
5/16
5/23
5/30
6/6
108
115
122
129
136
5:15
5:05
4:56
4:48
4:41
6:38
6:46
6.53
7:00
7:07
803
821
837
852
b. Do these values make sense based on the data? Why or why not?
The values make sense because the given data is numerie.
2. Since this is a sinusoidal function, the domain is all real numbers. However, this function has
a practical application. What is the practical domain of this function? Why?
X= Number of days of the year
OsxS 365
Domain: [0, 365]|
866
143
150
157
4:35
4:31
4:28
7:13
7:19
7:24
878
14.63
14.8
1493
888
896
901
903
4:27
427
7:28
7:30
15.02
15.05
15.03
6/13
164
6/20
171
From the domain we can calculate the number of hours of sunlight
Example: at x-118
f(118)- demonstratethe light on 118th day of the year.
3. Evaluate the model for July 4 and October 31. How close is the model to the actual data?
6/27
7/4
7/11
7/18
7/25
8/1
178
185
192
199
206
213
4:29
4:32
4:37
4:42
7:31
7:31
7:28
7:25
902
899
14.98
14.85
14.72
14.52
14.3
891
883
Round to the nearest hundredth.
July 4
f(x) = 2.836 sin(0.0172(x-80)) + 12.164
S(185) = 2.836 sin(0.0172(185-80) + 12.164
= 2.836 sin(1.806) + 12.164
= 2.75791612 + 12.164
7:19
7:13
871
858
4:48
4:55
220
5:01
7:05
14.07
8/8
8/15
8/22
8/29
9/5
844
227
234
241
5:08
5:15
5.22
5:28
6:56
6:46
635
828
811
793
13.8
13.52
13.22
12 93
14.92
248
6:24
776](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F83c8383a-4f6c-43c1-8de7-aa7faa557af8%2F86336301-aff4-4431-83e0-84641c56127b%2Fyk4cvcd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:9/12
255
5:35
6:13
1263
758
739
The following table gives the sunrise and sunset times for various dates of the year in Lincroft, NJ.
In order to create a function from this, we need to quantify the date into "day of year",
with January 1, which we will call "Day 1" (we can assume 1 corresponds to 11:59 PM on January
1, so we'll think of the New Year beginning at x- 0). This becomes our independent variable. We
also turn the amount of time between sunrise and sunset into "hours of daylight", which becomes
our dependent variable.
12.32
12.02
11.7
11.4
11.08
9/19
9/26
103
262
269
276
5:42
5:48
5:55
6:02
6:10
6:17
6:01
5:49
5:37
5:26
5:15
5:05
, beginning
721
702
684
283
10'10
10/17
1024
290
297
665
648
108
304
311
318
631
614
10/31
11/7
11/14
11/22
11/29
126
6:25
6:34
6:42
4:56
Day of
Year
10.52
10.23
10
Date
Sunrise
Sunset
Minutes of
Hours of
4:48
Daylight
563
569
580
Daylight
4:42
600
1/3
1/10
/17
7:20
7:20
7:17
4:43
4:49
4:57
9.38
9.48
326
333
340
6:50
6:58
7:05
4:34
588
573
565
560
558
558
9.8
10
17
4:31
4:30
9.55
9.42
9.67
1/24
1/31
2/7
2/14
2/21
24
31
38
45
52
7:13
7:08
7:01
6:52
6:43
6:33
5:05
5:14
5:22
531
5:39
5:47
592
606
621
639
656
674
9.87
10.1
1035
10.65
10.93
11.23
11.55
11.85
12.17
12.48
12.78
13.1
1338
13.68
13.95
14.2
14.43
12/13
12/20
12/27
347
354
361
7:12
7:16
7:19
4:32
4:34
4:37
9.33
9.3
9.3
1. The sinusoidal function that can be used to model this data is
2/28
59
3/7
3/14
3/21
3/28
4/4
f(x) = 2.836sin (0.0172(x-80)) +12.164 , where x is the day of the year, and f(x) gives the
number of hours of daylight on day x.
a. What are the amplitude, period, and midline of the function? Round to the nearest
thousandth.
6:22
5:55
6:02
66
693
711
730
749
767
786
73
6:11
6:00
5:48
5:37
5:25
80
6:10
6:17
6.24
6:31
87
Amplitude: 2.836
94
Period
T 365.301
4/11
0.0172
Midline: y-12.164
101
4/18
4/25
5/2
5/9
5/16
5/23
5/30
6/6
108
115
122
129
136
5:15
5:05
4:56
4:48
4:41
6:38
6:46
6.53
7:00
7:07
803
821
837
852
b. Do these values make sense based on the data? Why or why not?
The values make sense because the given data is numerie.
2. Since this is a sinusoidal function, the domain is all real numbers. However, this function has
a practical application. What is the practical domain of this function? Why?
X= Number of days of the year
OsxS 365
Domain: [0, 365]|
866
143
150
157
4:35
4:31
4:28
7:13
7:19
7:24
878
14.63
14.8
1493
888
896
901
903
4:27
427
7:28
7:30
15.02
15.05
15.03
6/13
164
6/20
171
From the domain we can calculate the number of hours of sunlight
Example: at x-118
f(118)- demonstratethe light on 118th day of the year.
3. Evaluate the model for July 4 and October 31. How close is the model to the actual data?
6/27
7/4
7/11
7/18
7/25
8/1
178
185
192
199
206
213
4:29
4:32
4:37
4:42
7:31
7:31
7:28
7:25
902
899
14.98
14.85
14.72
14.52
14.3
891
883
Round to the nearest hundredth.
July 4
f(x) = 2.836 sin(0.0172(x-80)) + 12.164
S(185) = 2.836 sin(0.0172(185-80) + 12.164
= 2.836 sin(1.806) + 12.164
= 2.75791612 + 12.164
7:19
7:13
871
858
4:48
4:55
220
5:01
7:05
14.07
8/8
8/15
8/22
8/29
9/5
844
227
234
241
5:08
5:15
5.22
5:28
6:56
6:46
635
828
811
793
13.8
13.52
13.22
12 93
14.92
248
6:24
776
![b. Use the first or second derivative tests to determine the maximum and minimum values
of the function, and on which dates they occur. Show and label all your written work.
Round to the ncarest hundredth.
fix)
October 31
f) = 2.836 sin(0.0172(x- 80)) + 12.164
f(304) = 2.836 sin(0.0172(304 – 80)) + 12.164
= 2.836 sin(3.8525) + 12.164
--1.8511958 + 12.164
= 10.31
2.836 sin(0.0172(x – 80) + 12.164
0.0487792 cos(0.0172(x – 80))
171.33, 353.98, 536 63
y = 2.836 sin(0.0172(171.33 – 80)) + 12.164
Given the results from the functions of July 4 and October 31", the results are very
close to the actual date.
y =15
Maximum (171.33, 15)
y = 2.836 sin(0.0172(353.98 – 80)) + 12.164
4. Find the first and second derivatives of the model without using technology. Show and label
all your work.
y =9.33
Minimum- (353.98, 9.33)
First Derivative:
f(x) = 2.836 sin(0.0172(x- 80)) + 12.164
d
= [2.836 sin(0.0172(x – 80))] +–12.164
e. Do the dates make sense based on what you know about the seasons?
Yes, it makes sense because the 172 day corresponds to June 21" because that is when
we have the most amount of sunlight since it is summertime, making it the maximum.
Meanwhile, on the 353 day corresponding to December 20 making it winter therefore
dx
dx
= 2.836 - 0.0172
f'(x) = 0.0487792 cos (0.0172(x – 80))
having minimum amount of sunlight.
6.
a Write the intervals (within the practical domain) when the first derivative is increasing
and decreasing.
b. What does this mean in terms of daylight at various times of the year?
Second Derivative
f'(x) = 0.0487792 cos(0.0172(x - 80))
*= (0.0487792 cos(0.0172(x- 80))]
dx
7.
- 0.0487792(- sin(0.0172(x – 80)) • 0.0172)
a For which value of x is the amount of daylight increasing the fastest? For which x-value
is it decreasing the fastest? Then find the dates that correspond to these x-values.
b. Determine the intervals where the function is concave up and concave down.
a.
Find the critical values of the function. Show and label your work. Round to the nearest
hundredth.
Given,
S(x) = 2.836 sin(0.0172(x - 80)) + 12.164
'(x) = 0.0487792 cos (0.0172(x – 80))
For critical values f'x) = 0
- 0.0487792 cos(0.0172(x - 80)) = 0
- cos (0.0172(x - 80)) = 0
0.0172(x - 80) = (2n + 1)
8.
a Use technology to obtain a printout of the function over its practical domain. Scale and
label the axes appropriately.
b. Label the maximum, minimum, and inflection points on the graph. You should label the
points both in the form (x, Ax)) and (date, f(x)).
9. Suppose the same analysis was done for Barrow, Alaska and Lima, Peru. How would you
expect the models to be different or the same regarding amplitude, maximum and minimum
values, and where these extrema occur? Hint: You should look up these locations on a map to
determine their latitudes and position in the Northern or Southern Hemisphere.
(2n + 1)m
- (x- 80) =
2(0.0172)
(2n + 1)m
0.0344
5.8936 + 6.2832n
x= 80 +
10. We can use the function to estimate the total amount of daylight for an entire year. Enter the
function into Maple and use Riemann sums (using midpoints for the height of the rectangles)
to estimate the total amount of daylight when:
0.0344
= 182.65n + 171.33
Where n is an integer, we plugged in 0, 1, 2. to get the critical points which are:
171.33, 353.98, 536.63, and so on.
a n- 50
b. n- 100
c. n= 365
Note: Write your final answer accurate to 6 decimal places.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F83c8383a-4f6c-43c1-8de7-aa7faa557af8%2F86336301-aff4-4431-83e0-84641c56127b%2Fjomat76_processed.jpeg&w=3840&q=75)
Transcribed Image Text:b. Use the first or second derivative tests to determine the maximum and minimum values
of the function, and on which dates they occur. Show and label all your written work.
Round to the ncarest hundredth.
fix)
October 31
f) = 2.836 sin(0.0172(x- 80)) + 12.164
f(304) = 2.836 sin(0.0172(304 – 80)) + 12.164
= 2.836 sin(3.8525) + 12.164
--1.8511958 + 12.164
= 10.31
2.836 sin(0.0172(x – 80) + 12.164
0.0487792 cos(0.0172(x – 80))
171.33, 353.98, 536 63
y = 2.836 sin(0.0172(171.33 – 80)) + 12.164
Given the results from the functions of July 4 and October 31", the results are very
close to the actual date.
y =15
Maximum (171.33, 15)
y = 2.836 sin(0.0172(353.98 – 80)) + 12.164
4. Find the first and second derivatives of the model without using technology. Show and label
all your work.
y =9.33
Minimum- (353.98, 9.33)
First Derivative:
f(x) = 2.836 sin(0.0172(x- 80)) + 12.164
d
= [2.836 sin(0.0172(x – 80))] +–12.164
e. Do the dates make sense based on what you know about the seasons?
Yes, it makes sense because the 172 day corresponds to June 21" because that is when
we have the most amount of sunlight since it is summertime, making it the maximum.
Meanwhile, on the 353 day corresponding to December 20 making it winter therefore
dx
dx
= 2.836 - 0.0172
f'(x) = 0.0487792 cos (0.0172(x – 80))
having minimum amount of sunlight.
6.
a Write the intervals (within the practical domain) when the first derivative is increasing
and decreasing.
b. What does this mean in terms of daylight at various times of the year?
Second Derivative
f'(x) = 0.0487792 cos(0.0172(x - 80))
*= (0.0487792 cos(0.0172(x- 80))]
dx
7.
- 0.0487792(- sin(0.0172(x – 80)) • 0.0172)
a For which value of x is the amount of daylight increasing the fastest? For which x-value
is it decreasing the fastest? Then find the dates that correspond to these x-values.
b. Determine the intervals where the function is concave up and concave down.
a.
Find the critical values of the function. Show and label your work. Round to the nearest
hundredth.
Given,
S(x) = 2.836 sin(0.0172(x - 80)) + 12.164
'(x) = 0.0487792 cos (0.0172(x – 80))
For critical values f'x) = 0
- 0.0487792 cos(0.0172(x - 80)) = 0
- cos (0.0172(x - 80)) = 0
0.0172(x - 80) = (2n + 1)
8.
a Use technology to obtain a printout of the function over its practical domain. Scale and
label the axes appropriately.
b. Label the maximum, minimum, and inflection points on the graph. You should label the
points both in the form (x, Ax)) and (date, f(x)).
9. Suppose the same analysis was done for Barrow, Alaska and Lima, Peru. How would you
expect the models to be different or the same regarding amplitude, maximum and minimum
values, and where these extrema occur? Hint: You should look up these locations on a map to
determine their latitudes and position in the Northern or Southern Hemisphere.
(2n + 1)m
- (x- 80) =
2(0.0172)
(2n + 1)m
0.0344
5.8936 + 6.2832n
x= 80 +
10. We can use the function to estimate the total amount of daylight for an entire year. Enter the
function into Maple and use Riemann sums (using midpoints for the height of the rectangles)
to estimate the total amount of daylight when:
0.0344
= 182.65n + 171.33
Where n is an integer, we plugged in 0, 1, 2. to get the critical points which are:
171.33, 353.98, 536.63, and so on.
a n- 50
b. n- 100
c. n= 365
Note: Write your final answer accurate to 6 decimal places.
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