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", beginning 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. 9/12 9/19 9/26 103 255 262 269 276 283 290 297 12.63 1232 1202 11.7 5:35 6:13 5:42 5:48 5:55 6:02 6:10 758 739 721 702 684 665 6.01 5.49 5:37 5:26 3:15 5:05 10'10 1017 1024 1031 117 I/14 114 11.08 10.8 10 52 10.23 10 6:17 648 304 311 318 625 6.34 6.42 4.56 631 614 600 Sunrise Hours of Day of Year Date Sunset Minutes of 4:48 Daylight 563 569 Daylight 938 4:42 720 7:20 7:17 1/3 3. 4:43 11/22 11/29 126 12/13 1220 326 6.50 4:34 588 573 565 98 9.55 9.42 17/10 1/17 1/24 131 27 2/14 2/21 2/28 10 4:49 9.48 333 6:58 4:31 17 4:57 580 24 31 38 45 7:13 7:08 701 652 6:43 5:05 5:14 5.22 531 5:39 9.67 9.87 10.1 340 347 354 361 7:05 7:12 7:16 430 4:32 434 592 560 558 9.33 93 606 621 639 656 10.35 10 65 10.93 1227 7:19 4:37 558 93 1. The sinusoidal function that can be used to model this data is f()-2836sin (00172( x -80)) +12.164, where x is the day of the year, and ftx) gives the mumber of hours of daylight on day x. . What are the amplitude, period, and midline of the function? Round to the nearest thousandth. 52 11 23 11.55 11.85 12.17 12.48 12.78 13.1 13.38 13.68 59 6:33 5:47 674 622 6:11 3/7 3/14 3/21 3/28 5:55 693 711 66 73 80 87 6.02 6:10 6:17 730 749 6.00 5.48 537 525 Amplitude 2836 Period =7= 365.301 Midline y-12 164 b. Do these values make sense based on the data? Why or why not? The valacs make sense because the given data is numeric 44 94 6.24 767 4/11 101 631 4/18 4/25 3/2 519 5/16 5/23 3/30 786 803 821 837 852 866 878 888 896 108 115 5:15 5:05 4.56 4:48 6:38 646 122 129 6.53 7.00 7:07 13.95 14.2 136 143 150 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 4.41 14.43 4:35 4:31 7:13 7:19 14.63 14.8 7:24 7:28 7:30 7:31 7:31 66 59E 5 50 157 428 427 427 429 4:32 14.93 15.02 15.05 15.03 14.98 Domain: (0, 365| 6/13 164 171 178 185 901 903 902 From the domain we can calculate the number of hours of sunlight| Example at x-118 F(118) - demonstratethe light an 118th day of the year. 3. Evaluate the model for July 4 and October 31. How close is the model to the actual data? 6/20 6/27 714 899 7/11 7/18 7/25 81 192 199 437 442 7.28 7.25 7:19 7:13 14.85 14.72 14.52 143 14.07 891 Round to the nearest hundredth. 883 871 July 4 4) = 2.836 sin(0.0172(x - 80)) + 12.164 r(185) = 2.836 sin(0.0172(185-80)) + 12164 - 2.836 sin(1806) + 12.164 =2.75791612 + 12.164 206 4:48 213 4.55 858 8/8 220 5.01 5.08 7:05 656 844 8/15 8/22 829 227 234 241 248 5:15 5.22 5.28 828 811 793 13.8 13.52 13.22 12.93 6:46 6.35 14.92 95 624 776
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", beginning 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. 9/12 9/19 9/26 103 255 262 269 276 283 290 297 12.63 1232 1202 11.7 5:35 6:13 5:42 5:48 5:55 6:02 6:10 758 739 721 702 684 665 6.01 5.49 5:37 5:26 3:15 5:05 10'10 1017 1024 1031 117 I/14 114 11.08 10.8 10 52 10.23 10 6:17 648 304 311 318 625 6.34 6.42 4.56 631 614 600 Sunrise Hours of Day of Year Date Sunset Minutes of 4:48 Daylight 563 569 Daylight 938 4:42 720 7:20 7:17 1/3 3. 4:43 11/22 11/29 126 12/13 1220 326 6.50 4:34 588 573 565 98 9.55 9.42 17/10 1/17 1/24 131 27 2/14 2/21 2/28 10 4:49 9.48 333 6:58 4:31 17 4:57 580 24 31 38 45 7:13 7:08 701 652 6:43 5:05 5:14 5.22 531 5:39 9.67 9.87 10.1 340 347 354 361 7:05 7:12 7:16 430 4:32 434 592 560 558 9.33 93 606 621 639 656 10.35 10 65 10.93 1227 7:19 4:37 558 93 1. The sinusoidal function that can be used to model this data is f()-2836sin (00172( x -80)) +12.164, where x is the day of the year, and ftx) gives the mumber of hours of daylight on day x. . What are the amplitude, period, and midline of the function? Round to the nearest thousandth. 52 11 23 11.55 11.85 12.17 12.48 12.78 13.1 13.38 13.68 59 6:33 5:47 674 622 6:11 3/7 3/14 3/21 3/28 5:55 693 711 66 73 80 87 6.02 6:10 6:17 730 749 6.00 5.48 537 525 Amplitude 2836 Period =7= 365.301 Midline y-12 164 b. Do these values make sense based on the data? Why or why not? The valacs make sense because the given data is numeric 44 94 6.24 767 4/11 101 631 4/18 4/25 3/2 519 5/16 5/23 3/30 786 803 821 837 852 866 878 888 896 108 115 5:15 5:05 4.56 4:48 6:38 646 122 129 6.53 7.00 7:07 13.95 14.2 136 143 150 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 4.41 14.43 4:35 4:31 7:13 7:19 14.63 14.8 7:24 7:28 7:30 7:31 7:31 66 59E 5 50 157 428 427 427 429 4:32 14.93 15.02 15.05 15.03 14.98 Domain: (0, 365| 6/13 164 171 178 185 901 903 902 From the domain we can calculate the number of hours of sunlight| Example at x-118 F(118) - demonstratethe light an 118th day of the year. 3. Evaluate the model for July 4 and October 31. How close is the model to the actual data? 6/20 6/27 714 899 7/11 7/18 7/25 81 192 199 437 442 7.28 7.25 7:19 7:13 14.85 14.72 14.52 143 14.07 891 Round to the nearest hundredth. 883 871 July 4 4) = 2.836 sin(0.0172(x - 80)) + 12.164 r(185) = 2.836 sin(0.0172(185-80)) + 12164 - 2.836 sin(1806) + 12.164 =2.75791612 + 12.164 206 4:48 213 4.55 858 8/8 220 5.01 5.08 7:05 656 844 8/15 8/22 829 227 234 241 248 5:15 5.22 5.28 828 811 793 13.8 13.52 13.22 12.93 6:46 6.35 14.92 95 624 776
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
6a
6b
7a
7b
8a
8b
![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", beginning
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.
9/12
255
5:35
6:13
758
12.63
9/19
262
5:42
5:48
5:55
6:01
5:49
739
721
12.32
12.02
9/26
269
10/3
276
5:37
702
11.7
10/10
10/17
10/24
283
6:02
5:26
684
11.4
290
6:10
5:15
665
11.08
10.8
297
304
5:05
4:56
6:17
648
6:25
631
10.52
10.23
10/31
Date
Day of
Year
Sunrise
Sunset
Minutes of
Hours of
11/7
311
6:34
4:48
614
Daylight
9.38
9.48
Daylight
11/14
318
6:42
6:50
4:42
600
10
1/3
1/10
3
10
7:20
7:20
4:43
4:49
563
569
326
333
4:34
4:31
11/22
9.8
9.55
588
11/29
6:58
7:05
7:12
573
1/17
17
7:17
4:57
580
9.67
126
340
347
354
4:30
4:32
4:34
565
1/24
1/31
24
31
7:13
7:08
5:05
5:14
592
606
9.42
9.33
9.3
9.87
12/13
560
10.1
12/20
7:16
558
2/7
2/14
38
45
7:01
6:52
5:22
5:31
621
639
10.35
10.65
12/27
361
7:19
4:37
558
93
10.93
11.23
1. The sinusoidal function that can be used to model this data is
S(4) = 2.836sin (0.0172(x- 80)) +12.164 , where x is the day of the year, and fix) 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
2/21
52
6:43
5:39
656
2/28
59
6:33
5:47
674
3/7
3/14
66
73
6:22
6:11
5:55
6:02
693
711
11.55
11.85
3/21
3/28
12.17
12.48
80
6:00
5:48
6:10
6:17
730
thousandth.
87
749
Amplitude: 2.836
T = 365.301
Midline: y-12.164
b. Do these values make sense based on the data? Why or why not?
The values make sense because the given data is numeric.
4/4
94
5:37
6:24
767
12.78
Period:
4/11
101
5:25
6:31
786
13.1
0.0172
13.38
13.68
4/18
108
4/25
5/2
5:15
5:05
6:38
6:46
6:53
803
821
115
122
4:56
837
852
13.95
5/9
129
4:48
7:00
14.2
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]|
5/16
5/23
136
4:41
7:07
866
14.43
143
4:35
7:13
878
14.63
14.8
5/30
150
4:31
7:19
888
6/6
157
4:28
7:24
896
14.93
4:27
7:28
901
15.02
15.05
15.03
14.98
6/13
164
6/20
171
4:27
7:30
903
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
178
4:29
7:31
7:31
902
7/4
185
4:32
899
7/11
192
4:37
7:28
891
14.85
7/18
199
4:42
7:25
883
Round to the nearest hundredth.
14.72
14.52
July 4
f(e) = 2.836 sin(0.0172(x - 80)) + 12.164
r(185) = 2.836 sin(0.0172(185 – 80)) + 12.164
= 2.836 sin(1.806) + 12.164
7/25
206
213
4:48
7:19
7:13
871
8/1
4:55
858
14.3
8/8
220
5:01
7:05
844
14.07
8/15
227
5:08
6:56
828
811
13.8
13.52
234
= 2.75791612 + 12.164
8/22
8/29
5:15
5:22
6:46
6:35
241
793
13.22
* 14.92
9/5
248
5:28
6:24
776
12.93](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F83c8383a-4f6c-43c1-8de7-aa7faa557af8%2Fcf8c30f6-6704-46d9-8f61-9fe671e1560a%2Fouh5pdm_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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", beginning
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.
9/12
255
5:35
6:13
758
12.63
9/19
262
5:42
5:48
5:55
6:01
5:49
739
721
12.32
12.02
9/26
269
10/3
276
5:37
702
11.7
10/10
10/17
10/24
283
6:02
5:26
684
11.4
290
6:10
5:15
665
11.08
10.8
297
304
5:05
4:56
6:17
648
6:25
631
10.52
10.23
10/31
Date
Day of
Year
Sunrise
Sunset
Minutes of
Hours of
11/7
311
6:34
4:48
614
Daylight
9.38
9.48
Daylight
11/14
318
6:42
6:50
4:42
600
10
1/3
1/10
3
10
7:20
7:20
4:43
4:49
563
569
326
333
4:34
4:31
11/22
9.8
9.55
588
11/29
6:58
7:05
7:12
573
1/17
17
7:17
4:57
580
9.67
126
340
347
354
4:30
4:32
4:34
565
1/24
1/31
24
31
7:13
7:08
5:05
5:14
592
606
9.42
9.33
9.3
9.87
12/13
560
10.1
12/20
7:16
558
2/7
2/14
38
45
7:01
6:52
5:22
5:31
621
639
10.35
10.65
12/27
361
7:19
4:37
558
93
10.93
11.23
1. The sinusoidal function that can be used to model this data is
S(4) = 2.836sin (0.0172(x- 80)) +12.164 , where x is the day of the year, and fix) 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
2/21
52
6:43
5:39
656
2/28
59
6:33
5:47
674
3/7
3/14
66
73
6:22
6:11
5:55
6:02
693
711
11.55
11.85
3/21
3/28
12.17
12.48
80
6:00
5:48
6:10
6:17
730
thousandth.
87
749
Amplitude: 2.836
T = 365.301
Midline: y-12.164
b. Do these values make sense based on the data? Why or why not?
The values make sense because the given data is numeric.
4/4
94
5:37
6:24
767
12.78
Period:
4/11
101
5:25
6:31
786
13.1
0.0172
13.38
13.68
4/18
108
4/25
5/2
5:15
5:05
6:38
6:46
6:53
803
821
115
122
4:56
837
852
13.95
5/9
129
4:48
7:00
14.2
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]|
5/16
5/23
136
4:41
7:07
866
14.43
143
4:35
7:13
878
14.63
14.8
5/30
150
4:31
7:19
888
6/6
157
4:28
7:24
896
14.93
4:27
7:28
901
15.02
15.05
15.03
14.98
6/13
164
6/20
171
4:27
7:30
903
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
178
4:29
7:31
7:31
902
7/4
185
4:32
899
7/11
192
4:37
7:28
891
14.85
7/18
199
4:42
7:25
883
Round to the nearest hundredth.
14.72
14.52
July 4
f(e) = 2.836 sin(0.0172(x - 80)) + 12.164
r(185) = 2.836 sin(0.0172(185 – 80)) + 12.164
= 2.836 sin(1.806) + 12.164
7/25
206
213
4:48
7:19
7:13
871
8/1
4:55
858
14.3
8/8
220
5:01
7:05
844
14.07
8/15
227
5:08
6:56
828
811
13.8
13.52
234
= 2.75791612 + 12.164
8/22
8/29
5:15
5:22
6:46
6:35
241
793
13.22
* 14.92
9/5
248
5:28
6:24
776
12.93
![b. Use the first or second derivative tests to determine the maximum and minimum values
October 31
f(x) = 2.836 sin(0.0172(x - 80)) + 12.164
f (304) = 2.836 sin(0.0172(304 – 80)) + 12.164
of the function, and on which dates they occur. Show and label all your written work.
Round to the nearest hundredth.
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
f(x)
= 2.836 sin(3.8525) + 12.164
--1.8511958 + 12.164
= 10.31
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)
4. Find the first and second derivatives of the model without using technology. Show and label
all your work.
y = 2.836 sin(0.0172(353.98 – 80)) + 12.164
y =9.33
Minimum- (353.98, 9.33)
First Derivative:
c. Do the dates make sense based on what you know about the seasons?
Yes, it makes sense because the 172d 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
f(x) = 2.836 sin(0.0172(x – 80)) + 12.164
d
d
= (2.836 sin(0.0172(x – 80))] +12.164
= 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
'(x) = 0.0487792 cos(0.0172(x – 80))
= (0.0487792 cos(0.0172(x – 80))|
0.0487792(- sin(0.0172(x – 80)) • 0.0172)
7.
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.
5.
a. Find the critical values of the function. Show and label your work. Round to the nearest
hundredth.
8.
Given,
f(2) = 2.836 sin(0.0172(x – 80)) + 12.164
r'(x) = 0.0487792 cos(0.0172(x – 80))
For critical values f'(x) = 0
- 0.0487792 cos(0.0172(x - 80)) = o
= cos(0.0172(x - 80)) = 0
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, (x)) and (date, fx)).
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.
- 0.0172(x - 80) = (2n + 1)
(2n + 1)m
- (x - 80) =-
2(0.0172)
(2n + 1)m
0.0344
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:
5.8936 + 6.2832n
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%2Fcf8c30f6-6704-46d9-8f61-9fe671e1560a%2Fr2k6lwd_processed.png&w=3840&q=75)
Transcribed Image Text:b. Use the first or second derivative tests to determine the maximum and minimum values
October 31
f(x) = 2.836 sin(0.0172(x - 80)) + 12.164
f (304) = 2.836 sin(0.0172(304 – 80)) + 12.164
of the function, and on which dates they occur. Show and label all your written work.
Round to the nearest hundredth.
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
f(x)
= 2.836 sin(3.8525) + 12.164
--1.8511958 + 12.164
= 10.31
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)
4. Find the first and second derivatives of the model without using technology. Show and label
all your work.
y = 2.836 sin(0.0172(353.98 – 80)) + 12.164
y =9.33
Minimum- (353.98, 9.33)
First Derivative:
c. Do the dates make sense based on what you know about the seasons?
Yes, it makes sense because the 172d 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
f(x) = 2.836 sin(0.0172(x – 80)) + 12.164
d
d
= (2.836 sin(0.0172(x – 80))] +12.164
= 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
'(x) = 0.0487792 cos(0.0172(x – 80))
= (0.0487792 cos(0.0172(x – 80))|
0.0487792(- sin(0.0172(x – 80)) • 0.0172)
7.
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.
5.
a. Find the critical values of the function. Show and label your work. Round to the nearest
hundredth.
8.
Given,
f(2) = 2.836 sin(0.0172(x – 80)) + 12.164
r'(x) = 0.0487792 cos(0.0172(x – 80))
For critical values f'(x) = 0
- 0.0487792 cos(0.0172(x - 80)) = o
= cos(0.0172(x - 80)) = 0
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, (x)) and (date, fx)).
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.
- 0.0172(x - 80) = (2n + 1)
(2n + 1)m
- (x - 80) =-
2(0.0172)
(2n + 1)m
0.0344
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:
5.8936 + 6.2832n
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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