Temperature. If the temperature in an aquarium (in degrees Celsius) is given by C ( t ) = t 3 − 2 t + 10 0 ≤ t ≤ 2 over a 2-hour period, what is the average temperature over this period?
Temperature. If the temperature in an aquarium (in degrees Celsius) is given by C ( t ) = t 3 − 2 t + 10 0 ≤ t ≤ 2 over a 2-hour period, what is the average temperature over this period?
Solution Summary: The author explains how the average temperature over the period left[0,2right] is 10°C.
A: Tan Latitude / Tan P
A = Tan 04° 30'/ Tan 77° 50.3'
A= 0.016960 803 S CA named opposite to latitude,
except when hour angle between 090° and 270°)
B: Tan Declination | Sin P
B Tan 052° 42.1'/ Sin 77° 50.3'
B = 1.34 2905601 SCB is alway named same as
declination)
C = A + B = 1.35 9866404 S CC correction, A+/- B:
if A and B have same name - add, If
different name- subtract)
=
Tan Azimuth 1/Ccx cos Latitude)
Tan Azimuth = 0.737640253
Azimuth
=
S 36.4° E CAzimuth takes combined
name of C correction and Hour Angle - If LHA
is between 0° and 180°, it is named "west", if
LHA is between 180° and 360° it is named "east"
True Azimuth= 143.6°
Compass Azimuth = 145.0°
Compass Error = 1.4° West
Variation 4.0 East
Deviation: 5.4 West
A: Tan Latitude / Tan P
A = Tan 04° 30'/ Tan 77° 50.3'
A= 0.016960 803 S CA named opposite to latitude,
except when hour angle between 090° and 270°)
B: Tan Declination | Sin P
B Tan 052° 42.1'/ Sin 77° 50.3'
B = 1.34 2905601 SCB is alway named same as
declination)
C = A + B = 1.35 9866404 S CC correction, A+/- B:
if A and B have same name - add, If
different name- subtract)
=
Tan Azimuth 1/Ccx cos Latitude)
Tan Azimuth = 0.737640253
Azimuth
=
S 36.4° E CAzimuth takes combined
name of C correction and Hour Angle - If LHA
is between 0° and 180°, it is named "west", if
LHA is between 180° and 360° it is named "east"
True Azimuth= 143.6°
Compass Azimuth = 145.0°
Compass Error = 1.4° West
Variation 4.0 East
Deviation: 5.4 West
Direction: Strictly write in 4 bond paper, because my activity
sheet is have 4 spaces. This is actually for maritime.
industry course, but I think geometry can do this.
use nautical almanac.
Sample Calculation (Amplitude- Sun):
On 07th May 2006 at Sunset, a vesel in position 10°00'N
0 10°00' W observed the sun bearing 288° by compass. Find
the
compass error.
LMT Sunset
07d
18h
13m
(+)00d
00h
40 м
LIT:
UTC Sunset:
07d
18h
53 m
added - since
longitude is
westerly
Declination Co7d 18h): N016° 55.5'
d(0.7):
(+)
00-6
N016 56.1'
Declination Sun:
Sin Amplitude Sin Declination (Los Latitude
- Sin 016° 56.1'/Cos 10°00'
= 0.295780189
Amplitude = WI. 2N (The prefix of amplitude is
named easterly if body is rising.
and westerly of body is setting.
The suffix is named came as
declination.)
True Bearing: 287.20
Compass Bearing
288.0°
Compass Error: 0.8' West
Chapter 5 Solutions
Pearson eText for Calculus for Business, Economics, Life Sciences, and Social Sciences, Brief Version -- Instant Access (Pearson+)
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