Explanation Given: The initial amount deposited is $5,000. The interest is compounded continuously at a rate of 7%. And then money is continuously deposited at a constant rate of $2,100 a year. Formula used: If money is continuously deposited into the account at the rate of $ m per year, then A satisfies the equation d A d t − r A = m ; where A ( 0 ) = P where d A d t is the rate of change of the amount A , rA is the rate of growth from continuous compounding and m is the rate of deposit. Integrating factor: The integrating factor of a function f ( x ) is given by I ( x ) = e ∫ f ( x ) d x . Calculation: Given that the initial amount deposited is $5,000 into an account at a rate of 7% which is compounded continuously. And then money is continuously deposited at a constant rate of $2,100 a year. Then, substitute r = 0.07 , m = 2 , 100 and P = 5 , 000 in the above equation. Therefore, the amount A at any time t must satisfy d A d t − 0.07 A = 2 , 100 ; A ( 0 ) = 5 , 000 . Thus, the integrating factor is, I ( t ) = e ∫ − 0.07 d t = e − 0.07 t Multiply both sides of the standard form by the integrating factor. e − 0.07 t d A d t − 0.07 e − 0.07 t A = 2 , 100 e − 0.07 t ( A e − 0.07 t ) ′ = 2 , 100 e − 0.07 t [ ∵ ( x y ) ′ = x y ′ + x ′ y ] Integrating on both sides, ∫ ( A e − 0

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Author: Raymond Barnett, Michael Ziegler

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Chapter 9.3, Problem 63E

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To find: The amount in the account at time t, and the amount in the account after 5 years.

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Chapter 9.3, Problem 62E

Chapter 9.3, Problem 64E

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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

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Ch. 9.1 - Show that y=Cx+1 is the general solution of the...Ch. 9.1 - Prob. 2MP Ch. 9.1 - Prob. 3MP Ch. 9.1 - Prob. 4MP Ch. 9.1 - Prob. 5MP Ch. 9.1 - Prob. 1ED Ch. 9.1 - Prob. 2ED Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E

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The amount in the account at time t , and the amount in the account after 5 years.

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