Explanation Given: The initial amount invested is $10,000. Additional investments are made at the rate of $1,000 per year. After 10 years the amount becomes $30,000. 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 and 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 invested is $10,000. Additional investments are made at the rate of $1,000 per year. After 10 years the amount becomes $30,000. Then substitute, m = 1 , 000 and P = 10 , 000 in the above equation. Therefore, the amount A at any time t must satisfy d A d t − r A = 1 , 000 ; A ( 0 ) = 10 , 000 . Now find the integrating factor. I ( t ) = e ∫ − r d t = e − r t Multiply both sides of the equation by the integrating factor. e − r t d A d t − r e − r t A = 1 , 000 e − r t ( A e − r t ) ′ = 1 , 000 e − r t [ ∵ ( x y ) ′ = x y ′ + x ′ y ] Integrating on both sides, ∫ ( A e − r t ) ′ d t = ∫ 1 , 000 e − r t d t A e − r t = 1 , 000 ( e − r t − r ) + C A e − r t = − 1 , 000 r e − r t + C A = − 1 , 000 r + C e r t The initial condition is A ( 0 ) = 10 , 000

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

14th Edition

ISBN: 9780134668574

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen, Christopher J. Stocker

Publisher: PEARSON

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Chapter 9.3, Problem 65E

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To find: The continuous compound rate of interest that has earned over 10-year period.

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Chapter 9.3, Problem 64E

Chapter 9.3, Problem 66E

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Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

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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