Explanation Given: Money is continuously drawn at a constant rate of $3,400 and the amount in the account becomes 0 after 5 years. Formula used: If money is continuously withdrawn from the account at the rate of $ m per year, then A satisfies the equation d A d t − r A = − m ; where A ( 0 ) = P , d A d t is the rate of change of the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 withdrawal. 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 $ P into an account at a rate of 4.25% which is compounded continuously. And then money is continuously drawn at a constant rate of $3,400. The amount in the account becomes 0 after 5 years. Then, substitute r = 0.0425 and m = 3 , 400 in the above equation. Therefore, the amount A at any time t must satisfy d A d t − 0.0425 A = − 3 , 400 ; A ( 0 ) = P . Now, the integrating factor is, I ( t ) = e ∫ − 0.0425 d t = e − 0.0425 t Multiply both sides of the standard form by the integrating factor. e − 0.0425 t d A d t − e − 0.0425 t 0.0425 A = − 3 , 400 e − 0.0425 t ( A e − 0.0425 t ) ′ = − 3 , 400 e − 0.0425 t [ ∵ ( x y ) ′ = x y ′ + x ′ y ] Integrate both sides. ∫ ( A e − 0.0425 t ) ′ d t = ∫ − 3 , 400 e − 0

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Chapter 9.3, Problem 62E

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To find: The initial deposit P which is deposited into an account that earns 4.25% compounded continuously

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Chapter 9.3, Problem 61E

Chapter 9.3, Problem 63E

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The initial deposit P which is deposited into an account that earns 4.25% compounded continuously

Solution Summary: The author calculates the initial deposit P, which is deposited into an account at a rate of 4.25% compounded continuously.

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