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Let A(t) represent the amount of money in a bank account at time t years. The rate at which the account is increasing is proportional to the amount of money in the account, which can be modeled by the differential equation ka, where k is a positive constant and t is time, in years. At time t = 0, there is $500 in the account. At the moment when the amount of money in the account is $1000, the amount is increasing at a rate of $50 per year. Which of the following is an expression for A(t)?
(A) A(t) = 500+0.051
(B) A(t) = 500(0.056)
(C) A(t) = 500005
(D) A(t) = 1000+501
(E) A(t)=10000051

Answer :

Answer:

(C)[tex]A(t)=500e^{0.05t}[/tex]

Step-by-step explanation:

If A(t) represent the amount of money in a bank account at time t years.

The rate at which the account is increasing is proportional to the amount of money in the account, which is modeled by the differential equation:

[tex]\dfrac{dA}{dt}=kA[/tex]

First, we solve this differential equation

[tex]\dfrac{dA}{A}=kdt\\$Taking Integrals\\ln A =kt+C, C a constant of integration\\Taking exponents of both sides\\A(t)=Ce^{kt}\\When t=0, A(t)=\$500\\500=C\\Therefore:\\A(t)=500e^{kt}[/tex]

At the moment when the amount of money in the account is $1000, the amount is increasing at a rate of $50 per year.

When A(t)=$1000, [tex]\dfrac{dA}{dt}=50[/tex]

Therefore:

[tex]50=1000k\\k=50/1000=0.05[/tex]

Substituting into our result from A(t) above:

[tex]A(t)=500e^{0.05t}[/tex]

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