Question

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 1

(C)
`A(t)=500e^(0.05t)`

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:

`(dA)/(dt)=kA`

First, we solve this differential equation

`(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)`

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,
`(dA)/(dt)=50`

Therefore:

`50=1000k\\k=50/1000=0.05`

Substituting into our result from A(t) above:

`A(t)=500e^(0.05t)`