Question

Two thousand dollars is deposited into a savings account at 8.5​% interest compounded continuously. ​(a) What is the formula for​ A(t), the balance after t​ years? ​(b) What differential equation is satisfied by​ A(t), the balance after t​ years? ​(c) How much money will be in the account after 5 ​years? ​(d) When will the balance reach ​$3000​? ​(e) How fast is the balance growing when it reaches ​$3000​?

Answer 1

a)
`A(t)=2000e^(0.085t)`

b)
`A'(t)=170e^(0.085t)`

c)$3059.1808

d)t=4.77 years

e) The balance growing is $254.99/year

Explanation:

We are given that Two thousand dollars is deposited into a savings account at 8.5​% interest compounded continuously.

Principal = $2000

Rate of interest = 8.5%

a) What is the formula for​ A(t), the balance after t​ years? ​

Formula
`A(t)=Pe^(rt)`

So,
`A(t)=2000e^(0.085t)`

B)What differential equation is satisfied by​ A(t), the balance after t​ years?

So,
`A'(t)=2000 * 0.085 e^(0.085t)`

`A'(t)=170e^(0.085t)`

c)How much money will be in the account after 5 ​years? ​

Substitute t = 5 in the formula "

`A(t)=2000e^(0.085t)\\A(5)=2000e^(0.085(5))\\A(5)=3059.1808`

d)When will the balance reach ​$3000​?

Substitute A(t)=3000

So,
`3000=2000e^(0.085t)`

t=4.77

The balance reach $3000 in 4.77 years

e)How fast is the balance growing when it reaches ​$3000​?

Substitute the value of t = 4.77 in derivative formula :

`A'(t)=170e^(0.085t)\\A'(t)=170e^(0.085 * 4.77)\\A'(t)=254.99`

Hence the balance growing is $254.99/year