Answer:
a)[tex]A(t)=2000e^{0.085t}[/tex]
b)[tex]A'(t)=170e^{0.085t}[/tex]
c)$3059.1808
d)t=4.77 years
e) The balance growing is $254.99/year
Step-by-step 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 [tex]A(t)=Pe^{rt}[/tex]
So,[tex]A(t)=2000e^{0.085t}[/tex]
B)What differential equation is satisfied by A(t), the balance after t years?
So, [tex]A'(t)=2000 \times 0.085 e^{0.085t}[/tex]
[tex]A'(t)=170e^{0.085t}[/tex]
c)How much money will be in the account after 5 years?
Substitute t = 5 in the formula "
[tex]A(t)=2000e^{0.085t}\\A(5)=2000e^{0.085(5)}\\A(5)=3059.1808[/tex]
d)When will the balance reach $3000?
Substitute A(t)=3000
So, [tex]3000=2000e^{0.085t}[/tex]
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 :
[tex]A'(t)=170e^{0.085t}\\A'(t)=170e^{0.085 \times 4.77}\\A'(t)=254.99[/tex]
Hence the balance growing is $254.99/year
