Question

Suppose that P0 is invested in a savings account in which interest is compounded continuously at 6.6% per year. That is, the balance P grows at the rate given by the following equation. dP/dt - = 0.066P(t)

A) Find the function P(t) that satisfies the equation.Write it in terms of P0 and 0.066.
B) Suppose that $1000 is invested. What is the balance after 2 years?
C) When will an investment of $1000 double itself?
A) Choose me correct answer below.
a) p(t) = P 0 e 0066t
b) p(t) = 0.066P 0 et
c) p(t) = P(t)e 0 066t
d) p(0) = P(t)e 0066t
B) The balance after 2 year is $ . (Type an integer or decimal rounded to two decimal places as needed.)
C) The doubling time is year. (Type an integer or decimal rounded to two decimal places as needed.)

Answer 1

a) p(t) = P 0 e 0066t

b)
`P(t)=1000 e^(0.066(2))=1141.108`

c)
`t=(ln(2))/(0.066)=10.502 years`

Explanation:

Part a

We have the following model:

`(dP)/(dt)=0.066P`

We can rewrite the expression like this:

`(dP)/(P)=0.066 dt`

If we integrate noth sides we got:

`ln(P)=0.066t +C`

And exponentiating both sides we got this:

`P(t)=P_o e^(0.066t)`

So the correct option would be:

a) p(t) = P 0 e 0066t

Part b

On this case we want to find the amount of money after two years if the initial investment is 1000 and the value of t=2. If we replace we got:

`P(t)=1000 e^(0.066(2))=1141.108`

Part c

On this case we need a value of time in order to duplicate the initial amount invested, so we need to solve the following equation:

`2P_o=P_o e^(0.066(t))`

`2=e^(0.066(t))`

If we apply natural log on both sides we got:

`ln(2)=0.066 t`

And if we solve for t we got:

`t=(ln(2))/(0.066)=10.502 years`