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.)

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dfbustos dfbustos · Answer 1 · 2019-12-16T20:37:47+01:00

Answer:

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

b) [tex]P(t)=1000 e^{0.066(2)}=1141.108[/tex]

c) [tex]t=\frac{ln(2)}{0.066}=10.502 years[/tex]

Step-by-step explanation:

Part a

We have the following model:

[tex]\frac{dP}{dt}=0.066P[/tex]

We can rewrite the expression like this:

[tex]\frac{dP}{P}=0.066 dt[/tex]

If we integrate noth sides we got:

[tex]ln(P)=0.066t +C[/tex]

And exponentiating both sides we got this:

[tex]P(t)=P_o e^{0.066t}[/tex]

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:

[tex]P(t)=1000 e^{0.066(2)}=1141.108[/tex]

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:

[tex]2P_o=P_o e^{0.066(t)}[/tex]

[tex]2=e^{0.066(t)}[/tex]

If we apply natural log on both sides we got:

[tex]ln(2)=0.066 t[/tex]

And if we solve for t we got:

[tex]t=\frac{ln(2)}{0.066}=10.502 years[/tex]