Answer:
(a) [tex]P(t) = P0*e^{0.064t}[/tex]
(b) P(22) = $2,043.88
(c) t=10.83 years
Step-by-step explanation:
The rate of change in the balance per year is given by:
[tex]\frac{dp}{dt}=0.064P(t)[/tex]
(a) Integrating this expression yields the function P(t) for the accumulated balance:
[tex]\int\frac{dp}{dt} \, dt =\int(0.064P(t)) dt\\P(t) = C*e^{0.064t}[/tex]
Where C is the initial invested amount P0:
[tex]P(t) = P0*e^{0.064t}[/tex]
(b) For P0 = $500 ad t =22 years
[tex]P(22) = 500*e^{0.064*22}\\P(22) = \$2,043.88[/tex]
(c) t for P(t) = $1,000 and P0 = $500
[tex]1,000 = 500*e^{0.064t}\\ln(2) = 0.064t\\t=10.83\ years[/tex]
