Question

suppose that $p_0$ is invested in a savings account in which interest is compounded continuously at 6.9% per year. that is, the balance $p$ grows at the rate given by the following equation
$\frac{dp}{dt} = 0.069p(t)$
(a) find the function $p(t)$ that satisfies the equation. write it in terms of $p_0$ and 0.069.
(b) suppose that $500 is invested. what is the balance after 2 years?
(c) when will an investment of $500 double itself?

(a) choose the correct answer below
\\(\bigcirc\\) a. $p(t) = 0.069p_0e^t$
\\(\bigcirc\\) b. $p(t) = p(t)e^{0.069t}$
\\(\bigcirc\\) c. $p_0 = p(t)e^{0.069t}$
\\(\bigcirc\\) d. $p(t) = p_0e^{0.069t}$

(b) the balance after 2 years is $\square$
(type an integer or decimal rounded to two decimal places as needed.)

(c) the doubling time is $\square$ years.
(type an integer or decimal rounded to two decimal places as needed.)

Explanation:

Step1: Solve the differential equation

The equation is $\frac{dP}{dt} = 0.009P$, separable. Integrate both sides: $\int \frac{1}{P}dP = \int 0.009 dt$ → $\ln|P| = 0.009t + C$. Exponentiate: $P(t) = P_0 e^{0.009t}$ (since $P(0)=P_0$).

Step2: Calculate balance after 2 years

Substitute $P_0=500$, $t=2$: $P(2)=500e^{0.009×2}=500e^{0.018}$. Compute $e^{0.018}≈1.01816$ → $500×1.01816≈509.08$.

Step3: Find doubling time

Set $P(t)=2P_0$: $2P_0=P_0e^{0.009t}$ → $2=e^{0.009t}$. Take ln: $\ln2=0.009t$ → $t=\frac{\ln2}{0.009}≈\frac{0.6931}{0.009}≈77.01$.

Answer:

(a) D. $P(t) = P_0e^{0.009t}$
(b) 509.08
(c) 77.01