suppose that $p_0$ is invested in a savings account in which interest is compounded continuously at 6.6% per…

suppose that $p_0$ 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 $\frac{dp}{dt}=0.066p(t)$ (a)find the function $p(t)$ that satisfies the equation. write it in terms of $p_0$ and 0.066 (b)suppose that $1000 is invested. what is the balance after 3 years? (c)when will an investment of $1000 double itself? (a) choose the correct answer below a. $p(t)=0.066p_0e^t$ b. $p(t)=p_0e^{0.066t}$ c. $p(t)=p(t)e^{0.066t}$ d. $p_0 = p(t)e^{0.066t}$ (b) the balance after 3 year is $ (type an integer or decimal rounded to two decimal places as needed )

suppose that $p_0$ 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 $\frac{dp}{dt}=0.066p(t)$ (a)find the function $p(t)$ that satisfies the equation. write it in terms of $p_0$ and 0.066 (b)suppose that $1000 is invested. what is the balance after 3 years? (c)when will an investment of $1000 double itself? (a) choose the correct answer below a. $p(t)=0.066p_0e^t$ b. $p(t)=p_0e^{0.066t}$ c. $p(t)=p(t)e^{0.066t}$ d. $p_0 = p(t)e^{0.066t}$ (b) the balance after 3 year is $ (type an integer or decimal rounded to two decimal places as needed )

Answer

Explanation:

Step1: Solve the differential equation

We have the differential equation $\frac{dP}{dt}=0.066P(t)$. This is a first - order separable differential equation. We can rewrite it as $\frac{dP}{P}=0.066dt$. Integrating both sides: $\int\frac{dP}{P}=\int0.066dt$. The integral of $\frac{1}{P}$ with respect to $P$ is $\ln|P|$ and the integral of $0.066$ with respect to $t$ is $0.066t + C$. So, $\ln|P| = 0.066t+C$. Exponentiating both sides gives $P(t)=Ce^{0.066t}$. When $t = 0$, $P(0)=P_0$, so $C = P_0$. Thus, $P(t)=P_0e^{0.066t}$.

Step2: Calculate the balance after 3 years

If $P_0 = 1000$ and $t = 3$, then $P(3)=1000e^{0.066\times3}=1000e^{0.198}$. Using a calculator, $P(3)\approx1000\times1.219977\approx1219.98$.

Step3: Find the time for the investment to double

We want to find $t$ when $P(t)=2P_0$. Substitute into the formula $P(t)=P_0e^{0.066t}$. So, $2P_0=P_0e^{0.066t}$. Divide both sides by $P_0$ (since $P_0\neq0$), we get $2 = e^{0.066t}$. Take the natural logarithm of both sides: $\ln(2)=\ln(e^{0.066t})$. Since $\ln(e^{x})=x$, we have $\ln(2)=0.066t$. Then $t=\frac{\ln(2)}{0.066}\approx\frac{0.693147}{0.066}\approx10.50$.

Answer:

(a) B. $P(t)=P_0e^{0.066t}$ (b) $1219.98$ (c) $10.50$ years