suppose that $p_0$ is invested in a savings account in which interest is compounded continuously at 6.1% per…

suppose that $p_0$ is invested in a savings account in which interest is compounded continuously at 6.1% per year. that is, the balance $p$ grows at the rate given by the following equation $\frac{dp}{dt}=0.061p(t)$ (a)find the function $p(t)$ that satisfies the equation. write it in terms of $p_0$ and 0.061 (b)suppose that $500 is invested. what is the balance after 1 years? (c)when will an investment of $500 double itself? (a) choose the correct answer below a. $p(t)=p_0e^{0.061t}$ b. $p(t)=0.061p_0e^{t}$ c. $p_0 = p(t)e^{0.061t}$ d. $p(t)=p(t)e^{0.061t}$ (b) the balance after 1 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 )

suppose that $p_0$ is invested in a savings account in which interest is compounded continuously at 6.1% per year. that is, the balance $p$ grows at the rate given by the following equation $\frac{dp}{dt}=0.061p(t)$ (a)find the function $p(t)$ that satisfies the equation. write it in terms of $p_0$ and 0.061 (b)suppose that $500 is invested. what is the balance after 1 years? (c)when will an investment of $500 double itself? (a) choose the correct answer below a. $p(t)=p_0e^{0.061t}$ b. $p(t)=0.061p_0e^{t}$ c. $p_0 = p(t)e^{0.061t}$ d. $p(t)=p(t)e^{0.061t}$ (b) the balance after 1 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

Explanation:

Step1: Solve the differential equation

The differential equation $\frac{dP}{dt}=0.061P(t)$ is a first - order separable differential equation. We can rewrite it as $\frac{dP}{P}=0.061dt$. Integrating both sides: $\int\frac{dP}{P}=\int0.061dt$. The left - hand side integral is $\ln|P|$ and the right - hand side is $0.061t + C$. So, $\ln|P| = 0.061t+C$. Exponentiating both sides gives $P(t)=Ce^{0.061t}$. When $t = 0$, $P(0)=P_0$, so $C = P_0$. Thus, $P(t)=P_0e^{0.061t}$.

Step2: Calculate balance after 1 year

If $P_0 = 500$ and $t = 1$, then $P(1)=500e^{0.061\times1}$. Using a calculator, $P(1)=500e^{0.061}\approx500\times1.0629=531.45$.

Step3: Find the doubling - time

We want to find $t$ when $P(t)=2P_0$. Substitute into $P(t)=P_0e^{0.061t}$, we get $2P_0=P_0e^{0.061t}$. Divide both sides by $P_0$ (since $P_0\neq0$), we have $2 = e^{0.061t}$. Take the natural logarithm of both sides: $\ln(2)=\ln(e^{0.061t})$. Since $\ln(e^{x})=x$, then $\ln(2)=0.061t$. Solving for $t$, we get $t=\frac{\ln(2)}{0.061}\approx\frac{0.6931}{0.061}\approx11.36$.

Answer:

(a) A. $P(t)=P_0e^{0.061t}$ (b) $531.45$ (c) $11.36$