mount p₀ is invested in the mandelbrot bond fund and interest is compounded continuously at 7.1% per year…
mount p₀ is invested in the mandelbrot bond fund and interest is compounded continuously at 7.1% per year, the balance p grows at the rate given by dp/dt = 0.071p. find the function that satisfies the equation. write it in terms of p₀ and 0.071. suppose that $1500 is invested. what is the balance after 1 year? after 3 years? what is the rate of change of the balance after 1 year? after 3 years?
Answer
Explanation:
Step1: Solve the differential equation
The differential equation $\frac{dP}{dt}=0.071P$ is a first - order separable differential equation. We can rewrite it as $\frac{dP}{P}=0.071dt$. Integrating both sides: $\int\frac{dP}{P}=\int0.071dt$. The left - hand side integral is $\ln|P|$ and the right - hand side is $0.071t + C$. So, $\ln|P| = 0.071t+C$. Exponentiating both sides gives $P = Ce^{0.071t}$. When $t = 0$, $P=P_0$, so $P_0 = C$. Thus, the function is $P(t)=P_0e^{0.071t}$.
Step2: Calculate the balance after 1 year
Given $P_0 = 1500$ and $t = 1$, substitute into $P(t)=P_0e^{0.071t}$. Then $P(1)=1500e^{0.071\times1}=1500e^{0.071}\approx1500\times1.0735 = 1610.25$.
Step3: Calculate the balance after 3 years
Given $P_0 = 1500$ and $t = 3$, substitute into $P(t)=P_0e^{0.071t}$. Then $P(3)=1500e^{0.071\times3}=1500e^{0.213}\approx1500\times1.2376=1856.4$.
Step4: Find the rate of change
The rate of change is given by $\frac{dP}{dt}=0.071P$. For $t = 1$, $P(1)\approx1610.25$, so $\frac{dP}{dt}\big|{t = 1}=0.071\times1610.25\approx114.33$. For $t = 3$, $P(3)\approx1856.4$, so $\frac{dP}{dt}\big|{t = 3}=0.071\times1856.4\approx131.80$.
Answer:
The function is $P(t)=P_0e^{0.071t}$. The balance after 1 year is approximately $$1610.25$. The balance after 3 years is approximately $$1856.4$. The rate of change after 1 year is approximately $$114.33$. The rate of change after 3 years is approximately $$131.80$.