Question

The rate of change of the value of an investment, S, with respect to time, t ≥ 0, is given by dS dt = e(rt/100)(5r), where r is the annual interest rate (assumed constant) and the principal of the investment is S(0) = 500.The rate of change of the value of an investment, S, with respect to time, t ≥ 0, is given by

Answer 1

$$\begin{aligned} \frac{dS}{d t}=5r \ e^{\frac{r t}{100}}, s(0)=500 \quad(\text { given })\\[2em] \therefore d S=5 r\ e^{\frac{r}{100} t} d t \quad \text { (integrate both sides) } \end{aligned}\\ \begin{aligned} \\ \int_{0}^{t} d S&=\int_{0}^{t} 5 r \ e^{\frac{r t}{100}} d t\\[1 em] S\bigg| _{0}^{t}& =5 r\left(\frac{100}{r}\right) e^{\frac{r}{100} t}\bigg| _{0}^{t} \\[1 em] S(t)-S(0) &=500\left(e^{\frac{r t}{100}}-1\right)\\[1 em] S(t) &=500\left(e^{\frac{r t}{100}}-1\right)+S(0) \\[1 em] &=500\left(e^{\frac{r t}{100}}-1\right)+500 \\[1 em] &=500\ e^{\frac{r t}{100}}-500+500 \\[1 em] \therefore S (t)&=500\ e^{\frac{r t}{100}}\\[1 em] \end{aligned}\\ \text{To verify the solution we find the derivative of }\\[1 em] \frac{dS}{d t}=500 (\frac{r}{100}) \ e^{\frac{r t}{100}}=5r \ e^{\frac{r t}{100}}$$