Science:Math Exam Resources/Courses/MATH104/December 2012/Question 01 (j)/Solution 1

Recall the formula for continuously compounded interest,

$A(t)=P{e}^{rt}$

where $A,P,r$ are the future value, principal and growth rate respectively and $t$ is the time since the investment was made. The rate of growth of the investment is simply the time derivative of the future value (i.e. $dA/dt$). Evaluating the time derivative of $A$ gives:

$\frac{dA}{dt}=\frac{d}{dt}\left(P{e}^{rt}\right)=P{e}^{rt}\cdot \frac{d}{dt}(rt)=rP{e}^{rt}.$

By the information given in the question, we want to solve for the time $t$ when the $dA/dt=10,000$ dollars/year, given $P=100,000$ dollars and $r=0.07$/year.

$\begin{array}{rl}10000& =(0.07)(100000)e^{0.07t}\\ 10000& =7000e^{\frac{7}{100}t}\\ \frac{10}{7}& =e^{\frac{7}{100}t}\\ \ln \left(\frac{10}{7}\right)& =\frac{7}{100}t\\ \to t& =\frac{100}{7}\ln \left(\frac{10}{7}\right)(\approx 5.10)\end{array}$

Therefore, we would need to wait (100/7)ln(10/7) years for our initial investment to be growing at a rate of $10,000 per year so we can retire.