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

Recall the formula for continuously compounded interest,

$A(t)=Pe^{rt}$

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:

${\frac {dA}{dt}}={\frac {d}{dt}}\left(Pe^{rt}\right)=Pe^{rt}\cdot {\frac {d}{dt}}(rt)=rPe^{rt}.$

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.

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

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.