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\noindent 9. Graduate borrows \$8000 at annual rate of 10\%, compounded
continuously. He pays continuously -- obviously spending a lot of time at his bank -- at constant rate of $k$ dollars per year. What should $k$ be to pay off
the loan in 3 years, and how much interest does he pay?
\medskip
\noindent Solution: This one is pretty straightforward. Let M(t)
be the money he owes after $t$ years. Then
$${dM\over dt}=(0.1)M-k \hbox{ or } {dM\over dt}-(0.1)M=-k.$$
The general solution to that is
$$M(t)=10k+CE^{t/10}\hbox{ and } M(0)=8000 \hbox{ implies }C=8000-10k.
$$
If he pays off the loan in three years, $M(3)=0$ and we have
$$0=10k +(8000 -10k)e^{0.3} \hbox{ or } k={8000e^{0.3}\over 10(e^{0.3}-1}=3086.64.$$
The total interest paid is just $(3)(3086.64)-8000=1259.91$.
\end
:wq
: