\input homework_macros.tex

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\def\dd#1#2{\frac{d#1}{d#2}}

\SN{6.5}
\pn3. The differential equation is $\dd yt=-0.6 y$ and the initial
value condition is $y(0)=100$ provided we measure time starting 
from now (and recall that it is already being measured in hours)
and $y$ denotes the number of grams of $\delta${--}glucono lactone acid.
The problem asks for $y(1)$.
Solve by finding $y(t)$ and then plug in $t=1$.
$y=y_0e^{kt}$, where $k=-0.6$ and $y_0=100$.
Hence $y(1)=y_0e^{k\cdot 1}=100 e^{-0.6}$.
Without a calculator, this is an acceptable form for the answer.


\pn5. Here $L(x)$ is measuring intensity of light in some unspecified
units.
The governing equation is $\dd Lx=-kL$ where $k$ depends on the water
into which you are diving.
You are given $L(18)=\frac 12 L(0)$, and you are asked to find $x$
so that $L(x)=\frac 1{10} L(0)$.
As usual, $L=L(0)e^{kx}$.
From the equation $L(18)=\frac 12 L(0)$ we find
$L(0)e^{18k}=\frac 12 L(0)$, so
$e^{18k}=\frac 12$.
Hence $L=L(0)(\frac 12)^{\frac x{18}}$.
Hence we need to solve
$L(0)(\frac 12)^{\frac x{18}}=\frac 1{10}L(0)$, or
$(\frac 12)^{\frac x{18}}=\frac 1{10}$, or
$2^{\frac x{18}}=10$, or
$\frac x{18}\>\ln2=\ln 10$, so
$x=18\frac{\ln 10}{\ln2}$.


\pn7. Here the governing equation is $\dd Bt=kB$ where $B$
is the number of bacteria and $t$ is the time measured in hours 
from now.
The initial condition is $B(0)=1$ and we are given the additional
information that $B(0.5)=2B(0)=2$.
We are asked for $B(24)$.
Now $B=B(0)e^{kt}=e^{kt}$ so our additional information gives us
$B(0.5)=e^{0.5k}=2$.
Hence $B=2^{2t}=4^t$ and
$B(24)=2^{48}$ or $4^{24}$.

\pn13. The differential equation is $A=A_0e^{rt}$ where $r$
is the rate of interest, $4\%$.
a) In 5 years you will have
$A=A_0e^{0.04\cdot 5}=A_0e^{0.2}$.
b) The doubling time is the time $t$ which satisfies the equation
$A(t)=2A_0$.
Hence $A_0e^{0.04 t}=2A_0$, so $e^{0.04 t}=2$ or
$0.04 t=\ln2$ or $t=\frac{\ln2}{0.04}$.
The tripling time is given by the equation $A(t)=3A_0$
and is solved the same way: $t=\frac{\ln3}{0.04}$.

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