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Section 4.3/4.4  Exponential growth/decay

Thm 8:  Suppose $c, k$ are constants.  Then
\v

\centerline{${dy \over dx} = ky$ if and only if $y = ce^{kx}$}

I.e., If the (instantaneous) rate of change of $y$ with respect to $x$ is 
proportional to $y$, then 

Section 4.3:  $k > 0$ implies exponential growth.

Section 4.4:  $k < 0$ implies exponential decay.

\l

For simplicity, take $k > 0$.  Then in section 4.4,
\u
Section 4.4 version of Thm 8:  Suppose $c, k$ are constants.  Then
\v
\u
\centerline{${dy \over dx} = -ky$ if and only if $y = ce^{-kx}$}
\u
I.e., If the (instantaneous) rate of change of $y$ with respect to $x$ is 
proportional to $y$, then 
we have exponential decay since $-k < 0$.

\vfill \vfill \vfill

Initial Value Problem (IVP):  ${dy \over dx} = ky$, $y(x_0) = y_0$  
\vskip 6pt
\rightline{if and only if $y = ce^{kx}$, $y_0 = ce^{kx_0}$  $\Rightarrow c 
= {y_0 \over e^{kx_0}}$~~~~~~~~}
\vskip 6pt
\centerline{Hence  $y = ce^{kx}$ where $c = {y_0 \over e^{kx_0}}$}

\eject

Precalculus:  

Let $P(0) = P_0$

{\bf Section 4.3, $k > 0$:} \hfil \break
 Doubling time = generation time:  If $P(t) = P_0e^{kt}$, then at what 
time $t$ is $P(t) = 2P_0$

$2P_0 = P_0e^{kt}$ $\Rightarrow$ $2 = e^{kt}$$\Rightarrow$  $ln(2) = 
ln(e^{kt}) = kt$ $\Rightarrow$  $t = {ln(2) \over k} $

{\bf Section 4.4, $-k < 0$:} \hfil \break
Half life: If $P(t) = P_0e^{-kt}$, then at what time $t$ is $P(t) = {1 
\over 2}P_0$

${1 \over 2}P_0 = P_0e^{-kt}$ $\Rightarrow$ ${1 \over 2} = e^{-kt}$     

\rightline{$\Rightarrow$ ${ 2} = e^{kt}$
$\Rightarrow$  $ln(2) = ln(e^{kt}) = kt$ $\Rightarrow$  $t = {ln(2) \over 
k} $~~~~~~~~~~~~~}
 
\l
\v

\line{You do NOT need to know the following for either exam 1b or exam 2:}
\u
Thm 11:  Newton's law of cooling
\u
The rate of change of temperature $T$ with respect to time $t$ is given by 
\u
\centerline{${dT \over dt} = -k(T-S)$}
\u
where $k > 0$ is the 
proportionality  constant, and $S$ is the constant temperature of the 
surrounding medium.
\u
Hence $T(t) = P_0e^{-kt} + S$  where $P_0 = T(0) - C$

 
Proof:  $T'= -k(T-S) \Rightarrow {T' \over T-S} = -k  \Rightarrow ln|T-S| 
= -kt + C_1$, for some constant $C_1$


 $|T-S| = e^{ln|T-S|} = e^{-kt + C_1} =  e^{-kt} e^{C_1} $

Hence $T-S = C_2e^{-kt}$ for some constant $C_2$

When $t = 0$,  $T(0)-S = C_2e^{0} = C_2$

Hence $T(t) = P_0e^{-kt} + S$  where $P_0 = T(0) - C$




 







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