# Question 125ce

Apr 9, 2017

$\textcolor{\mathmr{and} a n \ge}{L = - \frac{1}{K \left[t \ln \left(t\right) - t\right] + \frac{1}{5} + K}}$

$\textcolor{\mathmr{and} a n \ge}{y \left(45\right) = {137}^{\circ} F}$
color(orange)(t=90 min

$\textcolor{\mathmr{and} a n \ge}{p \left(t\right) = 100 {\left(2.9\right)}^{t}}$
$\textcolor{\mathmr{and} a n \ge}{= 7072.81}$ bacteria
$\textcolor{\mathmr{and} a n \ge}{p ' \left(4\right) = 7530.50}$
$\textcolor{\mathmr{and} a n \ge}{t = 4.3 h r s}$

#### Explanation:

$\frac{\mathrm{dL}}{\mathrm{dt}} = K {L}^{2} \ln t , L \left(1\right) = - 5$
$\frac{\mathrm{dL}}{\mathrm{dt}} = K {L}^{2} \ln t$
$\frac{\mathrm{dL}}{L} ^ 2 = K \ln t \mathrm{dt}$
$\int \frac{\mathrm{dL}}{L} ^ 2 = K \ln t \mathrm{dt}$
$- \frac{1}{L} = K \left[t \ln \left(t\right) - t\right] + C$
$- \frac{1}{-} 5 = K \left[t \ln \left(1\right) - 1\right] + C$
$\frac{1}{5} = - K + C$
$C = \frac{1}{5} + K$
$- \frac{1}{L} = K \left[t \ln \left(t\right) - t\right] + \frac{1}{5} + K$
$\textcolor{\mathmr{and} a n \ge}{L = - \frac{1}{K \left[t \ln \left(t\right) - t\right] + \frac{1}{5} + K}}$

color(red)(a))
Due to Newton's law of cooling: $\frac{\mathrm{dT}}{\mathrm{dt}} = k \left(T - 75\right)$
$\frac{\mathrm{dT}}{\mathrm{dt}} = k \left(T - 75\right)$
$y \left(t\right) = T \left(t\right) - 75$
$y \left(0\right) = T \left(0\right) - 75$
$y \left(0\right) = 185 - 75$
$y \left(0\right) = 110$
$\frac{\mathrm{dy}}{\mathrm{dt}} = k y$
$y \left(t\right) = 110 {e}^{k t}$
$y \left(30\right) = 110 {e}^{30 t}$
$y \left(30\right)$ because it is going in the oven for half an hour
$150 - 75 = 110 {e}^{30 t}$
$75 = 110 {e}^{30 t}$
$\frac{75}{100} = {e}^{30 t}$
$\ln \left(\frac{75}{110}\right) = 30 t$
$t = \left(\ln \frac{\frac{75}{110}}{30}\right)$
$y \left(t\right) = 110 {e}^{t \ln \frac{\frac{75}{110}}{30}}$
y(45) = 110e^(45(1/30)ln(75/110)
$= 62$
$y \left(45\right) = 62 + 75$
$\textcolor{\mathmr{and} a n \ge}{y \left(45\right) = {137}^{\circ} F}$

color(red)(b))
$y \left(t\right) = T \left(t\right) - 75$
$y \left(110\right) = 110 - 75 = 35$
$35 = 110 {e}^{t \ln \frac{\frac{75}{110}}{30}}$
$\frac{35}{110} = {e}^{t \ln \frac{\frac{75}{110}}{30}}$
$\ln \left(\frac{35}{110}\right) = t \ln \frac{\frac{75}{110}}{30}$
$30 \ln \left(\frac{35}{110}\right) = t \ln \left(\frac{75}{110}\right)$
$\frac{30 \ln \left(\frac{35}{110}\right)}{\ln} \left(\frac{75}{110}\right) = t$
color(orange)(t=90 min

color(red)(a))
$p \left(t\right) = 100 {e}^{k t}$
$290 = 100 {e}^{k \left(1\right)}$
$2.9 = {e}^{k}$
$\ln 2.9 = k$
$p \left(t\right) = 100 {e}^{\ln \left(2.9\right) t}$
$\textcolor{\mathmr{and} a n \ge}{p \left(t\right) = 100 {\left(2.9\right)}^{t}}$

color(red)(b))
$p \left(4\right) = 100 {\left(2.9\right)}^{4}$
$\textcolor{\mathmr{and} a n \ge}{= 7072.81}$ bacteria

color(red)(c))
$p \left(t\right) = 100 {e}^{k t}$ take derivative of this
$\frac{d \left(p \left(t\right)\right)}{\mathrm{dt}} = 100 \left(\frac{d}{\mathrm{dt}}\right) {e}^{k t}$
$p ' \left(t\right) = 100 k {e}^{k t}$
$\ln 2.9 = k$
$p ' \left(t\right) = 100 \left(\ln 2.9\right) {e}^{\ln 2.9 t}$
$p ' \left(t\right) = 100 \left(\ln 2.9\right) \left({2.9}^{t}\right)$
$p ' \left(4\right) = 100 \left(\ln 2.9\right) \left({2.9}^{4}\right)$
$\textcolor{\mathmr{and} a n \ge}{p ' \left(4\right) = 7530.50}$

color(red)(d))#
$10000 = 100 {\left(2.9\right)}^{t}$
$100 = {\left(2.9\right)}^{t}$
$\ln 100 = t \ln \left(2.9\right)$
$t = \ln \frac{100}{\ln} \left(2.9\right)$
$\textcolor{\mathmr{and} a n \ge}{t = 4.3 h r s}$