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\flushleft\underline{Quiz 2 --- MATH 1586 Spring 2018} \\
\begin{enumerate}
\item Recall Newton's law of cooling:
\[\dfrac{\mathrm{d}T}{\mathrm{d}t} = k(T-T_0),\]
where $T$ denotes the temperature at time $t$, $k$ is a proportionality constant, and $T_0$ is the ambient temperature of the surrounding space. 

When a cake is removed from an oven, its temperature is $280^{\circ} F$. 7 minutes later, it is $210^{\circ} F$. How long will it take for the cake to cool down to $80^{\circ} F$ in a room that is $74^{\circ} F$?  \\
\textit{Solution}: We are told that $T_0=74$, so the differential equation becomes
$$\dfrac{\mathrm{d}T}{\mathrm{d}t} = k(T-74).$$
Separate variables and integrate to get
$$\displaystyle\int \dfrac{1}{T-74} \mathrm{d}T = \displaystyle\int k \mathrm{d}t.$$
Calculate the integrals to get
$$\ln(T-74) = kt + C,$$
and plug both sides into $e^x$ to get
$$T-74 = e^{kt+C}=C_1 e^{kt}, \quad C_1=e^C,$$
hence our model is
$$(*) \hspace{35pt} T = 74 + C_1 e^{kt}.$$
We are told two initial conditions: the temperature of the cake leaving the oven yields $T(0)=280$ and the condition $7$ minutes later yields $T(7)=210$. Use the first condition to see
$$\underbrace{280}_{\mathrm{given}}=T(0)=\underbrace{74+C_1e^0}_{\mathrm{calculated}}.$$
Therefore we must solve $280=74+C_1$ for $C_1$, yielding
$$C_1 = 280-74 = 206.$$
Therefore we may update our model to get
$$(**) \hspace{35pt} T=74+206e^{kt}.$$
Now applying the second condition yields
$$\underbrace{210}_{\mathrm{given}} = T(7) = \underbrace{74+206e^{7k}}_{\mathrm{calculated}}.$$
Therefore we must solve $210=74+206e^{7k}$ for $k$. This means
$$k=\dfrac{1}{7}\ln \left(\dfrac{210-74}{206} \right)=\dfrac{1}{7} \ln \left( \dfrac{136}{206} \right).$$
So we may update our model again to get
$$(***) \hspace{35pt} T=74+206e^{\frac{1}{7}\ln \left( \frac{136}{206} \right)t}.$$
Finally, to answer the question of at what time the cake is at temperature $80^{\circ} F$, we see that we must solve
$$\underbrace{80}_{\mathrm{given}} = T(t) = \underbrace{74+206e^{\frac{t}{7}\ln\left(\frac{136}{206}\right)}}_{\mathrm{calculated}}.$$
This implies
$$6 = 206 e^{\frac{t}{7} \ln \left( \frac{136}{206} \right)},$$
so
$$\ln \left( \dfrac{6}{206} \right) = \dfrac{t}{7} \ln \left( \dfrac{136}{206} \right),$$
therefore
$$t = \dfrac{7\ln \left( \dfrac{6}{206} \right)}{\ln \left( \dfrac{136}{206} \right)} \approx 59.61.$$
\item Compute 
\[\displaystyle\int x e^{3x} \mathrm{d}x.\] \\
\textit{Solution}: Let $u=x$ and $\mathrm{d}v=e^{3x}$. Then $\mathrm{d}u=\mathrm{d}x$ and $v = \dfrac{1}{3} e^{3x}$. Therefore using integration by parts we see
$$\displaystyle\int xe^{3x} \mathrm{d}x = \dfrac{x}{3}e^{3x} - \dfrac{1}{3} \displaystyle\int e^{3x} \mathrm{d}x=\dfrac{x}{3} e^{3x} - \dfrac{1}{9} e^{3x} + C.$$

\item Compute
\[ \displaystyle\int_3^{\infty} e^{-5x} \mathrm{d}x.\] \\
\textit{Solution}: Compute
$$\begin{array}{ll}
\displaystyle\int_3^{\infty} e^{-5x} \mathrm{d}x &= \displaystyle\lim_{b \rightarrow \infty} \displaystyle\int_3^b e^{-5x} \mathrm{d}x \\
&= \displaystyle\lim_{b \rightarrow \infty} -\dfrac{1}{5} e^{-5x} \Bigg|_3^b \\
&= \displaystyle\lim_{b \rightarrow \infty} -\dfrac{e^{-5b}}{5} + \dfrac{1}{5}e^{-15} \\
&=\dfrac{1}{5}e^{-15}.
\end{array}$$
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