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Using \(\displaystyle T = T_0+(T_1-T_0)e^{kt}\) gives \(\displaystyle T = 57+(167-57)e^{kt}= 57+110e^{kt}\). Using the point \(\displaystyle (8, 160)\) gives \(\displaystyle 160= 57+110e^{k(8)}\). Isolating the exponential gives \(\displaystyle \frac{103}{110}=e^{8k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{103}{110} \right)}}{8}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle T = 57+110e^{\frac{\ln{\left(\frac{103}{110} \right)}}{8}t}\) and simplifying gives \(\displaystyle T = 110 \left(\frac{103}{110}\right)^{\frac{t}{8}} + 57\). Using \(\displaystyle t = 10\) gives \(\displaystyle T =110 \left(\frac{103}{110}\right)^{\frac{10}{8}} + 57\approx 158^\circ\)

Download \(\LaTeX\)

\begin{question}A coffee with temperature $167^\circ$ is left in a room with temperature $57^\circ$. After 8 minutes the temperature of the coffee is $160^\circ$, what is the temperature of the coffee after 10 minutes?
    \soln{9cm}{Using $T = T_0+(T_1-T_0)e^{kt}$ gives $T = 57+(167-57)e^{kt}= 57+110e^{kt}$. Using the point $(8, 160)$ gives $160= 57+110e^{k(8)}$. Isolating the exponential gives $\frac{103}{110}=e^{8k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{103}{110} \right)}}{8}$.  Substuting $k$ back into the equation gives $T = 57+110e^{\frac{\ln{\left(\frac{103}{110} \right)}}{8}t}$ and simplifying gives $T = 110 \left(\frac{103}{110}\right)^{\frac{t}{8}} + 57$. Using $t = 10$ gives $T =110 \left(\frac{103}{110}\right)^{\frac{10}{8}} + 57\approx 158^\circ$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\usepackage{tcolorbox}

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\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}

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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>A coffee with temperature  <img class="equation_image" title=" \displaystyle 167^\circ " src="/equation_images/%20%5Cdisplaystyle%20167%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 167^\circ " data-equation-content=" \displaystyle 167^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 57^\circ " src="/equation_images/%20%5Cdisplaystyle%2057%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 57^\circ " data-equation-content=" \displaystyle 57^\circ " /> . After 8 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 160^\circ " src="/equation_images/%20%5Cdisplaystyle%20160%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 160^\circ " data-equation-content=" \displaystyle 160^\circ " /> , what is the temperature of the coffee after 10 minutes?</p> </p>

HTML for Canvas

<p> <p>Using  <img class="equation_image" title=" \displaystyle T = T_0+(T_1-T_0)e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%20T_0%2B%28T_1-T_0%29e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = T_0+(T_1-T_0)e^{kt} " data-equation-content=" \displaystyle T = T_0+(T_1-T_0)e^{kt} " />  gives  <img class="equation_image" title=" \displaystyle T = 57+(167-57)e^{kt}= 57+110e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2057%2B%28167-57%29e%5E%7Bkt%7D%3D%2057%2B110e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = 57+(167-57)e^{kt}= 57+110e^{kt} " data-equation-content=" \displaystyle T = 57+(167-57)e^{kt}= 57+110e^{kt} " /> . Using the point  <img class="equation_image" title=" \displaystyle (8, 160) " src="/equation_images/%20%5Cdisplaystyle%20%288%2C%20160%29%20" alt="LaTeX:  \displaystyle (8, 160) " data-equation-content=" \displaystyle (8, 160) " />  gives  <img class="equation_image" title=" \displaystyle 160= 57+110e^{k(8)} " src="/equation_images/%20%5Cdisplaystyle%20160%3D%2057%2B110e%5E%7Bk%288%29%7D%20" alt="LaTeX:  \displaystyle 160= 57+110e^{k(8)} " data-equation-content=" \displaystyle 160= 57+110e^{k(8)} " /> . Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{103}{110}=e^{8k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B103%7D%7B110%7D%3De%5E%7B8k%7D%20" alt="LaTeX:  \displaystyle \frac{103}{110}=e^{8k} " data-equation-content=" \displaystyle \frac{103}{110}=e^{8k} " /> . Solving for  <img class="equation_image" title=" \displaystyle k " src="/equation_images/%20%5Cdisplaystyle%20k%20" alt="LaTeX:  \displaystyle k " data-equation-content=" \displaystyle k " />  gives  <img class="equation_image" title=" \displaystyle k=\frac{\ln{\left(\frac{103}{110} \right)}}{8} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B103%7D%7B110%7D%20%5Cright%29%7D%7D%7B8%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{103}{110} \right)}}{8} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{103}{110} \right)}}{8} " /> .  Substuting  <img class="equation_image" title=" \displaystyle k " src="/equation_images/%20%5Cdisplaystyle%20k%20" alt="LaTeX:  \displaystyle k " data-equation-content=" \displaystyle k " />  back into the equation gives  <img class="equation_image" title=" \displaystyle T = 57+110e^{\frac{\ln{\left(\frac{103}{110} \right)}}{8}t} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2057%2B110e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B103%7D%7B110%7D%20%5Cright%29%7D%7D%7B8%7Dt%7D%20" alt="LaTeX:  \displaystyle T = 57+110e^{\frac{\ln{\left(\frac{103}{110} \right)}}{8}t} " data-equation-content=" \displaystyle T = 57+110e^{\frac{\ln{\left(\frac{103}{110} \right)}}{8}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle T = 110 \left(\frac{103}{110}\right)^{\frac{t}{8}} + 57 " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%20110%20%5Cleft%28%5Cfrac%7B103%7D%7B110%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B8%7D%7D%20%2B%2057%20" alt="LaTeX:  \displaystyle T = 110 \left(\frac{103}{110}\right)^{\frac{t}{8}} + 57 " data-equation-content=" \displaystyle T = 110 \left(\frac{103}{110}\right)^{\frac{t}{8}} + 57 " /> . Using  <img class="equation_image" title=" \displaystyle t = 10 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%2010%20" alt="LaTeX:  \displaystyle t = 10 " data-equation-content=" \displaystyle t = 10 " />  gives  <img class="equation_image" title=" \displaystyle T =110 \left(\frac{103}{110}\right)^{\frac{10}{8}} + 57\approx 158^\circ " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D110%20%5Cleft%28%5Cfrac%7B103%7D%7B110%7D%5Cright%29%5E%7B%5Cfrac%7B10%7D%7B8%7D%7D%20%2B%2057%5Capprox%20158%5E%5Ccirc%20" alt="LaTeX:  \displaystyle T =110 \left(\frac{103}{110}\right)^{\frac{10}{8}} + 57\approx 158^\circ " data-equation-content=" \displaystyle T =110 \left(\frac{103}{110}\right)^{\frac{10}{8}} + 57\approx 158^\circ " /> </p> </p>