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Using \(\displaystyle T = T_0+(T_1-T_0)e^{kt}\) gives \(\displaystyle T = 82+(162-82)e^{kt}= 82+80e^{kt}\). Using the point \(\displaystyle (10, 138)\) gives \(\displaystyle 138= 82+80e^{k(10)}\). Isolating the exponential gives \(\displaystyle \frac{7}{10}=e^{10k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{7}{10} \right)}}{10}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle T = 82+80e^{\frac{\ln{\left(\frac{7}{10} \right)}}{10}t}\) and simplifying gives \(\displaystyle T = 80 \left(\frac{7}{10}\right)^{\frac{t}{10}} + 82\). Using \(\displaystyle t = 13\) gives \(\displaystyle T =80 \left(\frac{7}{10}\right)^{\frac{13}{10}} + 82\approx 132^\circ\)

Download \(\LaTeX\)

\begin{question}A coffee with temperature $162^\circ$ is left in a room with temperature $82^\circ$. After 10 minutes the temperature of the coffee is $138^\circ$, what is the temperature of the coffee after 13 minutes?
    \soln{9cm}{Using $T = T_0+(T_1-T_0)e^{kt}$ gives $T = 82+(162-82)e^{kt}= 82+80e^{kt}$. Using the point $(10, 138)$ gives $138= 82+80e^{k(10)}$. Isolating the exponential gives $\frac{7}{10}=e^{10k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{7}{10} \right)}}{10}$.  Substuting $k$ back into the equation gives $T = 82+80e^{\frac{\ln{\left(\frac{7}{10} \right)}}{10}t}$ and simplifying gives $T = 80 \left(\frac{7}{10}\right)^{\frac{t}{10}} + 82$. Using $t = 13$ gives $T =80 \left(\frac{7}{10}\right)^{\frac{13}{10}} + 82\approx 132^\circ$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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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 162^\circ " src="/equation_images/%20%5Cdisplaystyle%20162%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 162^\circ " data-equation-content=" \displaystyle 162^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 82^\circ " src="/equation_images/%20%5Cdisplaystyle%2082%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 82^\circ " data-equation-content=" \displaystyle 82^\circ " /> . After 10 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 138^\circ " src="/equation_images/%20%5Cdisplaystyle%20138%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 138^\circ " data-equation-content=" \displaystyle 138^\circ " /> , what is the temperature of the coffee after 13 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 = 82+(162-82)e^{kt}= 82+80e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2082%2B%28162-82%29e%5E%7Bkt%7D%3D%2082%2B80e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = 82+(162-82)e^{kt}= 82+80e^{kt} " data-equation-content=" \displaystyle T = 82+(162-82)e^{kt}= 82+80e^{kt} " /> . Using the point  <img class="equation_image" title=" \displaystyle (10, 138) " src="/equation_images/%20%5Cdisplaystyle%20%2810%2C%20138%29%20" alt="LaTeX:  \displaystyle (10, 138) " data-equation-content=" \displaystyle (10, 138) " />  gives  <img class="equation_image" title=" \displaystyle 138= 82+80e^{k(10)} " src="/equation_images/%20%5Cdisplaystyle%20138%3D%2082%2B80e%5E%7Bk%2810%29%7D%20" alt="LaTeX:  \displaystyle 138= 82+80e^{k(10)} " data-equation-content=" \displaystyle 138= 82+80e^{k(10)} " /> . Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{7}{10}=e^{10k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B7%7D%7B10%7D%3De%5E%7B10k%7D%20" alt="LaTeX:  \displaystyle \frac{7}{10}=e^{10k} " data-equation-content=" \displaystyle \frac{7}{10}=e^{10k} " /> . 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{7}{10} \right)}}{10} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B7%7D%7B10%7D%20%5Cright%29%7D%7D%7B10%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{7}{10} \right)}}{10} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{7}{10} \right)}}{10} " /> .  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 = 82+80e^{\frac{\ln{\left(\frac{7}{10} \right)}}{10}t} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2082%2B80e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B7%7D%7B10%7D%20%5Cright%29%7D%7D%7B10%7Dt%7D%20" alt="LaTeX:  \displaystyle T = 82+80e^{\frac{\ln{\left(\frac{7}{10} \right)}}{10}t} " data-equation-content=" \displaystyle T = 82+80e^{\frac{\ln{\left(\frac{7}{10} \right)}}{10}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle T = 80 \left(\frac{7}{10}\right)^{\frac{t}{10}} + 82 " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2080%20%5Cleft%28%5Cfrac%7B7%7D%7B10%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B10%7D%7D%20%2B%2082%20" alt="LaTeX:  \displaystyle T = 80 \left(\frac{7}{10}\right)^{\frac{t}{10}} + 82 " data-equation-content=" \displaystyle T = 80 \left(\frac{7}{10}\right)^{\frac{t}{10}} + 82 " /> . Using  <img class="equation_image" title=" \displaystyle t = 13 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%2013%20" alt="LaTeX:  \displaystyle t = 13 " data-equation-content=" \displaystyle t = 13 " />  gives  <img class="equation_image" title=" \displaystyle T =80 \left(\frac{7}{10}\right)^{\frac{13}{10}} + 82\approx 132^\circ " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D80%20%5Cleft%28%5Cfrac%7B7%7D%7B10%7D%5Cright%29%5E%7B%5Cfrac%7B13%7D%7B10%7D%7D%20%2B%2082%5Capprox%20132%5E%5Ccirc%20" alt="LaTeX:  \displaystyle T =80 \left(\frac{7}{10}\right)^{\frac{13}{10}} + 82\approx 132^\circ " data-equation-content=" \displaystyle T =80 \left(\frac{7}{10}\right)^{\frac{13}{10}} + 82\approx 132^\circ " /> </p> </p>