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Using \(\displaystyle T = T_0+(T_1-T_0)e^{kt}\) gives \(\displaystyle T = 66+(158-66)e^{kt}= 66+92e^{kt}\). Using the point \(\displaystyle (9, 131)\) gives \(\displaystyle 131= 66+92e^{k(9)}\). Isolating the exponential gives \(\displaystyle \frac{65}{92}=e^{9k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{65}{92} \right)}}{9}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle T = 66+92e^{\frac{\ln{\left(\frac{65}{92} \right)}}{9}t}\) and simplifying gives \(\displaystyle T = 92 \left(\frac{65}{92}\right)^{\frac{t}{9}} + 66\). Using \(\displaystyle t = 22\) gives \(\displaystyle T =92 \left(\frac{65}{92}\right)^{\frac{22}{9}} + 66\approx 105^\circ\)

Download \(\LaTeX\)

\begin{question}A coffee with temperature $158^\circ$ is left in a room with temperature $66^\circ$. After 9 minutes the temperature of the coffee is $131^\circ$, what is the temperature of the coffee after 22 minutes?
    \soln{9cm}{Using $T = T_0+(T_1-T_0)e^{kt}$ gives $T = 66+(158-66)e^{kt}= 66+92e^{kt}$. Using the point $(9, 131)$ gives $131= 66+92e^{k(9)}$. Isolating the exponential gives $\frac{65}{92}=e^{9k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{65}{92} \right)}}{9}$.  Substuting $k$ back into the equation gives $T = 66+92e^{\frac{\ln{\left(\frac{65}{92} \right)}}{9}t}$ and simplifying gives $T = 92 \left(\frac{65}{92}\right)^{\frac{t}{9}} + 66$. Using $t = 22$ gives $T =92 \left(\frac{65}{92}\right)^{\frac{22}{9}} + 66\approx 105^\circ$}

\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 158^\circ " src="/equation_images/%20%5Cdisplaystyle%20158%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 158^\circ " data-equation-content=" \displaystyle 158^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 66^\circ " src="/equation_images/%20%5Cdisplaystyle%2066%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 66^\circ " data-equation-content=" \displaystyle 66^\circ " /> . After 9 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 131^\circ " src="/equation_images/%20%5Cdisplaystyle%20131%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 131^\circ " data-equation-content=" \displaystyle 131^\circ " /> , what is the temperature of the coffee after 22 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 = 66+(158-66)e^{kt}= 66+92e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2066%2B%28158-66%29e%5E%7Bkt%7D%3D%2066%2B92e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = 66+(158-66)e^{kt}= 66+92e^{kt} " data-equation-content=" \displaystyle T = 66+(158-66)e^{kt}= 66+92e^{kt} " /> . Using the point  <img class="equation_image" title=" \displaystyle (9, 131) " src="/equation_images/%20%5Cdisplaystyle%20%289%2C%20131%29%20" alt="LaTeX:  \displaystyle (9, 131) " data-equation-content=" \displaystyle (9, 131) " />  gives  <img class="equation_image" title=" \displaystyle 131= 66+92e^{k(9)} " src="/equation_images/%20%5Cdisplaystyle%20131%3D%2066%2B92e%5E%7Bk%289%29%7D%20" alt="LaTeX:  \displaystyle 131= 66+92e^{k(9)} " data-equation-content=" \displaystyle 131= 66+92e^{k(9)} " /> . Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{65}{92}=e^{9k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B65%7D%7B92%7D%3De%5E%7B9k%7D%20" alt="LaTeX:  \displaystyle \frac{65}{92}=e^{9k} " data-equation-content=" \displaystyle \frac{65}{92}=e^{9k} " /> . 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{65}{92} \right)}}{9} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B65%7D%7B92%7D%20%5Cright%29%7D%7D%7B9%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{65}{92} \right)}}{9} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{65}{92} \right)}}{9} " /> .  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 = 66+92e^{\frac{\ln{\left(\frac{65}{92} \right)}}{9}t} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2066%2B92e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B65%7D%7B92%7D%20%5Cright%29%7D%7D%7B9%7Dt%7D%20" alt="LaTeX:  \displaystyle T = 66+92e^{\frac{\ln{\left(\frac{65}{92} \right)}}{9}t} " data-equation-content=" \displaystyle T = 66+92e^{\frac{\ln{\left(\frac{65}{92} \right)}}{9}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle T = 92 \left(\frac{65}{92}\right)^{\frac{t}{9}} + 66 " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2092%20%5Cleft%28%5Cfrac%7B65%7D%7B92%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B9%7D%7D%20%2B%2066%20" alt="LaTeX:  \displaystyle T = 92 \left(\frac{65}{92}\right)^{\frac{t}{9}} + 66 " data-equation-content=" \displaystyle T = 92 \left(\frac{65}{92}\right)^{\frac{t}{9}} + 66 " /> . Using  <img class="equation_image" title=" \displaystyle t = 22 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%2022%20" alt="LaTeX:  \displaystyle t = 22 " data-equation-content=" \displaystyle t = 22 " />  gives  <img class="equation_image" title=" \displaystyle T =92 \left(\frac{65}{92}\right)^{\frac{22}{9}} + 66\approx 105^\circ " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D92%20%5Cleft%28%5Cfrac%7B65%7D%7B92%7D%5Cright%29%5E%7B%5Cfrac%7B22%7D%7B9%7D%7D%20%2B%2066%5Capprox%20105%5E%5Ccirc%20" alt="LaTeX:  \displaystyle T =92 \left(\frac{65}{92}\right)^{\frac{22}{9}} + 66\approx 105^\circ " data-equation-content=" \displaystyle T =92 \left(\frac{65}{92}\right)^{\frac{22}{9}} + 66\approx 105^\circ " /> </p> </p>