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Using \(\displaystyle T = T_0+(T_1-T_0)e^{kt}\) gives \(\displaystyle T = 51+(179-51)e^{kt}= 51+128e^{kt}\). Using the point \(\displaystyle (3, 173)\) gives \(\displaystyle 173= 51+128e^{k(3)}\). Isolating the exponential gives \(\displaystyle \frac{61}{64}=e^{3k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{61}{64} \right)}}{3}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle T = 51+128e^{\frac{\ln{\left(\frac{61}{64} \right)}}{3}t}\) and simplifying gives \(\displaystyle T = 128 \left(\frac{61}{64}\right)^{\frac{t}{3}} + 51\). Using \(\displaystyle t = 12\) gives \(\displaystyle T =128 \left(\frac{61}{64}\right)^{\frac{12}{3}} + 51\approx 157^\circ\)

Download \(\LaTeX\)

\begin{question}A coffee with temperature $179^\circ$ is left in a room with temperature $51^\circ$. After 3 minutes the temperature of the coffee is $173^\circ$, what is the temperature of the coffee after 12 minutes?
    \soln{9cm}{Using $T = T_0+(T_1-T_0)e^{kt}$ gives $T = 51+(179-51)e^{kt}= 51+128e^{kt}$. Using the point $(3, 173)$ gives $173= 51+128e^{k(3)}$. Isolating the exponential gives $\frac{61}{64}=e^{3k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{61}{64} \right)}}{3}$.  Substuting $k$ back into the equation gives $T = 51+128e^{\frac{\ln{\left(\frac{61}{64} \right)}}{3}t}$ and simplifying gives $T = 128 \left(\frac{61}{64}\right)^{\frac{t}{3}} + 51$. Using $t = 12$ gives $T =128 \left(\frac{61}{64}\right)^{\frac{12}{3}} + 51\approx 157^\circ$}

\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 179^\circ " src="/equation_images/%20%5Cdisplaystyle%20179%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 179^\circ " data-equation-content=" \displaystyle 179^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 51^\circ " src="/equation_images/%20%5Cdisplaystyle%2051%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 51^\circ " data-equation-content=" \displaystyle 51^\circ " /> . After 3 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 173^\circ " src="/equation_images/%20%5Cdisplaystyle%20173%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 173^\circ " data-equation-content=" \displaystyle 173^\circ " /> , what is the temperature of the coffee after 12 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 = 51+(179-51)e^{kt}= 51+128e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2051%2B%28179-51%29e%5E%7Bkt%7D%3D%2051%2B128e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = 51+(179-51)e^{kt}= 51+128e^{kt} " data-equation-content=" \displaystyle T = 51+(179-51)e^{kt}= 51+128e^{kt} " /> . Using the point  <img class="equation_image" title=" \displaystyle (3, 173) " src="/equation_images/%20%5Cdisplaystyle%20%283%2C%20173%29%20" alt="LaTeX:  \displaystyle (3, 173) " data-equation-content=" \displaystyle (3, 173) " />  gives  <img class="equation_image" title=" \displaystyle 173= 51+128e^{k(3)} " src="/equation_images/%20%5Cdisplaystyle%20173%3D%2051%2B128e%5E%7Bk%283%29%7D%20" alt="LaTeX:  \displaystyle 173= 51+128e^{k(3)} " data-equation-content=" \displaystyle 173= 51+128e^{k(3)} " /> . Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{61}{64}=e^{3k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B61%7D%7B64%7D%3De%5E%7B3k%7D%20" alt="LaTeX:  \displaystyle \frac{61}{64}=e^{3k} " data-equation-content=" \displaystyle \frac{61}{64}=e^{3k} " /> . 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{61}{64} \right)}}{3} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B61%7D%7B64%7D%20%5Cright%29%7D%7D%7B3%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{61}{64} \right)}}{3} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{61}{64} \right)}}{3} " /> .  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 = 51+128e^{\frac{\ln{\left(\frac{61}{64} \right)}}{3}t} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2051%2B128e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B61%7D%7B64%7D%20%5Cright%29%7D%7D%7B3%7Dt%7D%20" alt="LaTeX:  \displaystyle T = 51+128e^{\frac{\ln{\left(\frac{61}{64} \right)}}{3}t} " data-equation-content=" \displaystyle T = 51+128e^{\frac{\ln{\left(\frac{61}{64} \right)}}{3}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle T = 128 \left(\frac{61}{64}\right)^{\frac{t}{3}} + 51 " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%20128%20%5Cleft%28%5Cfrac%7B61%7D%7B64%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B3%7D%7D%20%2B%2051%20" alt="LaTeX:  \displaystyle T = 128 \left(\frac{61}{64}\right)^{\frac{t}{3}} + 51 " data-equation-content=" \displaystyle T = 128 \left(\frac{61}{64}\right)^{\frac{t}{3}} + 51 " /> . Using  <img class="equation_image" title=" \displaystyle t = 12 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%2012%20" alt="LaTeX:  \displaystyle t = 12 " data-equation-content=" \displaystyle t = 12 " />  gives  <img class="equation_image" title=" \displaystyle T =128 \left(\frac{61}{64}\right)^{\frac{12}{3}} + 51\approx 157^\circ " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D128%20%5Cleft%28%5Cfrac%7B61%7D%7B64%7D%5Cright%29%5E%7B%5Cfrac%7B12%7D%7B3%7D%7D%20%2B%2051%5Capprox%20157%5E%5Ccirc%20" alt="LaTeX:  \displaystyle T =128 \left(\frac{61}{64}\right)^{\frac{12}{3}} + 51\approx 157^\circ " data-equation-content=" \displaystyle T =128 \left(\frac{61}{64}\right)^{\frac{12}{3}} + 51\approx 157^\circ " /> </p> </p>