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Newton's law of Cooling states that the change in temperature is directly proportional to the difference between the object's temperature and its surroundings. \begin{equation*} \frac{dT}{dt} = k(T(t)-T_{\text{room}})\end{equation*}Using the substitution \(\displaystyle y(t)=T(t)-71\) and calculating the derivative gives \(\displaystyle \frac{dy}{dt}=\frac{dT}{dt}\). Calculating the new initial condition using the point \(\displaystyle (4, 164)\) and the substition gives \(\displaystyle y(0) = T(0)-71 = 104\). The point \(\displaystyle (4, 164)\) must also be transformed to get \(\displaystyle y(4) = T(4)-71 = 164 - 71 = 93\). Substituting both of these into the equation gives the new equaiton \(\displaystyle \frac{dy}{dt}=ky\) which has the solution \(\displaystyle y(t) = y(0)e^{kt}=104e^{kt}\). Evaluating the function at the point gives \(\displaystyle 93=104e^{4k}\) and isolating the exponential gives \(\displaystyle \frac{93}{104}=e^{4k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{93}{104} \right)}}{4}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle y(t) = 104e^{\frac{\ln{\left(\frac{93}{104} \right)}}{4}t}\) and simplifying gives \(\displaystyle y(t) = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}}\). Substituting out \(\displaystyle y(t)\) gives \begin{equation*}T(t)-71 = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} \implies\, T(t)= 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} + 71 \end{equation*}Using \(\displaystyle t = 9\) gives \(\displaystyle T =104 \left(\frac{93}{104}\right)^{\frac{9}{4}} + 71\approx 152.00^\circ\)

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\begin{question}A coffee with temperature $175^\circ$ is left in a room with temperature $71^\circ$. After 4 minutes the temperature of the coffee is $164^\circ$, what is the temperature of the coffee after 9 minutes?
    \soln{9cm}{Newton's law of Cooling states that the change in temperature is directly proportional to the difference between the object's temperature and its surroundings. \begin{equation*} \frac{dT}{dt} = k(T(t)-T_{\text{room}})\end{equation*}Using the substitution $y(t)=T(t)-71$ and calculating the derivative gives $\frac{dy}{dt}=\frac{dT}{dt}$. Calculating the new initial condition using the point $(4, 164)$ and the substition gives $y(0) = T(0)-71 = 104$. The point $(4, 164)$ must also be transformed to get $y(4) = T(4)-71 = 164 - 71 = 93$. Substituting both of these into the equation gives the new equaiton $\frac{dy}{dt}=ky$ which has the solution $y(t) = y(0)e^{kt}=104e^{kt}$. Evaluating the function at the point gives $93=104e^{4k}$ and isolating the exponential gives $\frac{93}{104}=e^{4k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{93}{104} \right)}}{4}$.  Substuting $k$ back into the equation gives $y(t) = 104e^{\frac{\ln{\left(\frac{93}{104} \right)}}{4}t}$ and simplifying gives $y(t) = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}}$. Substituting out $y(t)$ gives 
\begin{equation*}T(t)-71 = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} \implies\, T(t)= 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} + 71 \end{equation*}Using $t = 9$ gives $T =104 \left(\frac{93}{104}\right)^{\frac{9}{4}} + 71\approx 152.00^\circ$}

\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.}

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HTML for Canvas

<p> <p>A coffee with temperature  <img class="equation_image" title=" \displaystyle 175^\circ " src="/equation_images/%20%5Cdisplaystyle%20175%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 175^\circ " data-equation-content=" \displaystyle 175^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 71^\circ " src="/equation_images/%20%5Cdisplaystyle%2071%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 71^\circ " data-equation-content=" \displaystyle 71^\circ " /> . After 4 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 164^\circ " src="/equation_images/%20%5Cdisplaystyle%20164%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 164^\circ " data-equation-content=" \displaystyle 164^\circ " /> , what is the temperature of the coffee after 9 minutes?</p> </p>

HTML for Canvas

<p> <p>Newton's law of Cooling states that the change in temperature is directly proportional to the difference between the object's temperature and its surroundings.  <img class="equation_image" title="  \frac{dT}{dt} = k(T(t)-T_{\text{room}}) " src="/equation_images/%20%20%5Cfrac%7BdT%7D%7Bdt%7D%20%3D%20k%28T%28t%29-T_%7B%5Ctext%7Broom%7D%7D%29%20" alt="LaTeX:   \frac{dT}{dt} = k(T(t)-T_{\text{room}}) " data-equation-content="  \frac{dT}{dt} = k(T(t)-T_{\text{room}}) " /> Using the substitution  <img class="equation_image" title=" \displaystyle y(t)=T(t)-71 " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%3DT%28t%29-71%20" alt="LaTeX:  \displaystyle y(t)=T(t)-71 " data-equation-content=" \displaystyle y(t)=T(t)-71 " />  and calculating the derivative gives  <img class="equation_image" title=" \displaystyle \frac{dy}{dt}=\frac{dT}{dt} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Cfrac%7BdT%7D%7Bdt%7D%20" alt="LaTeX:  \displaystyle \frac{dy}{dt}=\frac{dT}{dt} " data-equation-content=" \displaystyle \frac{dy}{dt}=\frac{dT}{dt} " /> . Calculating the new initial condition using the point  <img class="equation_image" title=" \displaystyle (4, 164) " src="/equation_images/%20%5Cdisplaystyle%20%284%2C%20164%29%20" alt="LaTeX:  \displaystyle (4, 164) " data-equation-content=" \displaystyle (4, 164) " />  and the substition gives  <img class="equation_image" title=" \displaystyle y(0) = T(0)-71 = 104 " src="/equation_images/%20%5Cdisplaystyle%20y%280%29%20%3D%20T%280%29-71%20%3D%20104%20" alt="LaTeX:  \displaystyle y(0) = T(0)-71 = 104 " data-equation-content=" \displaystyle y(0) = T(0)-71 = 104 " /> . The point  <img class="equation_image" title=" \displaystyle (4, 164) " src="/equation_images/%20%5Cdisplaystyle%20%284%2C%20164%29%20" alt="LaTeX:  \displaystyle (4, 164) " data-equation-content=" \displaystyle (4, 164) " />  must also be transformed to get  <img class="equation_image" title=" \displaystyle y(4) = T(4)-71 = 164 - 71 = 93 " src="/equation_images/%20%5Cdisplaystyle%20y%284%29%20%3D%20T%284%29-71%20%3D%20164%20-%2071%20%3D%2093%20" alt="LaTeX:  \displaystyle y(4) = T(4)-71 = 164 - 71 = 93 " data-equation-content=" \displaystyle y(4) = T(4)-71 = 164 - 71 = 93 " /> . Substituting both of these into the equation gives the new equaiton  <img class="equation_image" title=" \displaystyle \frac{dy}{dt}=ky " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdt%7D%3Dky%20" alt="LaTeX:  \displaystyle \frac{dy}{dt}=ky " data-equation-content=" \displaystyle \frac{dy}{dt}=ky " />  which has the solution  <img class="equation_image" title=" \displaystyle y(t) = y(0)e^{kt}=104e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20y%280%29e%5E%7Bkt%7D%3D104e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle y(t) = y(0)e^{kt}=104e^{kt} " data-equation-content=" \displaystyle y(t) = y(0)e^{kt}=104e^{kt} " /> . Evaluating the function at the point gives  <img class="equation_image" title=" \displaystyle 93=104e^{4k} " src="/equation_images/%20%5Cdisplaystyle%2093%3D104e%5E%7B4k%7D%20" alt="LaTeX:  \displaystyle 93=104e^{4k} " data-equation-content=" \displaystyle 93=104e^{4k} " />  and isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{93}{104}=e^{4k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B93%7D%7B104%7D%3De%5E%7B4k%7D%20" alt="LaTeX:  \displaystyle \frac{93}{104}=e^{4k} " data-equation-content=" \displaystyle \frac{93}{104}=e^{4k} " /> . 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{93}{104} \right)}}{4} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B93%7D%7B104%7D%20%5Cright%29%7D%7D%7B4%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{93}{104} \right)}}{4} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{93}{104} \right)}}{4} " /> .  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 y(t) = 104e^{\frac{\ln{\left(\frac{93}{104} \right)}}{4}t} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20104e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B93%7D%7B104%7D%20%5Cright%29%7D%7D%7B4%7Dt%7D%20" alt="LaTeX:  \displaystyle y(t) = 104e^{\frac{\ln{\left(\frac{93}{104} \right)}}{4}t} " data-equation-content=" \displaystyle y(t) = 104e^{\frac{\ln{\left(\frac{93}{104} \right)}}{4}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle y(t) = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20104%20%5Cleft%28%5Cfrac%7B93%7D%7B104%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B4%7D%7D%20" alt="LaTeX:  \displaystyle y(t) = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} " data-equation-content=" \displaystyle y(t) = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} " /> . Substituting out  <img class="equation_image" title=" \displaystyle y(t) " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20" alt="LaTeX:  \displaystyle y(t) " data-equation-content=" \displaystyle y(t) " />  gives 
 <img class="equation_image" title=" T(t)-71 = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} \implies\, T(t)= 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} + 71  " src="/equation_images/%20T%28t%29-71%20%3D%20104%20%5Cleft%28%5Cfrac%7B93%7D%7B104%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B4%7D%7D%20%5Cimplies%5C%2C%20T%28t%29%3D%20104%20%5Cleft%28%5Cfrac%7B93%7D%7B104%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B4%7D%7D%20%2B%2071%20%20" alt="LaTeX:  T(t)-71 = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} \implies\, T(t)= 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} + 71  " data-equation-content=" T(t)-71 = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} \implies\, T(t)= 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} + 71  " /> Using  <img class="equation_image" title=" \displaystyle t = 9 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%209%20" alt="LaTeX:  \displaystyle t = 9 " data-equation-content=" \displaystyle t = 9 " />  gives  <img class="equation_image" title=" \displaystyle T =104 \left(\frac{93}{104}\right)^{\frac{9}{4}} + 71\approx 152.00^\circ " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D104%20%5Cleft%28%5Cfrac%7B93%7D%7B104%7D%5Cright%29%5E%7B%5Cfrac%7B9%7D%7B4%7D%7D%20%2B%2071%5Capprox%20152.00%5E%5Ccirc%20" alt="LaTeX:  \displaystyle T =104 \left(\frac{93}{104}\right)^{\frac{9}{4}} + 71\approx 152.00^\circ " data-equation-content=" \displaystyle T =104 \left(\frac{93}{104}\right)^{\frac{9}{4}} + 71\approx 152.00^\circ " /> </p> </p>