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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 \(\displaystyle y(t)=T(t)-56\) and calculating the derivative gives \(\displaystyle \frac{dy}{dt}=\frac{dT}{dt}\). Calculating the new initial condition using the point \(\displaystyle (15, 156)\) and the substition gives \(\displaystyle y(0) = T(0)-56 = 106\). The point \(\displaystyle (15, 156)\) must also be transformed to get \(\displaystyle y(15) = T(15)-56 = 156 - 56 = 100\). Substituting both of these into the equation gives the new equaiton \(\displaystyle \frac{dy}{dt}=ky\) which has the solution \(\displaystyle y(t) = y(0)e^{kt}=106e^{kt}\). Evaluating the function at the point gives \(\displaystyle 100=106e^{15k}\) and isolating the exponential gives \(\displaystyle \frac{50}{53}=e^{15k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{50}{53} \right)}}{15}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle y(t) = 106e^{\frac{\ln{\left(\frac{50}{53} \right)}}{15}t}\) and simplifying gives \(\displaystyle y(t) = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}}\). Substituting out \(\displaystyle y(t)\) gives \begin{equation*}T(t)-56 = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} \implies\, T(t)= 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56 \end{equation*}Using \(\displaystyle T\) gives the equation \(\displaystyle 140=106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56\). Isolating the exponential gives \(\displaystyle \frac{42}{53}=\left(\frac{50}{53}\right)^{\frac{t}{15}}\). Taking the natural logarithm of both sides and solving for \(\displaystyle t\) gives \(\displaystyle t = \frac{15 \ln{\left(\frac{42}{53} \right)}}{\ln{\left(\frac{50}{53} \right)}}\approx 59.9\) minutes.

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\begin{question}A coffee with temperature $162^\circ$ is left in a room with temperature $56^\circ$. After 15 minutes the temperature of the coffee is $156^\circ$, how long until the coffee is $140^\circ$?
    \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 $y(t)=T(t)-56$ and calculating the derivative gives $\frac{dy}{dt}=\frac{dT}{dt}$. Calculating the new initial condition using the point $(15, 156)$ and the substition gives $y(0) = T(0)-56 = 106$. The point $(15, 156)$ must also be transformed to get $y(15) = T(15)-56 = 156 - 56 = 100$. Substituting both of these into the equation gives the new equaiton $\frac{dy}{dt}=ky$ which has the solution $y(t) = y(0)e^{kt}=106e^{kt}$. Evaluating the function at the point gives $100=106e^{15k}$ and isolating the exponential gives $\frac{50}{53}=e^{15k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{50}{53} \right)}}{15}$.  Substuting $k$ back into the equation gives $y(t) = 106e^{\frac{\ln{\left(\frac{50}{53} \right)}}{15}t}$ and simplifying gives $y(t) = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}}$. Substituting out $y(t)$ gives 
\begin{equation*}T(t)-56 = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} \implies\, T(t)= 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56 \end{equation*}Using $T$ gives the equation $140=106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56$.  Isolating the exponential gives $\frac{42}{53}=\left(\frac{50}{53}\right)^{\frac{t}{15}}$. Taking the natural logarithm of both sides and solving for $t$ gives $t = \frac{15 \ln{\left(\frac{42}{53} \right)}}{\ln{\left(\frac{50}{53} \right)}}\approx 59.9$ minutes. }

\end{question}

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\begin{document}\begin{question}(10pts) The question goes here!
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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 56^\circ " src="/equation_images/%20%5Cdisplaystyle%2056%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 56^\circ " data-equation-content=" \displaystyle 56^\circ " /> . After 15 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 156^\circ " src="/equation_images/%20%5Cdisplaystyle%20156%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 156^\circ " data-equation-content=" \displaystyle 156^\circ " /> , how long until the coffee is  <img class="equation_image" title=" \displaystyle 140^\circ " src="/equation_images/%20%5Cdisplaystyle%20140%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 140^\circ " data-equation-content=" \displaystyle 140^\circ " /> ?</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)-56 " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%3DT%28t%29-56%20" alt="LaTeX:  \displaystyle y(t)=T(t)-56 " data-equation-content=" \displaystyle y(t)=T(t)-56 " />  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 (15, 156) " src="/equation_images/%20%5Cdisplaystyle%20%2815%2C%20156%29%20" alt="LaTeX:  \displaystyle (15, 156) " data-equation-content=" \displaystyle (15, 156) " />  and the substition gives  <img class="equation_image" title=" \displaystyle y(0) = T(0)-56 = 106 " src="/equation_images/%20%5Cdisplaystyle%20y%280%29%20%3D%20T%280%29-56%20%3D%20106%20" alt="LaTeX:  \displaystyle y(0) = T(0)-56 = 106 " data-equation-content=" \displaystyle y(0) = T(0)-56 = 106 " /> . The point  <img class="equation_image" title=" \displaystyle (15, 156) " src="/equation_images/%20%5Cdisplaystyle%20%2815%2C%20156%29%20" alt="LaTeX:  \displaystyle (15, 156) " data-equation-content=" \displaystyle (15, 156) " />  must also be transformed to get  <img class="equation_image" title=" \displaystyle y(15) = T(15)-56 = 156 - 56 = 100 " src="/equation_images/%20%5Cdisplaystyle%20y%2815%29%20%3D%20T%2815%29-56%20%3D%20156%20-%2056%20%3D%20100%20" alt="LaTeX:  \displaystyle y(15) = T(15)-56 = 156 - 56 = 100 " data-equation-content=" \displaystyle y(15) = T(15)-56 = 156 - 56 = 100 " /> . 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}=106e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20y%280%29e%5E%7Bkt%7D%3D106e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle y(t) = y(0)e^{kt}=106e^{kt} " data-equation-content=" \displaystyle y(t) = y(0)e^{kt}=106e^{kt} " /> . Evaluating the function at the point gives  <img class="equation_image" title=" \displaystyle 100=106e^{15k} " src="/equation_images/%20%5Cdisplaystyle%20100%3D106e%5E%7B15k%7D%20" alt="LaTeX:  \displaystyle 100=106e^{15k} " data-equation-content=" \displaystyle 100=106e^{15k} " />  and isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{50}{53}=e^{15k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B50%7D%7B53%7D%3De%5E%7B15k%7D%20" alt="LaTeX:  \displaystyle \frac{50}{53}=e^{15k} " data-equation-content=" \displaystyle \frac{50}{53}=e^{15k} " /> . 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{50}{53} \right)}}{15} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B50%7D%7B53%7D%20%5Cright%29%7D%7D%7B15%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{50}{53} \right)}}{15} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{50}{53} \right)}}{15} " /> .  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) = 106e^{\frac{\ln{\left(\frac{50}{53} \right)}}{15}t} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20106e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B50%7D%7B53%7D%20%5Cright%29%7D%7D%7B15%7Dt%7D%20" alt="LaTeX:  \displaystyle y(t) = 106e^{\frac{\ln{\left(\frac{50}{53} \right)}}{15}t} " data-equation-content=" \displaystyle y(t) = 106e^{\frac{\ln{\left(\frac{50}{53} \right)}}{15}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle y(t) = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20106%20%5Cleft%28%5Cfrac%7B50%7D%7B53%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B15%7D%7D%20" alt="LaTeX:  \displaystyle y(t) = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} " data-equation-content=" \displaystyle y(t) = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} " /> . 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)-56 = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} \implies\, T(t)= 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56  " src="/equation_images/%20T%28t%29-56%20%3D%20106%20%5Cleft%28%5Cfrac%7B50%7D%7B53%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B15%7D%7D%20%5Cimplies%5C%2C%20T%28t%29%3D%20106%20%5Cleft%28%5Cfrac%7B50%7D%7B53%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B15%7D%7D%20%2B%2056%20%20" alt="LaTeX:  T(t)-56 = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} \implies\, T(t)= 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56  " data-equation-content=" T(t)-56 = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} \implies\, T(t)= 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56  " /> Using  <img class="equation_image" title=" \displaystyle T " src="/equation_images/%20%5Cdisplaystyle%20T%20" alt="LaTeX:  \displaystyle T " data-equation-content=" \displaystyle T " />  gives the equation  <img class="equation_image" title=" \displaystyle 140=106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56 " src="/equation_images/%20%5Cdisplaystyle%20140%3D106%20%5Cleft%28%5Cfrac%7B50%7D%7B53%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B15%7D%7D%20%2B%2056%20" alt="LaTeX:  \displaystyle 140=106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56 " data-equation-content=" \displaystyle 140=106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56 " /> .  Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{42}{53}=\left(\frac{50}{53}\right)^{\frac{t}{15}} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B42%7D%7B53%7D%3D%5Cleft%28%5Cfrac%7B50%7D%7B53%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B15%7D%7D%20" alt="LaTeX:  \displaystyle \frac{42}{53}=\left(\frac{50}{53}\right)^{\frac{t}{15}} " data-equation-content=" \displaystyle \frac{42}{53}=\left(\frac{50}{53}\right)^{\frac{t}{15}} " /> . Taking the natural logarithm of both sides and solving for  <img class="equation_image" title=" \displaystyle t " src="/equation_images/%20%5Cdisplaystyle%20t%20" alt="LaTeX:  \displaystyle t " data-equation-content=" \displaystyle t " />  gives  <img class="equation_image" title=" \displaystyle t = \frac{15 \ln{\left(\frac{42}{53} \right)}}{\ln{\left(\frac{50}{53} \right)}}\approx 59.9 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20%5Cfrac%7B15%20%5Cln%7B%5Cleft%28%5Cfrac%7B42%7D%7B53%7D%20%5Cright%29%7D%7D%7B%5Cln%7B%5Cleft%28%5Cfrac%7B50%7D%7B53%7D%20%5Cright%29%7D%7D%5Capprox%2059.9%20" alt="LaTeX:  \displaystyle t = \frac{15 \ln{\left(\frac{42}{53} \right)}}{\ln{\left(\frac{50}{53} \right)}}\approx 59.9 " data-equation-content=" \displaystyle t = \frac{15 \ln{\left(\frac{42}{53} \right)}}{\ln{\left(\frac{50}{53} \right)}}\approx 59.9 " />  minutes. </p> </p>