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Questions: Algebra BusinessCalculus

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Calculus
Applications of Derivatives
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A coffee with temperature \(\displaystyle 175^\circ\) is left in a room with temperature \(\displaystyle 77^\circ\). After 12 minutes the temperature of the coffee is \(\displaystyle 157^\circ\), how long until the coffee is \(\displaystyle 149^\circ\)?


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 \(\displaystyle y(t)=T(t)-77\) and calculating the derivative gives \(\displaystyle \frac{dy}{dt}=\frac{dT}{dt}\). Calculating the new initial condition using the point \(\displaystyle (12, 157)\) and the substition gives \(\displaystyle y(0) = T(0)-77 = 98\). The point \(\displaystyle (12, 157)\) must also be transformed to get \(\displaystyle y(12) = T(12)-77 = 157 - 77 = 80\). Substituting both of these into the equation gives the new equaiton \(\displaystyle \frac{dy}{dt}=ky\) which has the solution \(\displaystyle y(t) = y(0)e^{kt}=98e^{kt}\). Evaluating the function at the point gives \(\displaystyle 80=98e^{12k}\) and isolating the exponential gives \(\displaystyle \frac{40}{49}=e^{12k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{40}{49} \right)}}{12}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle y(t) = 98e^{\frac{\ln{\left(\frac{40}{49} \right)}}{12}t}\) and simplifying gives \(\displaystyle y(t) = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}}\). Substituting out \(\displaystyle y(t)\) gives \begin{equation*}T(t)-77 = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} \implies\, T(t)= 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77 \end{equation*}Using \(\displaystyle T\) gives the equation \(\displaystyle 149=98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77\). Isolating the exponential gives \(\displaystyle \frac{36}{49}=\left(\frac{40}{49}\right)^{\frac{t}{12}}\). Taking the natural logarithm of both sides and solving for \(\displaystyle t\) gives \(\displaystyle t = \frac{12 \ln{\left(\frac{36}{49} \right)}}{\ln{\left(\frac{40}{49} \right)}}\approx 18.2\) minutes.

Download \(\LaTeX\)

\begin{question}A coffee with temperature $175^\circ$ is left in a room with temperature $77^\circ$. After 12 minutes the temperature of the coffee is $157^\circ$, how long until the coffee is $149^\circ$?
    \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 $y(t)=T(t)-77$ and calculating the derivative gives $\frac{dy}{dt}=\frac{dT}{dt}$. Calculating the new initial condition using the point $(12, 157)$ and the substition gives $y(0) = T(0)-77 = 98$. The point $(12, 157)$ must also be transformed to get $y(12) = T(12)-77 = 157 - 77 = 80$. Substituting both of these into the equation gives the new equaiton $\frac{dy}{dt}=ky$ which has the solution $y(t) = y(0)e^{kt}=98e^{kt}$. Evaluating the function at the point gives $80=98e^{12k}$ and isolating the exponential gives $\frac{40}{49}=e^{12k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{40}{49} \right)}}{12}$.  Substuting $k$ back into the equation gives $y(t) = 98e^{\frac{\ln{\left(\frac{40}{49} \right)}}{12}t}$ and simplifying gives $y(t) = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}}$. Substituting out $y(t)$ gives 
\begin{equation*}T(t)-77 = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} \implies\, T(t)= 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77 \end{equation*}Using $T$ gives the equation $149=98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77$.  Isolating the exponential gives $\frac{36}{49}=\left(\frac{40}{49}\right)^{\frac{t}{12}}$. Taking the natural logarithm of both sides and solving for $t$ gives $t = \frac{12 \ln{\left(\frac{36}{49} \right)}}{\ln{\left(\frac{40}{49} \right)}}\approx 18.2$ minutes. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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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 77^\circ " src="/equation_images/%20%5Cdisplaystyle%2077%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 77^\circ " data-equation-content=" \displaystyle 77^\circ " /> . After 12 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 157^\circ " src="/equation_images/%20%5Cdisplaystyle%20157%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 157^\circ " data-equation-content=" \displaystyle 157^\circ " /> , how long until the coffee is  <img class="equation_image" title=" \displaystyle 149^\circ " src="/equation_images/%20%5Cdisplaystyle%20149%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 149^\circ " data-equation-content=" \displaystyle 149^\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)-77 " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%3DT%28t%29-77%20" alt="LaTeX:  \displaystyle y(t)=T(t)-77 " data-equation-content=" \displaystyle y(t)=T(t)-77 " />  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 (12, 157) " src="/equation_images/%20%5Cdisplaystyle%20%2812%2C%20157%29%20" alt="LaTeX:  \displaystyle (12, 157) " data-equation-content=" \displaystyle (12, 157) " />  and the substition gives  <img class="equation_image" title=" \displaystyle y(0) = T(0)-77 = 98 " src="/equation_images/%20%5Cdisplaystyle%20y%280%29%20%3D%20T%280%29-77%20%3D%2098%20" alt="LaTeX:  \displaystyle y(0) = T(0)-77 = 98 " data-equation-content=" \displaystyle y(0) = T(0)-77 = 98 " /> . The point  <img class="equation_image" title=" \displaystyle (12, 157) " src="/equation_images/%20%5Cdisplaystyle%20%2812%2C%20157%29%20" alt="LaTeX:  \displaystyle (12, 157) " data-equation-content=" \displaystyle (12, 157) " />  must also be transformed to get  <img class="equation_image" title=" \displaystyle y(12) = T(12)-77 = 157 - 77 = 80 " src="/equation_images/%20%5Cdisplaystyle%20y%2812%29%20%3D%20T%2812%29-77%20%3D%20157%20-%2077%20%3D%2080%20" alt="LaTeX:  \displaystyle y(12) = T(12)-77 = 157 - 77 = 80 " data-equation-content=" \displaystyle y(12) = T(12)-77 = 157 - 77 = 80 " /> . 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}=98e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20y%280%29e%5E%7Bkt%7D%3D98e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle y(t) = y(0)e^{kt}=98e^{kt} " data-equation-content=" \displaystyle y(t) = y(0)e^{kt}=98e^{kt} " /> . Evaluating the function at the point gives  <img class="equation_image" title=" \displaystyle 80=98e^{12k} " src="/equation_images/%20%5Cdisplaystyle%2080%3D98e%5E%7B12k%7D%20" alt="LaTeX:  \displaystyle 80=98e^{12k} " data-equation-content=" \displaystyle 80=98e^{12k} " />  and isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{40}{49}=e^{12k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B40%7D%7B49%7D%3De%5E%7B12k%7D%20" alt="LaTeX:  \displaystyle \frac{40}{49}=e^{12k} " data-equation-content=" \displaystyle \frac{40}{49}=e^{12k} " /> . 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{40}{49} \right)}}{12} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B40%7D%7B49%7D%20%5Cright%29%7D%7D%7B12%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{40}{49} \right)}}{12} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{40}{49} \right)}}{12} " /> .  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) = 98e^{\frac{\ln{\left(\frac{40}{49} \right)}}{12}t} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%2098e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B40%7D%7B49%7D%20%5Cright%29%7D%7D%7B12%7Dt%7D%20" alt="LaTeX:  \displaystyle y(t) = 98e^{\frac{\ln{\left(\frac{40}{49} \right)}}{12}t} " data-equation-content=" \displaystyle y(t) = 98e^{\frac{\ln{\left(\frac{40}{49} \right)}}{12}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle y(t) = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%2098%20%5Cleft%28%5Cfrac%7B40%7D%7B49%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B12%7D%7D%20" alt="LaTeX:  \displaystyle y(t) = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} " data-equation-content=" \displaystyle y(t) = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} " /> . 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)-77 = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} \implies\, T(t)= 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77  " src="/equation_images/%20T%28t%29-77%20%3D%2098%20%5Cleft%28%5Cfrac%7B40%7D%7B49%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B12%7D%7D%20%5Cimplies%5C%2C%20T%28t%29%3D%2098%20%5Cleft%28%5Cfrac%7B40%7D%7B49%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B12%7D%7D%20%2B%2077%20%20" alt="LaTeX:  T(t)-77 = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} \implies\, T(t)= 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77  " data-equation-content=" T(t)-77 = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} \implies\, T(t)= 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77  " /> 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 149=98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77 " src="/equation_images/%20%5Cdisplaystyle%20149%3D98%20%5Cleft%28%5Cfrac%7B40%7D%7B49%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B12%7D%7D%20%2B%2077%20" alt="LaTeX:  \displaystyle 149=98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77 " data-equation-content=" \displaystyle 149=98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77 " /> .  Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{36}{49}=\left(\frac{40}{49}\right)^{\frac{t}{12}} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B36%7D%7B49%7D%3D%5Cleft%28%5Cfrac%7B40%7D%7B49%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B12%7D%7D%20" alt="LaTeX:  \displaystyle \frac{36}{49}=\left(\frac{40}{49}\right)^{\frac{t}{12}} " data-equation-content=" \displaystyle \frac{36}{49}=\left(\frac{40}{49}\right)^{\frac{t}{12}} " /> . 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{12 \ln{\left(\frac{36}{49} \right)}}{\ln{\left(\frac{40}{49} \right)}}\approx 18.2 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20%5Cfrac%7B12%20%5Cln%7B%5Cleft%28%5Cfrac%7B36%7D%7B49%7D%20%5Cright%29%7D%7D%7B%5Cln%7B%5Cleft%28%5Cfrac%7B40%7D%7B49%7D%20%5Cright%29%7D%7D%5Capprox%2018.2%20" alt="LaTeX:  \displaystyle t = \frac{12 \ln{\left(\frac{36}{49} \right)}}{\ln{\left(\frac{40}{49} \right)}}\approx 18.2 " data-equation-content=" \displaystyle t = \frac{12 \ln{\left(\frac{36}{49} \right)}}{\ln{\left(\frac{40}{49} \right)}}\approx 18.2 " />  minutes. </p> </p>