Generate Math Exams and Quizzes in LaTeX and pdf formats

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 \(\displaystyle y(t)=T(t)-60\) and calculating the derivative gives \(\displaystyle \frac{dy}{dt}=\frac{dT}{dt}\). Calculating the new initial condition using the point \(\displaystyle (6, 140)\) and the substition gives \(\displaystyle y(0) = T(0)-60 = 90\). The point \(\displaystyle (6, 140)\) must also be transformed to get \(\displaystyle y(6) = T(6)-60 = 140 - 60 = 80\). Substituting both of these into the equation gives the new equaiton \(\displaystyle \frac{dy}{dt}=ky\) which has the solution \(\displaystyle y(t) = y(0)e^{kt}=90e^{kt}\). Evaluating the function at the point gives \(\displaystyle 80=90e^{6k}\) and isolating the exponential gives \(\displaystyle \frac{8}{9}=e^{6k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{8}{9} \right)}}{6}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle y(t) = 90e^{\frac{\ln{\left(\frac{8}{9} \right)}}{6}t}\) and simplifying gives \(\displaystyle y(t) = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}}\). Substituting out \(\displaystyle y(t)\) gives \begin{equation*}T(t)-60 = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} \implies\, T(t)= 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60 \end{equation*}Using \(\displaystyle T\) gives the equation \(\displaystyle 118=90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60\). Isolating the exponential gives \(\displaystyle \frac{29}{45}=\left(\frac{8}{9}\right)^{\frac{t}{6}}\). Taking the natural logarithm of both sides and solving for \(\displaystyle t\) gives \(\displaystyle t = \frac{6 \ln{\left(\frac{29}{45} \right)}}{\ln{\left(\frac{8}{9} \right)}}\approx 22.4\) minutes.

Download \(\LaTeX\)

\begin{question}A coffee with temperature $150^\circ$ is left in a room with temperature $60^\circ$. After 6 minutes the temperature of the coffee is $140^\circ$, how long until the coffee is $118^\circ$?
    \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 $y(t)=T(t)-60$ and calculating the derivative gives $\frac{dy}{dt}=\frac{dT}{dt}$. Calculating the new initial condition using the point $(6, 140)$ and the substition gives $y(0) = T(0)-60 = 90$. The point $(6, 140)$ must also be transformed to get $y(6) = T(6)-60 = 140 - 60 = 80$. Substituting both of these into the equation gives the new equaiton $\frac{dy}{dt}=ky$ which has the solution $y(t) = y(0)e^{kt}=90e^{kt}$. Evaluating the function at the point gives $80=90e^{6k}$ and isolating the exponential gives $\frac{8}{9}=e^{6k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{8}{9} \right)}}{6}$.  Substuting $k$ back into the equation gives $y(t) = 90e^{\frac{\ln{\left(\frac{8}{9} \right)}}{6}t}$ and simplifying gives $y(t) = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}}$. Substituting out $y(t)$ gives 
\begin{equation*}T(t)-60 = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} \implies\, T(t)= 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60 \end{equation*}Using $T$ gives the equation $118=90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60$.  Isolating the exponential gives $\frac{29}{45}=\left(\frac{8}{9}\right)^{\frac{t}{6}}$. Taking the natural logarithm of both sides and solving for $t$ gives $t = \frac{6 \ln{\left(\frac{29}{45} \right)}}{\ln{\left(\frac{8}{9} \right)}}\approx 22.4$ minutes. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)

\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}

\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}

\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}

\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue

\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 150^\circ " src="/equation_images/%20%5Cdisplaystyle%20150%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 150^\circ " data-equation-content=" \displaystyle 150^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 60^\circ " src="/equation_images/%20%5Cdisplaystyle%2060%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 60^\circ " data-equation-content=" \displaystyle 60^\circ " /> . After 6 minutes the temperature of 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 " /> , how long until the coffee is  <img class="equation_image" title=" \displaystyle 118^\circ " src="/equation_images/%20%5Cdisplaystyle%20118%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 118^\circ " data-equation-content=" \displaystyle 118^\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)-60 " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%3DT%28t%29-60%20" alt="LaTeX:  \displaystyle y(t)=T(t)-60 " data-equation-content=" \displaystyle y(t)=T(t)-60 " />  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 (6, 140) " src="/equation_images/%20%5Cdisplaystyle%20%286%2C%20140%29%20" alt="LaTeX:  \displaystyle (6, 140) " data-equation-content=" \displaystyle (6, 140) " />  and the substition gives  <img class="equation_image" title=" \displaystyle y(0) = T(0)-60 = 90 " src="/equation_images/%20%5Cdisplaystyle%20y%280%29%20%3D%20T%280%29-60%20%3D%2090%20" alt="LaTeX:  \displaystyle y(0) = T(0)-60 = 90 " data-equation-content=" \displaystyle y(0) = T(0)-60 = 90 " /> . The point  <img class="equation_image" title=" \displaystyle (6, 140) " src="/equation_images/%20%5Cdisplaystyle%20%286%2C%20140%29%20" alt="LaTeX:  \displaystyle (6, 140) " data-equation-content=" \displaystyle (6, 140) " />  must also be transformed to get  <img class="equation_image" title=" \displaystyle y(6) = T(6)-60 = 140 - 60 = 80 " src="/equation_images/%20%5Cdisplaystyle%20y%286%29%20%3D%20T%286%29-60%20%3D%20140%20-%2060%20%3D%2080%20" alt="LaTeX:  \displaystyle y(6) = T(6)-60 = 140 - 60 = 80 " data-equation-content=" \displaystyle y(6) = T(6)-60 = 140 - 60 = 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}=90e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20y%280%29e%5E%7Bkt%7D%3D90e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle y(t) = y(0)e^{kt}=90e^{kt} " data-equation-content=" \displaystyle y(t) = y(0)e^{kt}=90e^{kt} " /> . Evaluating the function at the point gives  <img class="equation_image" title=" \displaystyle 80=90e^{6k} " src="/equation_images/%20%5Cdisplaystyle%2080%3D90e%5E%7B6k%7D%20" alt="LaTeX:  \displaystyle 80=90e^{6k} " data-equation-content=" \displaystyle 80=90e^{6k} " />  and isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{8}{9}=e^{6k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B8%7D%7B9%7D%3De%5E%7B6k%7D%20" alt="LaTeX:  \displaystyle \frac{8}{9}=e^{6k} " data-equation-content=" \displaystyle \frac{8}{9}=e^{6k} " /> . 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{8}{9} \right)}}{6} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B8%7D%7B9%7D%20%5Cright%29%7D%7D%7B6%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{8}{9} \right)}}{6} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{8}{9} \right)}}{6} " /> .  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) = 90e^{\frac{\ln{\left(\frac{8}{9} \right)}}{6}t} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%2090e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B8%7D%7B9%7D%20%5Cright%29%7D%7D%7B6%7Dt%7D%20" alt="LaTeX:  \displaystyle y(t) = 90e^{\frac{\ln{\left(\frac{8}{9} \right)}}{6}t} " data-equation-content=" \displaystyle y(t) = 90e^{\frac{\ln{\left(\frac{8}{9} \right)}}{6}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle y(t) = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%2090%20%5Cleft%28%5Cfrac%7B8%7D%7B9%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B6%7D%7D%20" alt="LaTeX:  \displaystyle y(t) = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} " data-equation-content=" \displaystyle y(t) = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} " /> . 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)-60 = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} \implies\, T(t)= 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60  " src="/equation_images/%20T%28t%29-60%20%3D%2090%20%5Cleft%28%5Cfrac%7B8%7D%7B9%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B6%7D%7D%20%5Cimplies%5C%2C%20T%28t%29%3D%2090%20%5Cleft%28%5Cfrac%7B8%7D%7B9%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B6%7D%7D%20%2B%2060%20%20" alt="LaTeX:  T(t)-60 = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} \implies\, T(t)= 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60  " data-equation-content=" T(t)-60 = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} \implies\, T(t)= 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60  " /> 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 118=90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60 " src="/equation_images/%20%5Cdisplaystyle%20118%3D90%20%5Cleft%28%5Cfrac%7B8%7D%7B9%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B6%7D%7D%20%2B%2060%20" alt="LaTeX:  \displaystyle 118=90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60 " data-equation-content=" \displaystyle 118=90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60 " /> .  Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{29}{45}=\left(\frac{8}{9}\right)^{\frac{t}{6}} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B29%7D%7B45%7D%3D%5Cleft%28%5Cfrac%7B8%7D%7B9%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B6%7D%7D%20" alt="LaTeX:  \displaystyle \frac{29}{45}=\left(\frac{8}{9}\right)^{\frac{t}{6}} " data-equation-content=" \displaystyle \frac{29}{45}=\left(\frac{8}{9}\right)^{\frac{t}{6}} " /> . 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{6 \ln{\left(\frac{29}{45} \right)}}{\ln{\left(\frac{8}{9} \right)}}\approx 22.4 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20%5Cfrac%7B6%20%5Cln%7B%5Cleft%28%5Cfrac%7B29%7D%7B45%7D%20%5Cright%29%7D%7D%7B%5Cln%7B%5Cleft%28%5Cfrac%7B8%7D%7B9%7D%20%5Cright%29%7D%7D%5Capprox%2022.4%20" alt="LaTeX:  \displaystyle t = \frac{6 \ln{\left(\frac{29}{45} \right)}}{\ln{\left(\frac{8}{9} \right)}}\approx 22.4 " data-equation-content=" \displaystyle t = \frac{6 \ln{\left(\frac{29}{45} \right)}}{\ln{\left(\frac{8}{9} \right)}}\approx 22.4 " />  minutes. </p> </p>