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Drawing a diagram gives:
Identifing \(\displaystyle \frac{db}{dt}=405\), \(\displaystyle a=1.1\), and \(\displaystyle c=2.9\). Since the diagram is a right trinagle we can use the Pythagoren Theorem to get \(\displaystyle (1.1)^2 + b^2 = c^2\). Take the derivative with respect to time gives \(\displaystyle 0 + 2b\frac{db}{dt} = 2c\frac{dc}{dt}\). Solving for \(\displaystyle \frac{dc}{dt}\) gives \(\displaystyle \frac{dc}{dt} = \frac{b}{c}\frac{db}{dt}\)To find \(\displaystyle \frac{dc}{dt}\) we need to calculate \(\displaystyle b\) when \(\displaystyle c = 2.9\). Using the Pythagoren Theorem gives \(\displaystyle b = \sqrt{2.9^2 - 1.1^2}\). Finally calculating the value of the derivative \(\displaystyle \frac{dc}{dt}=\frac{ \sqrt{2.9^2 - 1.1^2} }{ 2.9 }\cdot 405 \approx 374.7\) kilometers per hour.

Download \(\LaTeX\)

\begin{question}A plane is flying horizontally at an altitude of 1.1 kilometers with a velocity of 405 kilometers per hour when it flies over a radar station. Find the rate at which the distance is changing when the plane is 2.9 kilometers from the station. Round to the nearest tenth. 
    \soln{9cm}{Drawing a diagram gives:\newline

\begin{center}
\begin{tikzpicture}
	 \draw(0, 0) -- (0, 3) node[midway, left]{a = 1.1 km};
	 \draw(0, 3) -- (5, 3) node[midway, above]{b};
	 \draw[dashed] (5, 3) -- (0, 0) node[midway, below]{c};
\end{tikzpicture}
\end{center}

Identifing $\frac{db}{dt}=405$, $a=1.1$, and $c=2.9$. Since the diagram is a right trinagle we can use the Pythagoren Theorem to get $(1.1)^2 + b^2 = c^2$. Take the derivative with respect to time gives $0 + 2b\frac{db}{dt} = 2c\frac{dc}{dt}$. Solving for $\frac{dc}{dt}$ gives $\frac{dc}{dt} = \frac{b}{c}\frac{db}{dt}$To find $\frac{dc}{dt}$ we need to calculate $b$ when $c = 2.9$. Using the Pythagoren Theorem gives $b = \sqrt{2.9^2 - 1.1^2}$. Finally calculating the value of the derivative $\frac{dc}{dt}=\frac{ \sqrt{2.9^2 - 1.1^2} }{ 2.9 }\cdot 405 \approx 374.7$ kilometers per hour.}

\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 plane is flying horizontally at an altitude of 1.1 kilometers with a velocity of 405 kilometers per hour when it flies over a radar station. Find the rate at which the distance is changing when the plane is 2.9 kilometers from the station. Round to the nearest tenth. </p> </p>

HTML for Canvas

<p> <p>Drawing a diagram gives:<br>
<?xml version="1.0" encoding="UTF-8"?>
<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" width="612pt" height="792pt" viewBox="0 0 612 792" version="1.1">
<g id="surface1">
</g>
</svg>
Identifing  <img class="equation_image" title=" \displaystyle \frac{db}{dt}=405 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdb%7D%7Bdt%7D%3D405%20" alt="LaTeX:  \displaystyle \frac{db}{dt}=405 " data-equation-content=" \displaystyle \frac{db}{dt}=405 " /> ,  <img class="equation_image" title=" \displaystyle a=1.1 " src="/equation_images/%20%5Cdisplaystyle%20a%3D1.1%20" alt="LaTeX:  \displaystyle a=1.1 " data-equation-content=" \displaystyle a=1.1 " /> , and  <img class="equation_image" title=" \displaystyle c=2.9 " src="/equation_images/%20%5Cdisplaystyle%20c%3D2.9%20" alt="LaTeX:  \displaystyle c=2.9 " data-equation-content=" \displaystyle c=2.9 " /> . Since the diagram is a right trinagle we can use the Pythagoren Theorem to get  <img class="equation_image" title=" \displaystyle (1.1)^2 + b^2 = c^2 " src="/equation_images/%20%5Cdisplaystyle%20%281.1%29%5E2%20%2B%20b%5E2%20%3D%20c%5E2%20" alt="LaTeX:  \displaystyle (1.1)^2 + b^2 = c^2 " data-equation-content=" \displaystyle (1.1)^2 + b^2 = c^2 " /> . Take the derivative with respect to time gives  <img class="equation_image" title=" \displaystyle 0 + 2b\frac{db}{dt} = 2c\frac{dc}{dt} " src="/equation_images/%20%5Cdisplaystyle%200%20%2B%202b%5Cfrac%7Bdb%7D%7Bdt%7D%20%3D%202c%5Cfrac%7Bdc%7D%7Bdt%7D%20" alt="LaTeX:  \displaystyle 0 + 2b\frac{db}{dt} = 2c\frac{dc}{dt} " data-equation-content=" \displaystyle 0 + 2b\frac{db}{dt} = 2c\frac{dc}{dt} " /> . Solving for  <img class="equation_image" title=" \displaystyle \frac{dc}{dt} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdc%7D%7Bdt%7D%20" alt="LaTeX:  \displaystyle \frac{dc}{dt} " data-equation-content=" \displaystyle \frac{dc}{dt} " />  gives  <img class="equation_image" title=" \displaystyle \frac{dc}{dt} = \frac{b}{c}\frac{db}{dt} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdc%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bb%7D%7Bc%7D%5Cfrac%7Bdb%7D%7Bdt%7D%20" alt="LaTeX:  \displaystyle \frac{dc}{dt} = \frac{b}{c}\frac{db}{dt} " data-equation-content=" \displaystyle \frac{dc}{dt} = \frac{b}{c}\frac{db}{dt} " /> To find  <img class="equation_image" title=" \displaystyle \frac{dc}{dt} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdc%7D%7Bdt%7D%20" alt="LaTeX:  \displaystyle \frac{dc}{dt} " data-equation-content=" \displaystyle \frac{dc}{dt} " />  we need to calculate  <img class="equation_image" title=" \displaystyle b " src="/equation_images/%20%5Cdisplaystyle%20b%20" alt="LaTeX:  \displaystyle b " data-equation-content=" \displaystyle b " />  when  <img class="equation_image" title=" \displaystyle c = 2.9 " src="/equation_images/%20%5Cdisplaystyle%20c%20%3D%202.9%20" alt="LaTeX:  \displaystyle c = 2.9 " data-equation-content=" \displaystyle c = 2.9 " /> . Using the Pythagoren Theorem gives  <img class="equation_image" title=" \displaystyle b = \sqrt{2.9^2 - 1.1^2} " src="/equation_images/%20%5Cdisplaystyle%20b%20%3D%20%5Csqrt%7B2.9%5E2%20-%201.1%5E2%7D%20" alt="LaTeX:  \displaystyle b = \sqrt{2.9^2 - 1.1^2} " data-equation-content=" \displaystyle b = \sqrt{2.9^2 - 1.1^2} " /> . Finally calculating the value of the derivative  <img class="equation_image" title=" \displaystyle \frac{dc}{dt}=\frac{ \sqrt{2.9^2 - 1.1^2} }{ 2.9 }\cdot 405 \approx 374.7 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdc%7D%7Bdt%7D%3D%5Cfrac%7B%20%5Csqrt%7B2.9%5E2%20-%201.1%5E2%7D%20%7D%7B%202.9%20%7D%5Ccdot%20405%20%5Capprox%20374.7%20" alt="LaTeX:  \displaystyle \frac{dc}{dt}=\frac{ \sqrt{2.9^2 - 1.1^2} }{ 2.9 }\cdot 405 \approx 374.7 " data-equation-content=" \displaystyle \frac{dc}{dt}=\frac{ \sqrt{2.9^2 - 1.1^2} }{ 2.9 }\cdot 405 \approx 374.7 " />  kilometers per hour.</p> </p>