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A car and a truck leave the same place at the same time. The car drives North at 75 miles per hour and the truck drives west at 70 miles per hour. How fast is the distance changing between the car and the truck after 3 hours?
Drawing a diagram gives:
\(\displaystyle c\) can be found using the Pythagorean theorem \(\displaystyle c=\sqrt{(225)^2+(210)^2}=15 \sqrt{421}\). Using the Pythagoren Theorem and implicit differentiation gives \(\displaystyle c^2=a^2+b^2 \iff 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}\). Solving for the unknown rate gives \(\displaystyle \frac{dc}{dt}=\frac{a}{c}\frac{da}{dt}+\frac{b}{c}\frac{db}{dt}\).Evaluating gives \(\displaystyle \frac{dc}{dt}=5 \sqrt{421}\)
\begin{tikzpicture}[scale=1.25, font=\Large]
\draw[-latex] (0,0)--(0,7.5) node[midway,left]{car $a$};
\draw[-latex] (0,0)--(7.0,0) node[midway,below]{truck $b$};
\draw[dashed] (7.0,0)--(0,7.5) node[midway,above]{$c$};
\draw (9.0,3.75) node[text width = 10cm, right]{\begin{tabular}{cc}
$a=225$ & $\frac{da}{dt}=75$ \\[5mm]
$b=210$ & $\frac{db}{dt}=70$ \\[5mm]
$c= $ & $\frac{dc}{dt}=$want \\[5mm]
\end{tabular}\newline
};
\end{tikzpicture}
\begin{question}A car and a truck leave the same place at the same time. The car drives North at 75 miles per hour and the truck drives west at 70 miles per hour. How fast is the distance changing between the car and the truck after 3 hours?
\soln{9cm}{Drawing a diagram gives:\newline\begin{tikzpicture}[scale=1.25, font=\Large]
\draw[-latex] (0,0)--(0,7.5) node[midway,left]{car $a$};
\draw[-latex] (0,0)--(7.0,0) node[midway,below]{truck $b$};
\draw[dashed] (7.0,0)--(0,7.5) node[midway,above]{$c$};
\draw (9.0,3.75) node[text width = 10cm, right]{\begin{tabular}{cc}
$a=225$ & $\frac{da}{dt}=75$ \\[5mm]
$b=210$ & $\frac{db}{dt}=70$ \\[5mm]
$c= $ & $\frac{dc}{dt}=$want \\[5mm]
\end{tabular}\newline
};
\end{tikzpicture}
$c$ can be found using the Pythagorean theorem $c=\sqrt{(225)^2+(210)^2}=15 \sqrt{421}$. Using the Pythagoren Theorem and implicit differentiation gives $c^2=a^2+b^2 \iff 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}$. Solving for the unknown rate gives $\frac{dc}{dt}=\frac{a}{c}\frac{da}{dt}+\frac{b}{c}\frac{db}{dt}$.Evaluating gives $\frac{dc}{dt}=5 \sqrt{421}$}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>A car and a truck leave the same place at the same time. The car drives North at 75 miles per hour and the truck drives west at 70 miles per hour. How fast is the distance changing between the car and the truck after 3 hours? </p> </p>
<p> <p>Drawing a diagram gives:<br><?xml version="1.0" encoding="UTF-8"?>
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<img class="equation_image" title=" \displaystyle c " src="/equation_images/%20%5Cdisplaystyle%20c%20" alt="LaTeX: \displaystyle c " data-equation-content=" \displaystyle c " /> can be found using the Pythagorean theorem <img class="equation_image" title=" \displaystyle c=\sqrt{(225)^2+(210)^2}=15 \sqrt{421} " src="/equation_images/%20%5Cdisplaystyle%20c%3D%5Csqrt%7B%28225%29%5E2%2B%28210%29%5E2%7D%3D15%20%5Csqrt%7B421%7D%20" alt="LaTeX: \displaystyle c=\sqrt{(225)^2+(210)^2}=15 \sqrt{421} " data-equation-content=" \displaystyle c=\sqrt{(225)^2+(210)^2}=15 \sqrt{421} " /> . Using the Pythagoren Theorem and implicit differentiation gives <img class="equation_image" title=" \displaystyle c^2=a^2+b^2 \iff 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt} " src="/equation_images/%20%5Cdisplaystyle%20c%5E2%3Da%5E2%2Bb%5E2%20%5Ciff%202c%5Cfrac%7Bdc%7D%7Bdt%7D%3D2a%5Cfrac%7Bda%7D%7Bdt%7D%2B2b%5Cfrac%7Bdb%7D%7Bdt%7D%20" alt="LaTeX: \displaystyle c^2=a^2+b^2 \iff 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt} " data-equation-content=" \displaystyle c^2=a^2+b^2 \iff 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt} " /> . Solving for the unknown rate gives <img class="equation_image" title=" \displaystyle \frac{dc}{dt}=\frac{a}{c}\frac{da}{dt}+\frac{b}{c}\frac{db}{dt} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdc%7D%7Bdt%7D%3D%5Cfrac%7Ba%7D%7Bc%7D%5Cfrac%7Bda%7D%7Bdt%7D%2B%5Cfrac%7Bb%7D%7Bc%7D%5Cfrac%7Bdb%7D%7Bdt%7D%20" alt="LaTeX: \displaystyle \frac{dc}{dt}=\frac{a}{c}\frac{da}{dt}+\frac{b}{c}\frac{db}{dt} " data-equation-content=" \displaystyle \frac{dc}{dt}=\frac{a}{c}\frac{da}{dt}+\frac{b}{c}\frac{db}{dt} " /> .Evaluating gives <img class="equation_image" title=" \displaystyle \frac{dc}{dt}=5 \sqrt{421} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdc%7D%7Bdt%7D%3D5%20%5Csqrt%7B421%7D%20" alt="LaTeX: \displaystyle \frac{dc}{dt}=5 \sqrt{421} " data-equation-content=" \displaystyle \frac{dc}{dt}=5 \sqrt{421} " /> </p> </p>