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A jet flies directly over a radar station at a constant height of 2km at a constant speed of 150km/hr. How fast is the angle of elevation of the jet changing when the horizontal distance from the radar station is 7km?
Let \(\displaystyle x\) be the plane's horizontal distance from the station and let \(\displaystyle h\) be the height of the plane. Drawing a diagram gives From the diagram \(\displaystyle \theta = \tan^{-1}\left( \frac{h}{x} \right)\). Taking the derivative gives: (don't forget that \(\displaystyle h\) is constant) \begin{equation*}\frac{ d\theta }{ dt } = \frac{1}{1+\left( \frac{h}{x}\right)^2}\cdot \left(-\frac{h}{x^2}\right)\left( \frac{dx}{dt} \right) = -\frac{h}{x^2+h^2}\frac{dx}{dt}\end{equation*}Evaluating at the given values gives \begin{equation*}\frac{d\theta}{dt} = -\frac{ 2 }{ 7^2+2^2}\cdot (150) = - \frac{300}{53} = -5.66\end{equation*} Note: The solution is negative because counter-clockwise is a positive rotation for angles and the units are radians per hour.
\begin{question}A jet flies directly over a radar station at a constant height of 2km at a constant speed of 150km/hr. How fast is the angle of elevation of the jet changing when the horizontal distance from the radar station is 7km?
\soln{9cm}{Let $x$ be the plane's horizontal distance from the station and let $h$ be the height of the plane. Drawing a diagram gives
\begin{center}
\begin{tikzpicture}
\draw(0, 0) -- (4, 0) node[midway, below]{$x$} -- (4, 3) node[midway, right]{$h$} -- (0, 0);
\draw (0.8, 0) arc (0:atan(3/4):0.8);
\draw (.8, .3) node[right]{$\theta$};
\end{tikzpicture}
\end{center}
From the diagram $\theta = \tan^{-1}\left( \frac{h}{x} \right)$. Taking the derivative gives: (don't forget that $h$ is constant)
\begin{equation*}\frac{ d\theta }{ dt } = \frac{1}{1+\left( \frac{h}{x}\right)^2}\cdot \left(-\frac{h}{x^2}\right)\left( \frac{dx}{dt} \right) = -\frac{h}{x^2+h^2}\frac{dx}{dt}\end{equation*}Evaluating at the given values gives \begin{equation*}\frac{d\theta}{dt} = -\frac{ 2 }{ 7^2+2^2}\cdot (150) = - \frac{300}{53} = -5.66\end{equation*}
Note: The solution is negative because counter-clockwise is a positive rotation for angles and the units are radians per hour. }
\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 jet flies directly over a radar station at a constant height of 2km at a constant speed of 150km/hr. How fast is the angle of elevation of the jet changing when the horizontal distance from the radar station is 7km?</p> </p>
<p> <p>Let <img class="equation_image" title=" \displaystyle x " src="/equation_images/%20%5Cdisplaystyle%20x%20" alt="LaTeX: \displaystyle x " data-equation-content=" \displaystyle x " /> be the plane's horizontal distance from the station and let <img class="equation_image" title=" \displaystyle h " src="/equation_images/%20%5Cdisplaystyle%20h%20" alt="LaTeX: \displaystyle h " data-equation-content=" \displaystyle h " /> be the height of the plane. Drawing a diagram gives<?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">
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From the diagram <img class="equation_image" title=" \displaystyle \theta = \tan^{-1}\left( \frac{h}{x} \right) " src="/equation_images/%20%5Cdisplaystyle%20%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7Bh%7D%7Bx%7D%20%5Cright%29%20" alt="LaTeX: \displaystyle \theta = \tan^{-1}\left( \frac{h}{x} \right) " data-equation-content=" \displaystyle \theta = \tan^{-1}\left( \frac{h}{x} \right) " /> . Taking the derivative gives: (don't forget that <img class="equation_image" title=" \displaystyle h " src="/equation_images/%20%5Cdisplaystyle%20h%20" alt="LaTeX: \displaystyle h " data-equation-content=" \displaystyle h " /> is constant)
<img class="equation_image" title=" \frac{ d\theta }{ dt } = \frac{1}{1+\left( \frac{h}{x}\right)^2}\cdot \left(-\frac{h}{x^2}\right)\left( \frac{dx}{dt} \right) = -\frac{h}{x^2+h^2}\frac{dx}{dt} " src="/equation_images/%20%5Cfrac%7B%20d%5Ctheta%20%7D%7B%20dt%20%7D%20%3D%20%5Cfrac%7B1%7D%7B1%2B%5Cleft%28%20%5Cfrac%7Bh%7D%7Bx%7D%5Cright%29%5E2%7D%5Ccdot%20%5Cleft%28-%5Cfrac%7Bh%7D%7Bx%5E2%7D%5Cright%29%5Cleft%28%20%5Cfrac%7Bdx%7D%7Bdt%7D%20%5Cright%29%20%3D%20-%5Cfrac%7Bh%7D%7Bx%5E2%2Bh%5E2%7D%5Cfrac%7Bdx%7D%7Bdt%7D%20" alt="LaTeX: \frac{ d\theta }{ dt } = \frac{1}{1+\left( \frac{h}{x}\right)^2}\cdot \left(-\frac{h}{x^2}\right)\left( \frac{dx}{dt} \right) = -\frac{h}{x^2+h^2}\frac{dx}{dt} " data-equation-content=" \frac{ d\theta }{ dt } = \frac{1}{1+\left( \frac{h}{x}\right)^2}\cdot \left(-\frac{h}{x^2}\right)\left( \frac{dx}{dt} \right) = -\frac{h}{x^2+h^2}\frac{dx}{dt} " /> Evaluating at the given values gives <img class="equation_image" title=" \frac{d\theta}{dt} = -\frac{ 2 }{ 7^2+2^2}\cdot (150) = - \frac{300}{53} = -5.66 " src="/equation_images/%20%5Cfrac%7Bd%5Ctheta%7D%7Bdt%7D%20%20%3D%20-%5Cfrac%7B%202%20%7D%7B%207%5E2%2B2%5E2%7D%5Ccdot%20%28150%29%20%3D%20-%20%5Cfrac%7B300%7D%7B53%7D%20%3D%20-5.66%20" alt="LaTeX: \frac{d\theta}{dt} = -\frac{ 2 }{ 7^2+2^2}\cdot (150) = - \frac{300}{53} = -5.66 " data-equation-content=" \frac{d\theta}{dt} = -\frac{ 2 }{ 7^2+2^2}\cdot (150) = - \frac{300}{53} = -5.66 " />
Note: The solution is negative because counter-clockwise is a positive rotation for angles and the units are radians per hour. </p> </p>