A plane flying horizontally at an altitude of 1 mi

Mabel Breault 2021-12-30 Answered
A plane flying horizontally at an altitude of 1 mi and a speed of \(\displaystyle{500}{\frac{{{m}{i}}}{{{h}}}}\) passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.
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Expert Answer

Jenny Bolton
Answered 2021-12-31 Author has 5446 answers
Note that y is the distance from the plane to the station the problem is referring to.
image
Differentiate with respect to time t the equation that relates everything.
\(\displaystyle{x}^{{{2}}}+{1}^{{{2}}}={y}^{{{2}}}\)
\(\displaystyle{2}{x}{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}+{0}={2}{y}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}\)
\(\displaystyle{x}{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}\)
\(\displaystyle{\frac{{{x}}}{{{y}}}}{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}\)
We are given that a \(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={500}{\frac{{{m}{i}}}{{{h}}}}\). We also need to find x at the moment \(\displaystyle{y}={2}{m}{i}\)
\(\displaystyle{x}^{{{2}}}+{1}^{{{2}}}={2}^{{{2}}}\)
\(\displaystyle{x}^{{2}}={4}-{1}={3}\)
\(\displaystyle{x}=\sqrt{{{3}}}\) (ignore negative root)
Plug the values into the differentiated equation.
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={\frac{{{x}}}{{{y}}}}{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}\)
\(\displaystyle{\frac{{\sqrt{{{3}}}}}{{{2}}}}\cdot{500}\)
\(\displaystyle{250}\sqrt{{{3}}}{\frac{{{m}{i}}}{{{h}}}}\)
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levurdondishav4
Answered 2022-01-01 Author has 5417 answers
P is the plane's position
R is the radar station's position
V is the point located vertically of the radar station at the plane's height
h is the plane's height
d is the distance between the plane and the radar station
x is the distance between the plane and the V point
Since the plane flies horizontally, we can conclude that PVR is a right triangle.
Therefore, the pythagorean theorem allows us to know that d is calculated:
\(\displaystyle{d}=\sqrt{{{h}^{{2}}+{x}^{{2}}}}\)
We are interested in the situation when \(\displaystyle{d}={2}{m}{i}\), and, since the plane flies horizontally, we know that \(\displaystyle{h}={1}{m}{i}\) regardless of the situation.
We are looking for
\(\displaystyle{\frac{{{d}{d}}}{{{\left.{d}{t}\right.}}}}={d}\)
\(\displaystyle{d}^{{2}}={h}^{{2}}+{x}^{{2}}\)
\(\displaystyle\Rightarrow{\frac{{{d}{\left({d}^{{2}}\right)}}}{{{\left.{d}{t}\right.}}}}={\frac{{{d}{\left({d}^{{2}}\right)}}}{{{\left.{d}{t}\right.}}}}{\frac{{{d}{d}}}{{{\left.{d}{t}\right.}}}}={\frac{{{d}{\left({h}^{{2}}\right)}}}{{{d}{h}}}}{\frac{{{d}{h}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{d}{\left({x}^{{2}}\right)}}}{{{\left.{d}{x}\right.}}}}{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}\)
\(\displaystyle={2}{d}\dot{{{d}}}={2}{x}\dot{{{x}}}\)
\(\displaystyle\Rightarrow\dot{{{d}}}={\frac{{{2}{x}\dot{{{x}}}}}{{{2}{d}}}}={\frac{{{x}\dot{{{x}}}}}{{{d}}}}\)
We can calculate that, when \(\displaystyle{d}={2}{m}{i}\):
\(\displaystyle{x}=\sqrt{{{d}^{{2}}-{h}^{{2}}}}=\sqrt{{{2}^{{2}}-{1}^{{2}}}}=\sqrt{{{3}}}{m}{i}\)
Knowing that the plane flies at a constant speed of \(\displaystyle{500}{\frac{{{m}{i}}}{{{h}}}}\), we can calculate:
\(\displaystyle{d}={\frac{{\sqrt{{{3}}}\cdot{500}}}{{{2}}}}={250}\sqrt{{{3}}}\approx{433}{\frac{{{m}{i}}}{{{h}}}}\)
0
Vasquez
Answered 2022-01-07 Author has 9055 answers

The appropriate diagram here is a right triangle with a short vertical leg of length 1 representing the distance between the station and a point 1 mi directly above it, a longer horizontal leg of variable length x representing the distance from the point above to the position of the plane at time t, and a hypotenuse connecting the plane’s
position with the station (call this distance y). We're told that \(\frac{dx}{dt}=500\). By the Pythagorean Theorem \(1+x^2=y^2\). Differentiating with respect to t and using our rules and formulas
\(2\frac{dy}{dt}=\frac{dy^2}{dy}\frac{dy}{dt}=\frac{dy^2}{dt}=\frac{d(1+x^2)}{dt}\)
\(=\frac{d1}{dt}+\frac{dx^2}{dt}=0+\frac{dx^2}{dx}\frac{dx}{dt}=2x\frac{dx}{dt}\)
Now setting y=2 in the equation \(1+x^2=y^2\) gives \(x=\sqrt{3}\). So at this point,
\(4\frac{dy}{dt}=2\sqrt{3}\frac{dx}{dt}=2\sqrt{3}500\)
so that \(\frac{dy}{dt}=250\sqrt{3}\) (in miles per hour).

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