# A plane flying horizontally at an altitude of 1 mi

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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Jenny Bolton
Note that y is the distance from the plane to the station the problem is referring to.

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.
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={\frac{{{x}}}{{{y}}}}{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}$$
$$\displaystyle{\frac{{\sqrt{{{3}}}}}{{{2}}}}\cdot{500}$$
$$\displaystyle{250}\sqrt{{{3}}}{\frac{{{m}{i}}}{{{h}}}}$$
###### Not exactly what you’re looking for?
levurdondishav4
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:
$$\displaystyle{d}={\frac{{\sqrt{{{3}}}\cdot{500}}}{{{2}}}}={250}\sqrt{{{3}}}\approx{433}{\frac{{{m}{i}}}{{{h}}}}$$
Vasquez

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).