# Rate and Distance Question Calc

How do I find the rate at which the distance from the plane to the station is increasing when it is 4 mi away from the station.

A plane flying horizontally at an altitude of 3 mi and a speed of 460 mi/h passes directly over a radar station

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It looks like the second sentence should precede the first? –  copper.hat Jun 25 '12 at 4:27

This is a related rates question. The general strategy:

1. Draw a picture. Invariably useful with these sort of problems. In the picture, label every quantity that is fixed, and every quantity that is changing with a name.

2. Write down the information you are given. Identify the rates you are told something about, and the rate you want to know something about.

3. Find an equation that relates the quantities whose rates you are being asked about. A relation among the quantities, not the rates.

4. Differentiate the equation you found in the previous step; this will give you an equation that relations the quantities and their rates of change.

5. Plug in all the information you have. Solve for the information you want to know.

Here, after you draw the picture, you'll see that it makes sense to think of the radar station as the origin, and the plane as flying on the line $y=3$. Let $p(t)$ be the horizontal position of the plane at time $t$ (so that the plane will be at the point $(p(t),3)$). Let $D(t)$ be the distance from the plane to the radar station.

You are told how $p(t)$ is changing: that is, you are given information about $\frac{dp}{dt}$. You are being asked about how the Distance is changing; that is, you are being asked to find $$\frac{dD}{dt}\Bigl|_{D=4}$$ (Actually, it's unclear if you want the derivative when $D$, the straight line from the plane to the radar, is $4$, or if you want it when $p(t)$ is $4$; I think it is the former, though).

So you wan to find some equation that relates $p$ and $D$. After you do that, taking the derivative of the equation with respect to $t$ will give you an equation that relates $p$, $D$, $\frac{dp}{dt}$, and $\frac{dD}{dt}$. You know the value of $\frac{dp}{dt}$, and you know the value of $D$ you want. So you should figure out what $p$ is for that $D$ (if necessary), plug everything in, and solve for $\frac{dD}{dt}$.

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Let $x$ denote the horizontal distance that the plane travels from a point directly overhead. Let $y$ be the slant distance from the radar station to the plane.

By the Pythagorean Theorem, $x^2 + 3^2 = y^2$. Differentiating this leads us to:

$$2x \frac{dx}{dt} = 2y\frac{dy}{dt} \tag{1}$$

Note that $\frac{dx}{dt} = 460$ miles per hour.

Simplifying equation (1) and plugging in the value for $\frac{dx}{dt}$ leads us to:

$$460x = y\frac{dy}{dt} \tag{2}$$

When $x = 4, y = \sqrt {25} = 5$ because of the Pythagorean Theorem. So, we now have:

$$460\cdot 4 = 5 \frac{dy}{dt} \tag{3}$$

Thus,

$$\frac{dy}{dt} = 368 \space \text{miles per hour}$$

So, the distance between the plane and the radar station is increasing at $368$ mph.

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Don't use $=$ when you are approximating. If you are using a decimal approximation of an inexact square root, then use $\approx$; better yet, leave the radical indicated! –  Arturo Magidin Jun 25 '12 at 4:40
@ArturoMagidin Yeah, you're right. –  Joe Jun 25 '12 at 4:42
It seems I made a mistake by letting $x=3$ instead of $x=4$. The numbers are nice and friendly now - no need for $\approx.$ I think I'm going to take a look at this tomorrow morning when I have a fresh mind to make sure this work is correct. I don't wish to steer the OP down the wrong path. –  Joe Jun 25 '12 at 4:49

More succinctly (and lower sodium too):

Let $x(t) = 4 + 460 t$, $y(t) = 3$. $\delta(t) = \sqrt{x(t)^2+y(t)^2}$. We want to compute $\dot{\delta}(0)$.

$\dot{\delta(0)} = \frac{1}{2 \sqrt{x(0)^2+y(0)^2}} (2 x(0) \dot{x}(0)+2 y(0) \dot{y}(0)) = \frac{4 \cdot 460}{\sqrt{4^2+3^2}} = 368.$

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I think Arturo's answer is great and this is merely meant to try to explicitly show some of the point he brings up.

As he mentioned, one of the key points to solving this problem is drawing a picture and writing down the information.

With an image drawn one can begin to notice things which were not necessarily apparent before. The first important piece of information that should be noticed in this case is that $(1)$ the altitude of the plane, $(2)$ the distance along the ground from the radar station to the plane and $(3)$ the direct distance from the radar station and the plane are all related by the Pythagorean theorem i.e. $$x^2+y^2=z^2$$ When dealing with speeds we are always looking at how distances change with respect to time i.e. $$\dfrac{dD}{dt}$$ Where $D$ is distance and $t$ is time. In the current problem all variables $x,y,z$ are distances which are subject to the passing of time. As a result we differentiate both sides with respect to $t$ to understand how these distances vary as time passes. We thus get: \begin{align*}\frac{d}{dt}\left(x^2+y^2\right)&=\frac{d}{dt}\left(z^2\right) \\ \frac{d}{dt}\left(x^2\right)+\frac{d}{dt}\left(y^2\right)&=\frac{d}{dt}\left(z^2\right) \\ 2x\frac{dx}{dt}+2y\frac{dy}{dt}&=2z\frac{dz}{dt} \\ x\frac{dx}{dt}+y\frac{dy}{dt}&=z\frac{dz}{dt} \\ \end{align*}

Now, we are given the quantities for $x=4$ and $y=3$ at a specific instant. We can also figure out $z$ using the Pythagorean theorem: \begin{align*}z^2&=x^2+y^2& \\ z&=\sqrt{x^2+y^2} \\ z&=\sqrt{4^2+3^2}=5 \end{align*} We also know that the rate that the plane is moving away from the radar station is $460$ mph $\left(\text{aka.} \dfrac{dx}{dt}=460\right)$ and that rate that the altitude is changing is $0$ mph $\left(\dfrac{dy}{dt}=0\right)$. Plugging all of this into our above equation gives: \begin{align*}x\frac{dx}{dt}+y\frac{dy}{dt}&=z\frac{dz}{dt} \\ (4)\frac{dx}{dt}+(3)\frac{dy}{dt}&=(5)\frac{dz}{dt} \\ (4)(460)+(3)(0)&=(5)\frac{dz}{dt} \\ 1840&=5\frac{dz}{dt} \\ \frac{dz}{dt}&=368 \end{align*}

Note that $\dfrac{dz}{dt}$ is indeed the rate at which the distance between the radar station and the plane is changing and so we have our final answer: $368$ mph.

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