Gina's question at Yahoo! Answers regarding rate of change of plane's distance from radar station

MarkFL

Staff member
Here is the question:

related rates; Find the rate at which the distance from the plane to the station is increasing when it is..?

A plane flying horizontally at an altitude of 1 mi and a speed of 470 mi/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 5 mi away from the station. (Round your answer to the nearest whole number.)
__ mi/h

I have posted a link there to this topic so the OP can see my work.

MarkFL

Staff member
Hello Gina,

First, let's draw a diagram:

The plane is at $P$, the radar station is at $R$, $h$ is the altitude of the plane (which is constant since its flight is said to be horizontal) and $x$ is the distance from the radar station to the point on the ground (or at the same level as the radar station) directly below the plane. $s$ is the distance from the plane the the radar station.

Using the Pythagorean theorem, we may state:

(1) $$\displaystyle x^2+h^2=s^2$$

Implicitly differentiating (1) with respect to time $t$, we find:

$$\displaystyle 2x\frac{dx}{dt}=2s\frac{ds}{dt}$$

We are interested in solving for $$\displaystyle \frac{ds}{dt}$$ since we are asked to find the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station. So, we find:

$$\displaystyle \frac{ds}{dt}=\frac{x}{s}\frac{dx}{dt}$$

Now, we do not know $x$ but we know $h$ and $s$, and so solving (1) for $x$ (and taking the positive root since it represents a distance), we find:

$$\displaystyle x=\sqrt{s^2-h^2}$$

And so we have:

$$\displaystyle \frac{ds}{dt}=\frac{\sqrt{s^2-h^2}}{s}\frac{dx}{dt}$$

Now, the speed $v$ of the plane represents the time rate of change of $x$, hence:

$$\displaystyle v=\frac{dx}{dt}$$

and so we may write:

$$\displaystyle \frac{ds}{dt}=\frac{v\sqrt{s^2-h^2}}{s}$$

Now, we may plug in the given data:

$$\displaystyle v=470\frac{\text{mi}}{\text{hr}},\,s=5\text{ mi},\,h=1\text{ mi}$$

and we then find:

$$\displaystyle \frac{ds}{dt}=\frac{470\sqrt{5^2-1^2}}{5} \frac{\text{mi}}{\text{hr}}=188\sqrt{6}\frac{\text{mi}}{\text{hr}}\approx460.504071643238 \text{ mph}$$

And so we have found that the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station is about 461 mph.