# A plane flying horizontally at an altitude of 1 mi and speed of 500mi/hr passes directly over a radar station. How do you find the rate at which the distance from the plane to the station is increasing when it is 2 miles away from the station?

Jul 5, 2015

When the plane is 2mi away from the radar station, its distance's increase rate is approximately 433mi/h.

#### Explanation:

The following image represents our problem:

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:

$d = \sqrt{{h}^{2} + {x}^{2}}$

We are interested in the situation when d=2mi, and, since the plane flies horizontally, we know that h=1mi regardless of the situation.

We are looking for $\frac{\mathrm{dd}}{\mathrm{dt}} = \dot{d}$

${d}^{2} = {h}^{2} + {x}^{2}$

$\rightarrow \frac{d \left({d}^{2}\right)}{\mathrm{dt}} = \frac{d \left({d}^{2}\right)}{\mathrm{dd}} \frac{\mathrm{dd}}{\mathrm{dt}} = \cancel{\frac{d \left({h}^{2}\right)}{\mathrm{dh}} \frac{\mathrm{dh}}{\mathrm{dt}}} + \frac{d \left({x}^{2}\right)}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}}$

$= 2 d \dot{d} = 2 x \dot{x}$

$\rightarrow \dot{d} = \frac{2 x \dot{x}}{2 d} = \frac{x \dot{x}}{d}$

We can calculate that, when d=2mi:

$x = \sqrt{{d}^{2} - {h}^{2}} = \sqrt{{2}^{2} - {1}^{2}} = \sqrt{3}$ mi

Knowing that the plane flies at a constant speed of 500mi/h, we can calculate:

$\dot{d} = \frac{\sqrt{3} \cdot 500}{2} = 250 \sqrt{3} \approx 433$ mi/h