Question

g A plane flying horizontally at an altitude of 2 miles and a speed of 550 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 miles away from the station. (Round your answer to the nearest whole number.)

Answer 1

Answer:

The rate at which the distance from the plane to the station is increasing is when it is 5 miles away from the station is 504 mi/h

Step-by-step explanation:

An illustrative diagram for the scenario is shown in the attachment below.

In the diagram, y represent the altitude, z is the horizontal distance from the plane to the station and x is the distance from the plane to the station.

From the Pythagorean theorem, we can write that

x² = y² + z²

Differentiate this with respect to time t

That is,

$\frac{d}{dt}(x^{2}) = \frac{d}{dt}(y^{2}) + \frac{d}{dt}(z^{2})$

= $2x(\frac{dx}{dt}) = 2y(\frac{dy}{dt}) + 2z(\frac{dz}{dt})$

$x(\frac{dx}{dt}) = y(\frac{dy}{dt}) + z(\frac{dz}{dt})$

The rate at which the distance from the plane to the station is increasing is $\frac{dx}{dt}$

$\frac{dy}{dt}$ is the rate at which the altitude is increasing, since the altitude is 2, that is constant, $\frac{dy}{dt} = 0$.

$\frac{dz}{dt}$ is the rate at which the horizontal distance is increasing which is the speed, that is, $\frac{dz}{dt} = 550 mi/h$

$y = 2$ and $x = 5$

Now, we will determine z when x = 5.

From x² = y² + z²

5² = 2² + z²

25 = 4 + z²

z² = 25-4

z² = 21

z =√21

Putting all the values into the equation

$x(\frac{dx}{dt}) = y(\frac{dy}{dt}) + z(\frac{dz}{dt})$

$5(\frac{dx}{dt}) = 2(0) + \sqrt{21} (550)$

$5(\frac{dx}{dt}) = \sqrt{21} (550)$

$\frac{dx}{dt} = \frac{\sqrt{21} (550)}{5}$

$\frac{dx}{dt} = 504 mi/h$

Hence, the rate at which the distance from the plane to the station is increasing is when it is 5 miles away from the station is 504 mi/h.