Respuesta :
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,
[tex]\frac{d}{dt}(x^{2}) = \frac{d}{dt}(y^{2}) + \frac{d}{dt}(z^{2})[/tex]
= [tex]2x(\frac{dx}{dt}) = 2y(\frac{dy}{dt}) + 2z(\frac{dz}{dt})[/tex]
[tex]x(\frac{dx}{dt}) = y(\frac{dy}{dt}) + z(\frac{dz}{dt})[/tex]
The rate at which the distance from the plane to the station is increasing is [tex]\frac{dx}{dt}[/tex]
[tex]\frac{dy}{dt}[/tex] is the rate at which the altitude is increasing, since the altitude is 2, that is constant, [tex]\frac{dy}{dt} = 0[/tex].
[tex]\frac{dz}{dt}[/tex] is the rate at which the horizontal distance is increasing which is the speed, that is, [tex]\frac{dz}{dt} = 550 mi/h[/tex]
[tex]y = 2[/tex] and [tex]x = 5[/tex]
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
[tex]x(\frac{dx}{dt}) = y(\frac{dy}{dt}) + z(\frac{dz}{dt})[/tex]
[tex]5(\frac{dx}{dt}) = 2(0) + \sqrt{21} (550)[/tex]
[tex]5(\frac{dx}{dt}) = \sqrt{21} (550)[/tex]
[tex]\frac{dx}{dt} = \frac{\sqrt{21} (550)}{5}[/tex]
[tex]\frac{dx}{dt} = 504 mi/h[/tex]
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.
