Answer:
The rate at which the distance from the plane to the station is increasing is 331 miles per hour.
Step-by-step explanation:
We can find the rate at which the distance from the plane to the station is increasing by imaging the formation of a right triangle with the following dimensions:
a: is one side of the triangle = altitude of the plane = 3 miles
b: is the other side of the triangle = the distance traveled by the plane when it is 4 miles away from the station and an altitude of 3 miles
h: is the hypotenuse of the triangle = distance between the plane and the station = 4 miles
First, we need to find b:
[tex] a^{2} + b^{2} = h^{2} [/tex] (1)
[tex] b = \sqrt{h^{2} - a^{2}} = \sqrt{(4 mi)^{2} - (3 mi)^{2}} = \sqrt{7} miles [/tex]
Now, to find the rate we need to find the derivative of equation (1) with respect to time:
[tex]\frac{d}{dt}(a^{2}) + \frac{d}{dt}(b^{2}) = \frac{d}{dt}(h^{2})[/tex]
[tex] 2a\frac{da}{dt} + 2b\frac{db}{dt} = 2h\frac{dh}{dt} [/tex]
Since "da/dt" is constant (the altitude of the plane does not change with time), we have:
[tex] 0 + 2b\frac{db}{dt} = 2h\frac{dh}{dt} [/tex]
And knowing that the plane is moving at a speed of 500 mi/h (db/dt):
[tex]\sqrt{7} mi*500 mi/h = 4 mi*\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{\sqrt{7} mi*500 mi/h}{4 mi} = 331 mi/h[/tex]
Therefore, the rate at which the distance from the plane to the station is increasing is 331 miles per hour.
I hope it helps you!
