A plane flying horizontally at an altitude of 1 mile and a speed of 510 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 2 miles away from the station. (Round your answer to the nearest whole number.) mi/h

Answer:

442 miles

Step-by-step explanation:

Given

To properly solve this question, I illustrate some given parameters using attached image

From the image, apply Pythagoras theorem

[tex]x^2 + 1^2 = y^2[/tex]

Differentiate w.r.t time (t)

[tex]2x\frac{dx}{dt} + 0 = 2y\frac{dy}{dt}[/tex]

[tex]2x\frac{dx}{dt} = 2y\frac{dy}{dt}[/tex]

Divide both sides by 2

[tex]x\frac{dx}{dt} = y\frac{dy}{dt}[/tex]

From the question, we have that the plan travels are 510mi/h.

This implies that:

[tex]\frac{dx}{dt} = 510mi/h[/tex]

So, we then calculate the value of x when the distance (y) is 2mi i.e.:

[tex]y = 2mi[/tex]

Apply Pythagoras theorem

[tex]x^2 + 1^2 = y^2[/tex]

[tex]x^2 + 1^2 = 2^2[/tex]

[tex]x^2 + 1 = 4[/tex]

[tex]x^2 = 4-1[/tex]

[tex]x^2 = 3[/tex]

[tex]x = \sqrt 3[/tex]

So, the expression becomes:

[tex]x\frac{dx}{dt} = y\frac{dy}{dt}[/tex]

[tex]\sqrt 3 * 510 = 2* \frac{dy}{dt}[/tex]

[tex]\frac{\sqrt 3 * 510}{2} = \frac{dy}{dt}[/tex]

[tex]\sqrt 3 * 255 = \frac{dy}{dt}[/tex]

[tex]\frac{dy}{dt} = 255\sqrt 3[/tex]

[tex]\frac{dy}{dt} = 255 * 1.7321[/tex]

[tex]\frac{dy}{dt} = 441.655[/tex]

[tex]\frac{dy}{dt} = 442[/tex]

Hence, the distance is 442 miles

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