A plane flying horizontally at an attitude of 1 mi and a speed of 510 mi/h passe

Question

A plane flying horizontally at an attitude of 1 mi 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 is 3mi away from the station (Round your answer to the nearest whole number) please keep in that mind that distance = squareroot (attitude)^2 + (horizontal distance)^2 (or y^2 = x^2 + h^2) differentiate with respect to t on both sides of the equation using the chain rule to solve for dy/dt the given speed of the plane is dx/dt

Explanation / Answer

3) We have that the surface area formula is:

A= 4r2    and diameter is 2r. Also, we know that dA/dt = 9 and the diameter is 12 cm,

meaning that r = 6.

To determine the rate at which the diameter decreases, which will be 2 dr/dt .

Differentiating the expression for A yields

dA/dt = 8r dr/dt

-9 = 8*6 dr/dt

dr/dt = -3/16

so the diameter is decreasing at a rate of 3/16 i.e 0.0597 cm/min.

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