t a height of h meters above the surface of the earth, the atmospheric pressure in torr is approximately: (ℎ) = 760−ℎ/7000 A plane ascends from an altitude of 4000 meters at 550 meters per minute at t = 0 minutes. How fast is the atmospheric pressure outside of the plane decreasing at this time?

Question

23-10-2025
Mathematics

College

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t a height of h meters above the surface of the earth, the atmospheric pressure in torr is approximately: (ℎ) = 760−ℎ/7000 A plane ascends from an altitude of 4000 meters at 550 meters per minute at t = 0 minutes. How fast is the atmospheric pressure outside of the plane decreasing at this time?

Answer :

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MattZayn5 MattZayn5 · Answer 1 · 2024-01-02T09:41:43-05:00

Final answer:

To find how fast the atmospheric pressure is decreasing as the plane ascends, the rate of change of pressure with respect to height is multiplied by the rate of height change with respect to time using the chain rule. The pressure is found to be decreasing at approximately 0.07857 torr per minute at the altitude of 4000 meters when ascending at 550 meters per minute.

Explanation:

The question concerns the rate at which atmospheric pressure decreases as a plane ascends. The atmospheric pressure at a height h meters above the surface of the earth is given by the formula P(h) = 760 - h/7000, where P is in torr. The plane's ascent rate is 550 meters per minute at t = 0 minutes, and we are interested in how fast the pressure is decreasing at this time, which is the rate of change of pressure with respect to time.

Using calculus, specifically the chain rule, we can find the rate of change of pressure with respect to time (dP/dt) by relating it to the rate of change of pressure with respect to height (dP/dh) and the rate of height change with respect to time (dh/dt).

The given ascent rate of the plane is dh/dt = 550 meters per minute. The derivative of the pressure with respect to height, dP/dh, is -1/7000. Therefore, the rate of change of pressure with respect to time as the plane ascends at this rate is:

dP/dt = (dP/dh) imes (dh/dt)

Substituting the given rates, we get:

dP/dt = (-1/7000) imes 550 = -0.07857 torr per minute

So, the atmospheric pressure is decreasing at a rate of approximately 0.07857 torr per minute at the time when the plane is at an altitude of 4000 meters and climbing at 550 meters per minute.