Solve Air Pressure Problem: Find y(x) Given y'(x)

[zetshield21]

Problem:
the efficiency of turbofan engines of commercial airplanes depend on air pressure and usually is maximum near about 10973 m, above mean sea level. Find the air pressure y(x) at this height given the rate of change y'(x) is proportional to the pressure, and at 5486 m the pressure has decrease to half its value y0 at mean sea level.

note: please solve my problem please!

 

[SteamKing]

Only you can solve this problem. If you show some work, you might get some comments about whether it is correct.

 

[Chestermiller]

Problem:
the efficiency of turbofan engines of commercial airplanes depend on air pressure and usually is maximum near about 10973 m, above mean sea level. Find the air pressure y(x) at this height given the rate of change y'(x) is proportional to the pressure, and at 5486 m the pressure has decrease to half its value y0 at mean sea level.

note: please solve my problem please!

The words in bold imply that

dp/dz = kp, where k is a constant, p is the pressure, and z is the altitude

 

[berkeman]

This thread is closed. zetshield21 -- check your PMs. You may only repost if you follow the Homework Help Template and rules.

And guys, when you see somebody post like this with zero effort shown, please click the Report button on the OP's post to have a Mentor deal with it. Thanks.

 

[rezeew]

I would approach this problem by first understanding the relationship between air pressure and altitude. Air pressure decreases as altitude increases, with a known rate of change.

Using the information provided, we know that at 5486 m, the pressure has decreased to half its value at sea level. This means that at 5486 m, the pressure is equal to y0/2. We can use this information to set up an equation:

y(x) = y0/2

Next, we are given that the rate of change of pressure, y'(x), is proportional to the pressure itself. This can be written as:

y'(x) ∝ y(x)

We can rewrite this using a constant of proportionality, k:

y'(x) = k * y(x)

Now, we can substitute our equation for y(x) into this equation:

y'(x) = k * (y0/2)

We are also given that at 10973 m, the efficiency of the turbofan engines is maximum. This means that at this altitude, the pressure is at its maximum value, which we can represent as ymax. We can use this information to set up another equation:

y(10973) = ymax

Now we have two equations with two unknowns (k and ymax). We can solve for k by plugging in the values we know:

k = y'(10973) / (y0/2)

Next, we can use this value of k to solve for ymax:

ymax = y(10973) = k * (y0/2)

Finally, we can substitute this value of ymax into our original equation for y(x):

y(x) = (y0/2) * (y(10973)/y0)

This gives us the air pressure at any altitude x, given the known rate of change y'(x) and the pressure at sea level, y0.

In conclusion, the air pressure at an altitude of 10973 m is equal to (y0/2) * (y(10973)/y0), where y(10973) is the maximum pressure at this altitude and y0 is the pressure at sea level. This solution takes into account the given information about the rate of change of pressure and the maximum efficiency of the turbofan engines at a specific altitude.