Interpreting Constant A & Deriving k in Air Pressure Decay

[fuzz95]

hey guys! I'm really confused as to what this question is trying to ask me!
can someone help me out :)

Air pressure decays approximately exponentially at about 0.4 per cent for each rise of 30 metres above sea level. If we let p = p(h) denote air pressure (measured in some appropriate units) at h metres above sea level, then we can model this phenomenon using the equation: p = p(h) = Aekh for some appropriate constants A and k.

Give an interpretation for the constant A. (We never need to know the actual numerical value of A to do the rest of this exercise.)
?and how do i derive this k = ln(0.996) / 30 ??thanks!

 

[Evgeny.Makarov]

Hi, and welcome to the forum!

If we let p = p(h) denote air pressure (measured in some appropriate units) at h metres above sea level, then we can model this phenomenon using the equation: p = p(h) = Aekh for some appropriate constants A and k.

This must be $p(h)=Ae^{kh}$. In plain text, it is customary to write exponentiation using ^, and don't forget the order of operations: exponentiation is done before multiplication. Therefore, in plain text p(h) can be written as A*e^(kh).

Give an interpretation for the constant A.

Hint: At what height the pressure is $A$?

and how do i derive this k = ln(0.996) / 30 ??

The statement "Air pressure decays approximately exponentially at about 0.4 per cent for each rise of 30 metres above sea level" amounts to
\[
p(h+30)=0.996p(h).
\]
Substitute the definition of $p$ and see if you can solve this equation for $k$.

 

[fuzz95]

okay so i was able to do the first part!

however still have no idea how to go about doing the second half??!

 

[Evgeny.Makarov]

however still have no idea how to go about doing the second half??!

The statement "Air pressure decays approximately exponentially at about 0.4 per cent for each rise of 30 metres above sea level" amounts to
\[
p(h+30)=0.996p(h).
\]
Substitute the definition of $p$ and see if you can solve this equation for $k$.

Well, let me do this for you. Substituting $p(h)=Ae^{kh}$ into the equation above gives
\[
Ae^{k(h+30)}=0.996Ae^{kh}.
\]
Can you solve it for $k$? If not, do you know basic properties of exponeniation? Do you know that $\ln(x)$ is the inverse of $e^x$? Have you seen any exponential equations solved? Have you solved any yourself? You may need to do some preparatory work.

 

[CellCoree]

I can help clarify the question and provide an interpretation for the constant A and the derivation of k in air pressure decay.

The equation given, p = p(h) = Aekh, is known as the exponential decay model. It is used to describe how a quantity decreases over time or distance. In this case, it is used to model the decrease in air pressure as the altitude increases.

The constant A in this equation represents the initial air pressure at sea level. In other words, it is the air pressure at h = 0 meters. The value of A is not important for the rest of the exercise, as it is a constant that remains the same throughout the model. However, it helps us understand the starting point for the air pressure decay.

To derive the value of k, we need to use the given information that air pressure decays by 0.4% for every 30 meters increase in altitude. This means that when h = 30 meters, the air pressure decreases by 0.4% or 0.004. Plugging this into the equation, we get:

p(30) = Aek(30) = Ae^30k

But we also know that p(30) = 0.996A, as the air pressure decreases by 0.4%. Therefore, we can set these two equations equal to each other and solve for k:

0.996A = Ae^30k
0.996 = e^30k
ln(0.996) = 30k
k = ln(0.996) / 30

This value of k represents the rate of decay of air pressure with increasing altitude. It tells us how much the air pressure decreases for every unit increase in altitude. In this case, it is approximately 0.0011, or 0.11%.

I hope this helps clarify the question and the interpretation of constant A and derivation of k in air pressure decay. Let me know if you have any further questions.