MHB Interpreting Constant A & Deriving k in Air Pressure Decay

AI Thread Summary
The discussion revolves around understanding the exponential decay of air pressure with altitude, modeled by the equation p(h) = A e^(kh). The constant A represents the air pressure at sea level, while the challenge lies in deriving the constant k, which is linked to the decay rate of 0.4 percent for every 30 meters. To find k, participants are advised to substitute the pressure equation into the decay relationship, leading to the equation p(h+30) = 0.996 p(h). The solution involves recognizing the properties of exponentiation and logarithms, particularly that k can be derived as k = ln(0.996) / 30.
fuzz95
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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!
 
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Hi, and welcome to the forum!

fuzz95 said:
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).

fuzz95 said:
Give an interpretation for the constant A.
Hint: At what height the pressure is $A$?

fuzz95 said:
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 $k$.
 
okay so i was able to do the first part!

however still have no idea how to go about doing the second half??!
 
fuzz95 said:
however still have no idea how to go about doing the second half??!

Evgeny.Makarov said:
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.
 
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