Diff-EQ interpretation: The answer in the back substitutes into expression for dp/dh?

[apple2357]

The last part of this question is confusing me!
My reading of the last part is i substitute 0 and 1 and into the expression for p and use this to find s.
The answer in the back substitutes into the expression for dp/dh ?
Can someone help please?

 

[Deleted member 4993]

The last part of this question is confusing me!
My reading of the last part is i substitute 0 and 1 and into the expression for p and use this to find s.
The answer in the back substitutes into the expression for dp/dh ?
Can someone help please?

The (d) part of the question is asking for (pressure change)/(kilometer) at a particular height (h) →

which is equivalent of dp/dh

 

[Dr.Peterson]

The last part of this question is confusing me!
My reading of the last part is i substitute 0 and 1 and into the expression for p and use this to find s.
The answer in the back substitutes into the expression for dp/dh ?
Can someone help please?

There are two ways you could interpret the question, which give slightly different answers.

First, taking it literally, "the atmospheric pressure will change by s% for every kilometre gained in height" would ask for the percent change going from height h to height h+1. Plug those in, and you will find that the result is independent of h, which is effectively what they are saying. Find that number.

But they may also mean this in a calculus sense, the percent change from height h to h + ∆h, divided by ∆h. This ends up being p'/p, and the differential equation (model) dp/dh = kp says exactly that p'/p is constant. So they may want you to find that value, p'/p, which is trivial based on what I just said. You will find that the two numbers are close.

The difference between the two is closely related to the nominal vs effective interest rate in compound interest, and to the "rate constant" of exponential growth.

Since you say the book gives an answer, please quote the entire explanation and result so we can tell you why they did that. I suspect it is like my second answer. Possibly they have also discussed this in an example before this exercise. Can you find one?

 

[apple2357]

This is the solution to part (d).
Any thoughts?

 

[Dr.Peterson]

This is the solution to part (d).
Any thoughts?

Wow! Their work is utterly wrong, though their answer is right!

What they've done is to calculate the percentage change over one kilometer in the rate of change of pressure with height, not in the pressure itself, as the question is asking. But since that rate of change is proportional to the pressure (kp), the percentage change is the same as what they asked for. So if someone checked whether the answer was correct without looking at the work, they would have missed the error.

What they should be doing is the first approach I suggested, since they are interpreting the question as referring to the change over an entire kilometer, not an instantaneous rate. They chose to use 0 and 1 arbitrarily, perhaps realizing that for exponential change the percent change will be constant; since you lack that experience, you should really do it as I said, letting the starting point be an unknown h and looking at the percentage change from h to h+1. But you would get the same result using 0 and 1; you just wouldn't be sure you were doing the right thing.

 

[apple2357]

Wow! Their work is utterly wrong, though their answer is right!

What they've done is to calculate the percentage change over one kilometer in the rate of change of pressure with height, not in the pressure itself, as the question is asking. But since that rate of change is proportional to the pressure (kp), the percentage change is the same as what they asked for. So if someone checked whether the answer was correct without looking at the work, they would have missed the error.

What they should be doing is the first approach I suggested, since they are interpreting the question as referring to the change over an entire kilometer, not an instantaneous rate. They chose to use 0 and 1 arbitrarily, perhaps realizing that for exponential change the percent change will be constant; since you lack that experience, you should really do it as I said, letting the starting point be an unknown h and looking at the percentage change from h to h+1. But you would get the same result using 0 and 1; you just wouldn't be sure you were doing the right thing.

Thanks! This was confusing me. If the question said 'rate of change' would this be ok?

 

[Dr.Peterson]

Thanks! This was confusing me. If the question said 'rate of change' would this be ok?

I suppose; but I don't think anyone would actually ask that question. In particular, that is not what "the model predicts", as the question says, except in an indirect way, so they would have to change the whole question if they meant that. It is the answer that is wrong, not the question.

If by "model" they mean the given equation p = e^(-0.13h), that says that we have exponential decrease, which implies that the percentage decrease over a given distance is constant, something you have perhaps been taught about exponential models. The idea that the percentage change in rate of change is constant, is a secondary inference, not something you'd normally say it "predicts".

If by "model" they mean the differential equation, that more or less directly says is that the rate of change in the pressure relative to height (dp/dh) is proportional to the pressure (p), so that the percentage change in pressure (change in pressure over pressure) is constant. That constant, in the limit, is k = -0.13 = 13% decrease. The percentage change over a kilometer is close to that, about 12.2%; if you did the calculation over a smaller change in height, it would approach 13% per km. For example, from h=0 to 0.1, p changes from p = e^(-0.13*0) = 1 to p = e^(-0.13*0.1) = 0.987, a percentage change of (0.987 - 1)/1 = -0.013 = -1.3%, which is -13% per km.