I've stumbled upon a direct proof on Reddit, which uses differentiation to solve an equation.

 

See https://medium.com/criminal-clouds/hydrostatic-lapse-part-1-of-3-e8a1534cd12d

 

At particular point we reach this equation:

 

This equation is then "logged" on both sides:

 

then differentiated on both sides:

 

then rearranged:

 

then differentiated on both sides again and solved:

 

 

Is this the correct way to solve this equation?

 

I'm no maths expert, but this differentiation of both sides seems like far too much of a short cut.

 

No. Two expressions may be equal at one point, but the derivatives could be quite different.

 

Simple example: f(x)=x, g(x)=x^2. f(x)=g(x) has two solutions, x=0 and x=1.

f'(x)=1, g'(x)=2x. f'(x)=g'(x) has solution x=1/2.

 

In your case, La=Lp is trivially obvious solution by inspection.

La=Lp seems obvious to me too, I just don't how you would actually go about solving this.

La=Lp seems obvious to me too, I just don't how you would actually go about solving this.

 

 

It is a brutal rearrangement leading to a huge right hand side involving logs and the product log function

 

http://www.wolframalpha.com/input/?i=%28a-bx%29%5Ey%3D%28a-by%29%5Ex

Further to Mathematics excellent point - I would also note that derivatives can be the same whilst the functions are different

 

y=x^2+x+c this is a infinite group of curves - yet they all have the same derivative. Think of it this way - the function includes the information of where the curve is at every point AND the slope - but the derivative only includes the slope; you lose information when you differentiate.

 

It is a brutal rearrangement leading to a huge right hand side involving logs and the product log function

 

 

But isn't L_a=L_{p ?}

La=Lp seems obvious to me too, I just don't how you would actually go about solving this.

It then becomes an identity.

It then becomes an identity.

I'm not sure what your getting at here.

 

I would of thought that there would be a way to solve La in terms of Lp.

 

If you look at this spreadsheet you will see that if La = Lp then the top equality holds.

I'm not sure what your getting at here.

 

I would of thought that there would be a way to solve La in terms of Lp.

 

If you look at this spreadsheet you will see that if La = Lp then the top equality holds.

I wan't trying to be profound. If La=Lp, both sides of the equation are the same symbolically - i,.e. identical.

I'm not sure what your getting at here.

 

I would of thought that there would be a way to solve La in terms of Lp.

 

If you look at this spreadsheet you will see that if La = Lp then the top equality holds.

on the spreadsheet E3 = D3 and then you are surprised that B6 = C6! They are calculating with the same values to start with aren't they!

Edited June 27, 201510 yr by Robittybob1

I'm not sure what your getting at here.

 

I would of thought that there would be a way to solve La in terms of Lp.

 

If you look at this spreadsheet you will see that if La = Lp then the top equality holds.

 

An identity is different from an equations. Put simply an equation has a LHS and a RHS that are equal - we re-arrange to find out what values are solutions, or how one variable looks when shown as a combination of all the others. An identity (which is designated with a three bar equals sign) is just two ways of writing the same expression - ie both sides are always equal no matter what values the variables take.

 

y = sin(x) is an equation

sin²(x) [latex]\equiv[/latex] 1 - cos²(x) is an identity - no matter what value x takes the LHS equals the RHS

 

The solution set to your initial question is large and complex - one single solution is that L_a=L_p but as Mathematic said this is trivial. The guy on Reddit is saying that L_a must equal Lp and drawing conclusions from an assertion that they must be equivalent. This is not the whole answer; the two sides of the equation balance when L_a=Lp, but also in an infinite number of other cases

This is not the whole answer; the two sides of the equation balance when L_a=Lp, but also in an infinite number of other cases

Can you provide another solution that also allows the equality to work? I cannot get the left side result to match the right side result when La differs from Lp.

Assuming L_a, L_p, and h are held constant while T_s changes, and this equation remains accurate, this "proof" is fine. In other words, think of both sides as functions of the time t; if the functions are equal during some period, then the derivatives will also be equal.

 

It's a bit overkill, though; asymptotic behavior at infinite T_s should be enough...