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Can differentiation be used to solve an equation?


king kyle

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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:

1*EAeu87gVDbyQZXW6_1sTtA.png

 

This equation is then "logged" on both sides:

1*T_R30Dx8cUTqc8ThlXfUiw.png

 

then differentiated on both sides:

1*ij7uqXOnhYwF1Juqa8QPMA.png

 

then rearranged:

1*CdNsvndTWakZPzQSW-1C8g.png

 

then differentiated on both sides again and solved:

 

1*m2RHA2Q2tcdUIBPVQexwVg.png

 

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.

 

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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.

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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.

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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.

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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 by Robittybob1
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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 La=Lp but as Mathematic said this is trivial. The guy on Reddit is saying that La 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 La=Lp, but also in an infinite number of other cases

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This is not the whole answer; the two sides of the equation balance when La=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.

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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...

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