Is it valid to split the derivative in deriving the equation for e_s(T)?

[KingBigness]

Homework Statement

From
$\frac{de_{s}}{dT} = \frac{L_{v}e_{s}}{R_{v}T^{2}}$

derive

$e_{s}(T) = 6.11 e^{\frac{L}{RV}(\frac{1}{T}-\frac{1}{273})}$

Homework Equations

The Attempt at a Solution

The way my lecturer derived it was he 'split' the derivative and took them to their respective sides and integrated.

So he got

$\frac{de_{s}}{e_{s}} = \frac{LdT}{R_{v}T^{2}}$

However I was under the impression that you can't 'split' a derivative like that. Is this just a shortcut some physicists take to make the maths more simple? If it is what is the correct way of deriving this?

 

[Curious3141]

http://en.wikipedia.org/wiki/Separation_of_variables

It's not just "physicist's shorthand", it's a well-worn technique called Separation of Variables. Some might balk at the notation, but it's generally accepted and makes life easier.

 

[KingBigness]

http://en.wikipedia.org/wiki/Separation_of_variables

It's not just "physicist's shorthand", it's a well-worn technique called Separation of Variables. Some might balk at the notation, but it's generally accepted and makes life easier.

How accepted is it? I am doing a double degree in pure mathematics and physics. If I was to do this in my pure maths classes would I be stoned?

 

[iRaid]

How accepted is it? I am doing a double degree in pure mathematics and physics. If I was to do this in my pure maths classes would I be stoned?

I've taken calculus 1 and I've seen it before? :s

 

[KingBigness]

I've taken calculus 1 and I've seen it before? :s

Sorry I wasn't very clear. I have seen it before and done it countless times I just always have a memory of a teacher saying don't tell a pure mathematician about it.

 

[SammyS]

Homework Statement

From
$\displaystyle \frac{de_{s}}{dT} = \frac{L_{v}e_{s}}{R_{v}T^{2}}$

derive

$\displaystyle e_{s}(T) = 6.11 e^{\frac{L}{RV}(\frac{1}{T}-\frac{1}{273})}$

Homework Equations

The Attempt at a Solution

The way my lecturer derived it was he 'split' the derivative and took them to their respective sides and integrated.

So he got

$\displaystyle \frac{de_{s}}{e_{s}} = \frac{LdT}{R_{v}T^{2}}$

However, I was under the impression that you can't 'split' a derivative like that. Is this just a shortcut some physicists take to make the maths more simple? If it is what is the correct way of deriving this?

In my experience, physicists & engineers are notorious for treating Leibniz's notation for the derivative as if it were a fraction.

A more rigorous handling of this derivative equation might be as follows:

Treating e_s as a function of T we have

$\displaystyle \frac{d}{dT}(e_{s}) = \frac{L_{v}e_{s}}{R_{v}T^{2}}$

Rewriting this equation gives us

$\displaystyle \frac{1}{e_{s}}\frac{d}{dT}(e_{s}) = \frac{L}{R_{v}T^{2}}$

Integrating w.r.t. T gives

$\displaystyle \int {\frac{1}{e_{s}}\frac{d}{dT}(e_{s})}dT = \int{\frac{L}{R_{v}T^{2}}}dT$

We can rewrite the integral on the left hand side.

$\displaystyle \int {\frac{1}{e_{s}}\frac{d}{dT}(e_{s})}dT = \int {\frac{1}{e_{s}}}\,de_{s}$​

Alternatively, if $\displaystyle \frac{d}{dT}(e_{s}) = \frac{L_{v}e_{s}}{R_{v}T^{2}}\,,$
then the differential of e_s is given by $\displaystyle d\,e_{s} = \frac{L_{v}e_{s}}{R_{v}T^{2}}\,dT$

 

[KingBigness]

In my experience, physicists & engineers are notorious for treating Leibniz's notation for the derivative as if it were a fraction.

A more rigorous handling of this derivative equation might be as follows:

Treating e_s as a function of T we have

$\displaystyle \frac{d}{dT}(e_{s}) = \frac{L_{v}e_{s}}{R_{v}T^{2}}$

Rewriting this equation gives us

$\displaystyle \frac{1}{e_{s}}\frac{d}{dT}(e_{s}) = \frac{L}{R_{v}T^{2}}$

Integrating w.r.t. T gives

$\displaystyle \int {\frac{1}{e_{s}}\frac{d}{dT}(e_{s})}dT = \int{\frac{L}{R_{v}T^{2}}}dT$

We can rewrite the integral on the left hand side.

$\displaystyle \int {\frac{1}{e_{s}}\frac{d}{dT}(e_{s})}dT = \int {\frac{1}{e_{s}}}\,de_{s}$​

Alternatively, if $\displaystyle \frac{d}{dT}(e_{s}) = \frac{L_{v}e_{s}}{R_{v}T^{2}}\,,$
then the differential of e_s is given by $\displaystyle d\,e_{s} = \frac{L_{v}e_{s}}{R_{v}T^{2}}\,dT$

exactly what I wanted. thank you