Help with probably basic integral?

[BeeKay]

In some reading I came across this equation

P/P1=(T/T1)^-g/aR which can apparently be derived from dp/P=(-g/RT)dh when T=T1 + a(h-h1).

I don't really understand how they could have taken the a out of the integral in order for it to be part of the exponent once you get rid the the ln on both sides. It should end up as -g/aR * (integral of dt/T) before you integrate both sides and then get rid of the ln on both sides, correct? I guess I am mostly confused about the fact that both temperature t is changing based on height h, but both are changing. If I just substituted T as T1+a(h-h1) I have no idea how to proceed from there. Does this require any more advanced types of integrals or am I just missing something obvious? Sorry about my lack of LaTeX, I am going to learn that soon but for now I hope you don't mind this mess I presented. Thanks in advance.

 

[andrewkirk]

I agree with their result, but I can't explain it without giving the whole thing away.

The first step is to substitute the expression for T in terms of T1 and H (3rd formula) into the second formula and then use separation of variables to integrate both sides. Then when you exponentiate to get rid of logs, your constant of integration becomes a coefficient, which you can get rid of by dividing by P1, expressed in terms of the formula you just got for P (replacing h by h1 and T by T1).

Set out your working (if you're not up to Latex yet, at least space it out clearly with one formula per line) and if you think you hit an obstacle, point it out and no doubt somebody will help you.

 

[BeeKay]

The part where I get stuck is pretty much once I substitute T in terms of T1 and H. Once I get to

-g/(R(T1+a(h-h1))*dh

I have no idea where to go from there. I don't even see how I could separate the variables. From what I remember when I did limited stuff with differential equations, it was only with x, dx, y, and dy. In this case there is P, dP, T, dt (because temperature changes with height), h and dh (height is changing). How do you go about doing it with three variables? On the left side, I easily see how they got the integral and how ln(p)-ln(p1) turned into P/P1 due to log rules. I assume I would have to do something on the right similar to this to get T/T1, but I do not see how I can integrate the right side with respects to both t and h. Or is this my mistake? I know you want to see my work but up until this point I haven't really had any clue as to where to go from here.

 

[andrewkirk]

Once I get to

-g/(R(T1+a(h-h1))*dh

I have no idea where to go from there. I don't even see how I could separate the variables... but I do not see how I can integrate the right side with respects to both t and h.

Fortunately, you don't have to integrate it with respect to T. The items in that RHS are g, R, T1, a, h1, h.

Assuming a and g are constants and R is Avogadro's constant, and that T1 and h1 are values of T and h at some base time, the only variable amongst the RHS items is h. So you just integrate with respect to that.

 

[BeeKay]

Okay, that makes a little bit more sense. But as for a being a constant, I know it is. But I cannot just bring it outside the integral because it is not multiplied by the T1 term, right? Once you pull out g and R, how do you integrate 1/(T1+a(h-h1)) and somehow get a to end up as a constant on the outside? I thought in order to do that both terms in the denominator would need to be multiplied by a?

 

[andrewkirk]

Multiply by a inside the integral and divide by a outside the integral. Then your integrand will be of the form 1/(h+const).

 

[BeeKay]

So I'll have -g/aR*Integral of (a*dh/(T1+a(h-h1)). I don't see how that is comparable to 1/(h+const). Where did the T go? Is "h" the T in this case? Doesn't the integrand have to end up as something that will integrate to ln(t)-ln(t1) so I can use log rules to make that ln(t/t1) and then exponentiate from there? Where does the constant part on the bottom go? Does the dh when integrated turn into (h-h1) and when that is multiplied by a it somehow cancels on the bottom?

 

[andrewkirk]

So I'll have -g/aR*Integral of (a*dh/(T1+a(h-h1)). I don't see how that is comparable to 1/(h+const).

Let's write that integrand out neatly for you:

$$\frac{a}{T1+a(h-h_1)}$$

How can you get rid of $a$ as a coefficient of $h$ in the denominator without changing the value of the fraction? (This is just algebra, not calculus).

 

[BeeKay]

You can divide it by a, making it 1/(t1/a+(h-h1)). I just don't see how that h-h1 in the denominator gets canceled with something. If I look at it as 1/x+1 and try to integrate, it is ln(x+1). So what makes the h-h1 cancel?

 

[andrewkirk]

You can divide it by a, making it 1/(t1/a+(h-h1)).

So now collect terms in the denominator to express it as $\frac{1}{h+(TermsNotInvolvingh)}$.

Then integrate wrt h (what is $\int\frac{1}{x+b}dx$)?

 

[BeeKay]

Integral of 1/x+b = ln(x+b). Terms collected in the denominator is

1/(h-h1)+((t1/a))?

That means the integral is ln(h+(t1/a)), right?

 

[andrewkirk]

Why are you grouping the constant h1 with the variable h? All constants belong in the (TermsNotInvolvingh) bit.

 

[BeeKay]

I have been under the impression that you could set both bounds on the integral, but it makes more sense that h1 would always be a constant, I'd assume 0? But even then, I don't see what you do with the constant portion of the integral because none of that is involved in the derived equation.

 

[andrewkirk]

Try following the steps in post 2. If what you end up with doesn't look like what they got, try using your third formula in the OP to perform some substitutions.

 

[BeeKay]

I can't see any way to get 1/T by itself in the integrand and still end up getting the right final formula. You at some point said multiply by a on the inside of the integral and divide on the outside, but doesn't this just cancel out and do nothing because you end up bringing out the inside constant anyway? When I integrate with respect to h, that means that the dh turns into h-h1 correct? Does this have anything to do with cancelling? I honestly don't see how these people got this proof, but I've seen the equation before so I guess it has to work somehow.

 

[BeeKay]

Just curious, you said you agreed with their result in your first post and that you wanted to lead me to it rather than just give me the answer, but did you make a mistake in the calculations or are you just sick of helping me haha. I have just been staring at it trying to figure it out but I think I should probably just trust people smarter than me that they did it right.

 

[andrewkirk]

I can't see any way to get 1/T by itself in the integrand and still end up getting the right final formula

You're getting ahead of yourself. 1/T has nothing to do with the step you are up to, which was that in post 11 you have an integrand that you want to integrate wrt h.
By the way, your denominator is missing enclosing parentheses in post 11, so you'll need to provide them before proceeding. So do that, express the denominator as (h+(TermsNotInvolvingh)), perform the integration and then carry on with the plan as outlined in post 2.

When I integrate with respect to h, that means that the dh turns into h-h1 correct?

No. When you integrate the dh disappears and you get a constant of integration. Post 2 suggests how you can deal with that constant.

 

[BeeKay]

Okay. Back to post 11 with better parentheses. I am trying to integrate

1/(h-(T1/a)-h1) wrt h

This, as far as I know, should be solved as if it were just like

constant/ (x+constant)

because t1, a, and h1 are all constants because we would be given them in the problem. This integrates to

ln(h-(t1/a-h1)

Where does the dh go/what does it turn into? Back when I was in school and doing integrals I always just kind of ignored it because it wasn't always important I guess. Is this where I exponentiate? That would make it

(P/P1) = h-(T1/a-h1)^-(g/aR)

I don't understand what dividing by P1 would do here.

 

[andrewkirk]

The general approach of your integration is OK but the result is wrong because of a simple sign error. You need to be much careful with your signs and parentheses, which are jumping around all over the place - randomly appearing, disappearing and switching sign.

Set your deduction out line by line and write a justification below each line of how you got that from the preceding line; and take care with parentheses and signs. Check it over twice before posting, to make sure that every step is valid. You can't expect to get the result if you don't keep your parentheses and signs in order.

The dh disappears when you do the integration.

 

[BeeKay]

Equations given.
1) (dP/P) = (-g/(RT))dh
2) T=T1+a(h-h1)

First substitute T from equation 2 into equation 1.

(dP/P)=(-g/(R(T1+a(h-h1)))) dh

Multiply by a on inside of integrand, while dividing on the outside. I will also take constants out of integrand here.

(dP/P) = (-g/(aR)) * Integral (a/(T1+a(h-h1))) dh

Then divide the integrand by a to get it into form of 1/(h + constants)

(dP/P)= (-g/(aR)) * Integral (1/(h+(T1/a)-h1) dh

Integrate both sides wrt h

ln(P)-ln(P1)= (-g/(aR)) * ln(h+(T1/a)-h1) (would this be integrated with bounds h to h1?)

I don't want to exponentiate before you confirm I am on the right track. Sorry about the formatting, is there a way I can transfer over Microsoft Word Equation Editor?

 

[BeeKay]

I think I'm nearly there, but I seem to have an extra a somewhere. I took it from the ln(h+(T1/a) -h1) and then substituted T1=T-a(h-h1). This cancels out the h and h1, but I still end up with T/a. Is this right? Or do I not have to multiply the inside by a and divide the outside?

 

[BeeKay]

I'll clarify in case you ask exactly to see that work.

h+(T1/a)-h1-------->h+((T-a(h-h1))/a)-h1

h+(T/a)-h+h1-h1

This leaves me with 1/a in the denominator, which when brought out of the integrand cancels out the a that was there from when I divided the outside by a. When I divided the integrand by a to change the form without changing the expression algebraically, was I supposed to divide the outside as well?

 

[andrewkirk]

In post 20, it's all good up to here:

Integrate both sides wrt h

ln(P)-ln(P1)= (-g/(aR)) * ln(h+(T1/a)-h1) (would this be integrated with bounds h to h1?)

In answer to your question, Yes the right side has to be written as the value of your right-side above for h=h minus the value for h=h1. The latter gives you some cancellation.
Then you exponentiate and the differences turn into ratios.

The last step is to manipulate your right-side denominator into a form where you can substitute using your last equation in the OP to get the desired result.

is there a way I can transfer over Microsoft Word Equation Editor?

I don't know for sure, but I expect not. I suggest you put the investment into learning latex. It is the only serious option for writing maths on a computer and it's the sort of thing you fall in love with once you get the hang of it. It's incredibly liberating.

There's a latex primer for the site here. Most people find it quicker to type two consecutive #s to begin and end in-line latex and two $s to begin and end separate-line ('display') latex rather than the longer escape commands given in that primer. But other than that it's a good way to start. If you right-click a piece of latexed math symbols on this site you get the option of displaying the code that was used to generate it.

Reasons why the Microsoft version is no good:

1. You can't search and replace in formulas, so if you want to change a variable name in a long doc you're up for a very long hunting and manual replacing session.

2. It is buggy and does things like randomly delete parts of formulas without warning or suddenly change all of the non-formula parts of the doc to italics.

3. It often hits a major error if you use it in a document that also has mark-up in it, where it will refuse to save unless you accept all marked-up changes - thereby losing track of who has contributed what.

4. Because it converts the code you write into its internal format of non-printable symbols it can become difficult to trouble-shoot formulas with formatting problems.

I had to use MS at work until I persuaded them to install MiKtex and Texmaker, and I had all of the above problems, which were most aggravating.