## 15 Reputation

1 years, 134 days

## @Kitonum sorry just had another loo...

@Kitonum sorry just had another look at this, and I will need to define an alias (I will have alot more functions and caluations) but when I do this the second line 'convert(%,diff') no longer converts:

chainrule3.pdf

many thanks

## @Kitonum  and is that just the...

and is that just the notation, i.e the way maple writes things, so it is just spelling out to us that V= V(h(x))?( for the chain rule to have been carried out here)

## @Kitonum   ok many thanks so by y...

@Kitonum

ok many thanks so by your second line, D(V) in this case is equivalent to dV/dh?

## " The non-uniqueness of the an...

" The non-uniqueness of the answer is because of Maths, not because of Maple" I know... but thanks for the insult

## @acer    i find this way too...

i find this way too heavy to read, can you reply wiht the specific example i gave maybe instead? or can someone else reply? as i said i found the first reply above much simpler and easier to understand, such as the explanation ' ... is equivalent to R', a click glimpse at thtat statement and i got it at that stage, something simple please..

## okay many thanks for the reply. i prefer...

okay many thanks for the reply. i prefer the first reply and took this style.

(re the second reply...yes in hindsight wiht your knowledge acer it woudl have been better to present the more genreal case in my oriinal post, but , wanting to play around and figure out things myself... ofc now after the intial reply I would have realsied this myself, but without the knowledge of maple you wouldnt, and i definitely wouldnt say overall for every single example in maple it is better to post a more general case, leaving the persona sking the question less figuring out and playing around to do themselves, so to conclude stop trying to find ways to complain about newbies posts... :))) ).

yes this was a trivial example, here is the one that I am after:

mu/To*Ho

in terms of any combination of these pre-defined variables :

(R,Fr defined previously and the following..)

Bo:= rho *Go*Ho^2/ sigma=W

E:= To*Uo/Ho=St

B:=mu*Uo/sigma=Ca

A:= rho*Ho*Uo/mu=Re

So sticking with this example given in the FIRST reply, I use this 'trick' of re-taining the equality for A,B,E.. etc )the dimensionless variables), ofc then there is the quesiton of which variable you choose and having a little play around, that's not what she said, I see that which you choose affects the output, but I have not worked out exactly what the rule is or what I need to do/ how to work out which is the better variable to use.( in the case of the first example I see that I can equally do Uo=solve(R,Uo) or Ho=solve(R,Ho) howver not rho=solve(R,rho) or mu=..., so I guess that it has to be a variable in the expression I am trying to solve for in terms of the dimensionless numbers, but, when there's more than one dimesnionless number, does each need to be solving for a unique variable, is it bad if i repeat a variable that is in the final expression amongst the dimensionless variables, e.g in the first example below, A and R both I have done the equilvalent relationship via Uo...

subs({Uo=solve(A,Uo),sigma=solve(B,sigma),To=solve(E,To), Uo=solve(R,Uo), Go=solve(W,Go)}, mu/To*Ho)

yields

\mu*Uo/St*Ho^2

subs({mu=solve(A,mu),sigma=solve(B,sigma),To=solve(E,To), Uo=solve(R,Uo), Go=solve(W,Go)}, mu/To*Ho)

yields

rho*Ho*Uo*Uo/Re *St*Ho^2

Many thanks !!

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