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Author Topic: Mathematics Help Thread  (Read 228756 times)

Dutchling

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Re: Mathematics Help Thread
« Reply #1350 on: November 21, 2013, 04:13:14 am »

Let Km = a
[S] = b
Vmax = c

Because typing all that out in an expression is annoying. Anyways,

(a + b)/bc = a/bc + b/bc = a/c * 1/b + 1/c

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Yes. Thank you!
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Dutchling

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Re: Mathematics Help Thread
« Reply #1351 on: November 21, 2013, 05:43:30 am »

New problem :S



I just don't get how to get y on both sides of the equation >.<. Don't you just get an endless loop this way...

(In case you think anyone thinks I'm letting Bay12 make my homework, most of the work is chemistry related to the formulas and they are no problem for me :P)
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Skyrunner

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Re: Mathematics Help Thread
« Reply #1352 on: November 21, 2013, 05:52:23 am »

But if chemistry is applied physics, and physics is applied math, how can you do chemistry without math?!

More seriously:

Assume that x != 0.

Working backwards:

y = a - by/x
xy = ax - by
xy + by = ax
y(x + b) = ax
y = ax / (x + b)


It's a pretty non-intuitive way to go from the first equation to the second, at least for me.
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i2amroy

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Re: Mathematics Help Thread
« Reply #1353 on: November 21, 2013, 05:54:32 pm »

It's a pretty non-intuitive way to go from the first equation to the second, at least for me.
I think this is probably how I would have done it (which seems a bit more intuitive to me):
y = a - y(b/x)
y + y(b/x) = a
x(y + y(b/x)) = ax
yx + yb = ax
y(x + b) = ax
y = ax / (x+b)
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Skyrunner

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Re: Mathematics Help Thread
« Reply #1354 on: November 21, 2013, 09:25:27 pm »

Personally I was unable to go from 1 to 2 and had to work backwards >_<

Also: why isn't (-3)1/2 equal to sqrt(3)*i? It sounds so logical~
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ZetaX

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Re: Mathematics Help Thread
« Reply #1355 on: November 21, 2013, 09:48:22 pm »

All you know in your case is that squaring both terms makes them equal, thus up to sign, they indeed are the same. Unless carefully defining everything, nothing more can be said. So this is mostly a question of definitions. Until you very accurately define what x^y, i and sqrt are meaning, it is quite hard to compare them. But you _could_ define (-x)^(1/2) to mean i·sqrt(x) for positive reals x, which obviously makes it true.

Also consider that i and -i are completely indistinguishable in the following sense: if you swap any occurence of one with the other, including all the definitions, it wont make any difference on mathematics. This is _not_ the same as saying that they are equal, though, just that there is no algebraic nor analytic way to really say one is "better" than the other. Or still another way, you could always define stuff with -i instead and nothing would change, so setting (-x)^(1/2) to mean -i·sqrt(x) is fine, too.
To give a real world example of this effect that is even related: assume I replace absolutely every occurance of left with right and vice versa in language (but not meaning); then you wouldn't be able to tell the difference at all, as there is no special property the one has but the other doesn't. This is different with up/down or front/back.
(@possible nitpickers: yeah, I know that unless I also change charges and some other physical properties along CPT symmetry, you actually could tell right and left apart; which by the way is exactly according to the i<->-i symmetry)
« Last Edit: November 21, 2013, 10:00:13 pm by ZetaX »
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Mego

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Re: Mathematics Help Thread
« Reply #1356 on: November 21, 2013, 10:39:28 pm »

Personally I was unable to go from 1 to 2 and had to work backwards >_<

Also: why isn't (-3)1/2 equal to sqrt(3)*i? It sounds so logical~

Umm... Why wouldn't it be?

MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1357 on: November 22, 2013, 02:15:38 am »

Personally I was unable to go from 1 to 2 and had to work backwards >_<

Also: why isn't (-3)1/2 equal to sqrt(3)*i? It sounds so logical~

Umm... Why wouldn't it be?
It technically would have equal rights to be -sqrt(3)*i, the problem is just that there is no continuous extension of noninteger exponentiation on the complex numbers that satisfies (a*b)^q = +a^q*b^q all the time.
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Mego

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Re: Mathematics Help Thread
« Reply #1358 on: November 22, 2013, 02:27:31 am »

Personally I was unable to go from 1 to 2 and had to work backwards >_<

Also: why isn't (-3)1/2 equal to sqrt(3)*i? It sounds so logical~

Umm... Why wouldn't it be?
It technically would have equal rights to be -sqrt(3)*i, the problem is just that there is no continuous extension of noninteger exponentiation on the complex numbers that satisfies (a*b)^q = +a^q*b^q all the time.

Maybe I'm just not understanding this (complex analysis is not a subject that I expect to ever get into, considering my degree is in computer science), but, taking the definition of the square root function to be that which returns the set of values which, when squared, reproduce the argument (sqrt(4) = +/- 2), I'm having issues understanding where the problem comes from.

sqrt(-3) would then return +/- ~1.732i, the two points on the imaginary axis whose distance to the real axis is ~1.732. This is no different than i*sqrt(3), which would be i*(+/- ~1.732).

Another

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Re: Mathematics Help Thread
« Reply #1359 on: November 22, 2013, 04:52:39 am »

In complex analysis simple Sqrt(z) is a two-valued function both on complex and on real numbers. (a^x would be countably-infinite valued for non-zero a, non-rational real part of x.) The usual way around it is to define a variant of sqrt(z) that returns a single value z1 in the complex half-plane with 0<=Arg(z1)<pi.

According to that narrowed definition sqrt(-3)=+i*sqrt(3).
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Skyrunner

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Re: Mathematics Help Thread
« Reply #1360 on: December 09, 2013, 08:16:24 am »

I wonder if you can derive the volume of a right (circular) cone (V) with a known height and bottom side radius this way? Bear in mind I haven't formally learned integration yet, only up to limits and differentiation in high school calc. Also I've only taken Korean calc so my vocabulary will be all over the place.

Constants r = radius of bottom side, hM = height of cone.

If h = distance from tip of cone, cutting a plane parallel to the bottom side, you could say that the radius of the intersect is rh / hM, since if you cut the cone in a plane perpendicular to the bottom side that also includes the 'axis' of the cone simple triangle similarity shows you that.

You could also say that the area of the intersection is (rh / hM)2π = r2h2/hM2π

Now, let's integrate for h! If I get basic integration correct, since A(h) = r2h2/hM2π, the integrated function which we shall call V(h) = r2h3/(3hM2)π

If I do V(hM) - V(0), that should effectively be the volume from the tip of the cone to the bottom side, since I added all the cross-sections. Therefore V = r2hM3/(3hM2)π = r2hM/3π; I can change that to 1/3hr2π, the conic volume formula we know. o_O



Did I do this right? Or did I make some fundamental error somewhere? :o
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Virex

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Re: Mathematics Help Thread
« Reply #1361 on: December 09, 2013, 08:48:50 am »


Nope, no errors. That's exactly how you can derive the volume of an object starting from it's mathematical definition.

New problem :S



I just don't get how to get y on both sides of the equation >.<. Don't you just get an endless loop this way...

(In case you think anyone thinks I'm letting Bay12 make my homework, most of the work is chemistry related to the formulas and they are no problem for me :P )


y = ax/(b+x)
    Multiply by b+x on both sides:
yb + yx = ax
    Assume x =/= 0, divide by x
b*y/x + y = a
    Bring b*y/x to the right-hand side:
y = a - b*y/x
« Last Edit: December 09, 2013, 08:52:05 am by Virex »
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1362 on: December 09, 2013, 01:59:56 pm »

I wonder if you can derive the volume of a right (circular) cone (V) with a known height and bottom side radius this way? Bear in mind I haven't formally learned integration yet, only up to limits and differentiation in high school calc. Also I've only taken Korean calc so my vocabulary will be all over the place.

Constants r = radius of bottom side, hM = height of cone.

If h = distance from tip of cone, cutting a plane parallel to the bottom side, you could say that the radius of the intersect is rh / hM, since if you cut the cone in a plane perpendicular to the bottom side that also includes the 'axis' of the cone simple triangle similarity shows you that.

You could also say that the area of the intersection is (rh / hM)2π = r2h2/hM2π

Now, let's integrate for h! If I get basic integration correct, since A(h) = r2h2/hM2π, the integrated function which we shall call V(h) = r2h3/(3hM2)π

If I do V(hM) - V(0), that should effectively be the volume from the tip of the cone to the bottom side, since I added all the cross-sections. Therefore V = r2hM3/(3hM2)π = r2hM/3π; I can change that to 1/3hr2π, the conic volume formula we know. o_O



Did I do this right? Or did I make some fundamental error somewhere? :o
Your reasoning is totally correct. It even has a name, it's called Fubini's theorem. Congrats on finding it yourself :P
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scrdest

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Re: Mathematics Help Thread
« Reply #1363 on: December 10, 2013, 03:31:09 am »

It's more of a physics things I'm doing for chemistry, but hey, I'll just throw it out here and maybe someone will figure this out. I've spent 12 A4 pages on it so far, so I'm getting somewhat desperate.


If you check the tables, the density of Hydrogen Chloride is, rounded up to two digits after decimal point, equal to 1 + half of the mass concentration. i.e. 36% acid will have the density of, roughly, 1,18 g/cm3, 18% - 1,09 and so on.

So, I'm trying to derive a formula that would prove that.

I suspect that the fact that molar mass of water is approximately half the molar mass of acid (18 vs 36 g/mol) is somewhat involved.

D - density
m - mass
M - molar mass
C - molar concentration
Cp - mass concentration
s - of substance (HCl)
w - of water
no index - of solution

So, givens are:

Dw = 1 g/cm3
Mw = 18 g/mol
Ms = 36 g/mol


Variables:
V
ms
mw

Any ideas?
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MagmaMcFry

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Re: Mathematics Help Thread
« Reply #1364 on: December 10, 2013, 05:08:53 am »

We know that an acid mass concentration of x means that every 1 gram of solution has x grams of acid and (1-x) grams of water. Therefore D(x) = V(x)/m = (x*Vs+(1-x)*Vw)/m = x*Ds + (1-x)*Dw. For Dw = 1 and Ds = 1.5, we get D(x) = 1g/cm³*(1+x/2), which is exactly what you wanted to show.
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