Math(s)
 sum yun gai
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Re: Math(s)
ok, one more yay!
∫ 1/[x(x²1)^(3/2)] dx by trig substitution.
i have this one broken down to ∫ dθ/tan² θ i know this can be rewritten as cot² θ dθ, which i can look up in an integral table. i am having problems getting to the solution given by the integral table. i think the problem is it's been at least 5 years since i did any maths of this level so i'm not remembering the little tweaks to get to where they want me to go...
∫ 1/[x(x²1)^(3/2)] dx by trig substitution.
i have this one broken down to ∫ dθ/tan² θ i know this can be rewritten as cot² θ dθ, which i can look up in an integral table. i am having problems getting to the solution given by the integral table. i think the problem is it's been at least 5 years since i did any maths of this level so i'm not remembering the little tweaks to get to where they want me to go...
ntw3001 wrote:Sass has to come from the heart, not from the shirt.
traubster wrote:I find it irritating whenever I walk through a cemetery and there's not one gravestone that reads something like, "We're all grateful that he's dead. Sorry if he owed you money."
 Saltine
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Re: Math(s)
So you're trying to come up with a nondivinelyinspired way of getting that
∫cot²θdθ = θcotθ?
Well, I would probably do something like
u=cotθ=cosθ/sinθ,
du= (1cot²θ)dθ = (1+u²)dθ
∫cot²θdθ becomes
∫u²/(1+u²) du
∫(1+u²1)/(1+u²) du
∫1/(1+u²) du∫(1+u²)/(1+u²) du
tan⁻¹u ∫ du
tan⁻¹u  u
tan⁻¹cotθ  cotθ
tan⁻¹(tan(π/2 θ))  cotθ
π/2 θ  cotθ
Which is what we were looking for, apart from a constant of π/2 which doesn't matter because it gets absorbed into the constant of integration the end
∫cot²θdθ = θcotθ?
Well, I would probably do something like
u=cotθ=cosθ/sinθ,
du= (1cot²θ)dθ = (1+u²)dθ
∫cot²θdθ becomes
∫u²/(1+u²) du
∫(1+u²1)/(1+u²) du
∫1/(1+u²) du∫(1+u²)/(1+u²) du
tan⁻¹u ∫ du
tan⁻¹u  u
tan⁻¹cotθ  cotθ
tan⁻¹(tan(π/2 θ))  cotθ
π/2 θ  cotθ
Which is what we were looking for, apart from a constant of π/2 which doesn't matter because it gets absorbed into the constant of integration the end
Saltine
 sum yun gai
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Re: Math(s)
ok, i get the principle of this but i can't seem to set up the integral to save my life
"Find the fluid force on the side of the tank (in pic) where the dimensions are given in feet. Assume the tank is full of water."
ok, so i know the principle is W*h(y)*L(y), where w = 62.4. i have for my L(y) = 2x. since the side is a line, i have y = 3(x1). rearranging gives x = (y+3)/3, so L(y) = (2/3)(y+3). and since we're going from bottom to top (instead of a submerged tank) then h(y) = y?
but when i set up my integral i am not able to get the answer given in the back of the book. i don't see where i'm going wrong with the setup. help plz?
"Find the fluid force on the side of the tank (in pic) where the dimensions are given in feet. Assume the tank is full of water."
ok, so i know the principle is W*h(y)*L(y), where w = 62.4. i have for my L(y) = 2x. since the side is a line, i have y = 3(x1). rearranging gives x = (y+3)/3, so L(y) = (2/3)(y+3). and since we're going from bottom to top (instead of a submerged tank) then h(y) = y?
but when i set up my integral i am not able to get the answer given in the back of the book. i don't see where i'm going wrong with the setup. help plz?
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ntw3001 wrote:Sass has to come from the heart, not from the shirt.
traubster wrote:I find it irritating whenever I walk through a cemetery and there's not one gravestone that reads something like, "We're all grateful that he's dead. Sorry if he owed you money."
 loofah
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Re: Math(s)
Oops. I picked the wrong side.
Last edited by loofah on Wed Oct 13, 2010 11:04 pm, edited 1 time in total.
Saltine wrote:This is all logically consistent, but the artist does not go on to explain that you love Hitler. See, this is why logicians don't write popular music.
 sum yun gai
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Re: Math(s)
i believe the picture is supposed to represent the endon view of the tank but i could be wrong about that. since this came from an odd numbered problem i do have the answer i just can't find a way to get there. the answer is supposed to be 748.8lbs
ntw3001 wrote:Sass has to come from the heart, not from the shirt.
traubster wrote:I find it irritating whenever I walk through a cemetery and there's not one gravestone that reads something like, "We're all grateful that he's dead. Sorry if he owed you money."
 Saltine
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Re: Math(s)
syg... I think your problem is that h(y)=3y and not y. h(y) is the depth of the water as a function of y, and it looks like you're measuring y to be 0 at the bottom of the tank and 3 feet at the top. The depth of the water is 3 feet at the bottom of the tank and 0 at the top. So 3y. (The sanity check is that the pressure needs to increase as you go down.)
Everything else looks right to me, so try again.
Everything else looks right to me, so try again.
Saltine
 Saltine
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Re: Math(s)
Here's a good one you may have heard before: A mother is 21 years older than her son. In six years, she will be five times as old as her son. What is she doing right now?
Saltine
 sum yun gai
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Re: Math(s)
is she.... makking babby?
ntw3001 wrote:Sass has to come from the heart, not from the shirt.
traubster wrote:I find it irritating whenever I walk through a cemetery and there's not one gravestone that reads something like, "We're all grateful that he's dead. Sorry if he owed you money."
Re: Math(s)
Well, that's a somewhat indelicate question!
I have an additional problem for you. To be fun, I suggest a couple of restrictions:
1) don't look up this problem on the internet (it's got a few very complete write ups)
2) think about it for a while prior to writing a program to find the answer for you. If you solve the problem this way, indicate how many calculations were necessary to get your answer.
The problem:
A customer walks into a local 711 in Oregon (or any other mystical land where there is no sales tax) and grabs four items from the shelf.
Right as he is about to check out, the power goes off. The cashier indicates that it's no problem, as he has a calculator somewhere.
The cashier dusts off the calculator, and sets about totaling the four items. He finishes with a flourish, and then pauses and says, "Now that's odd... the total is $7.11. Huh!"
He thoughtfully provides exact change, and as he is about to leave, the cashier gets a panicked look on his face and stops him.
"I'm really sorry, but I didn't total the products correctly! I accidentally multiplied all the prices together instead of adding them!"
The customer sighs, and put the items back on the counter. The cashier again totals them up (making sure to use the "+" key this time). The cashier
finishes and then, looks distressed again. He says "Now that's just strange! The total is $7.11 again!".
Assuming that no additional errors were made in these two calculations, what are the prices of the four items?
I have an additional problem for you. To be fun, I suggest a couple of restrictions:
1) don't look up this problem on the internet (it's got a few very complete write ups)
2) think about it for a while prior to writing a program to find the answer for you. If you solve the problem this way, indicate how many calculations were necessary to get your answer.
The problem:
A customer walks into a local 711 in Oregon (or any other mystical land where there is no sales tax) and grabs four items from the shelf.
Right as he is about to check out, the power goes off. The cashier indicates that it's no problem, as he has a calculator somewhere.
The cashier dusts off the calculator, and sets about totaling the four items. He finishes with a flourish, and then pauses and says, "Now that's odd... the total is $7.11. Huh!"
He thoughtfully provides exact change, and as he is about to leave, the cashier gets a panicked look on his face and stops him.
"I'm really sorry, but I didn't total the products correctly! I accidentally multiplied all the prices together instead of adding them!"
The customer sighs, and put the items back on the counter. The cashier again totals them up (making sure to use the "+" key this time). The cashier
finishes and then, looks distressed again. He says "Now that's just strange! The total is $7.11 again!".
Assuming that no additional errors were made in these two calculations, what are the prices of the four items?
Last edited by efnar on Sat Nov 06, 2010 5:53 pm, edited 1 time in total.
"Keep in mind that it's hard to rob someone who has taken the precautionary measure of setting him or herself ablaze."
 Saltine
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Re: Math(s)
SHOW YOUR WORK!sum yun gai wrote:is she.... makking babby?
I messed around with the numbers and couldn't figure it out in ten minutes. I thought about it before I wrote the program. I don't know what you mean by a "calculation" exactly but it was a recursive program which hit the degenerate case about 2700 times before returning the answer. I can be more descriptive here in the future. I don't know if anyone else will try to solve it so I'll wait to post anything for a few.efnar wrote:think about it for a while prior to writing a program to find the answer for you. If you solve the problem this way, indicate how many calculations were necessary to get your answer.
Saltine
Re: Math(s)
Saltine wrote:I don't know what you mean by a "calculation" exactly.
I admit, I was thinking about counting machine adds and multiplies, but this is a somewhat boring task. Counting the total number of possibilities checked (I'll call this a "round") is a fine metric; let's go with that.
For additional metrics:
The sillybruteforce method that I wrote 12 years ago (an approach that one could think of as the worst case computationally) took 1,240,043,927 rounds to find an answer.
With some thought, an alternate (stillbruteforce flavored) approach runs in 136 <= rounds <= 5,550. (The ordering of certain choices is arbitrary, but using this approach wouldn't get the answer in less than 136 rounds, or more than 5550 rounds).
It is possible to solve this problem using paperandpencil only, though it involves some number theory...
Last edited by efnar on Sun Oct 17, 2010 2:19 pm, edited 1 time in total.
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 Saltine
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Re: Math(s)
My program looks something like this pseudocode:
The output is the number of cents in each product. It runs through the (n==1) case 2,707 times. If it were to run through all the rounds without finding anything it would take 49,512 rounds.
Code: Select all
findem (N, S, P2, P3, P5, P79) {
// Find N integers whose sum is S and whose product is 2^P2 * 3^P3 * 5^P5 * 79^P79
// returns a multiset of N such integers or null if no such set exists
// If you're only looking for one number, this is an easy check:
if (N==1) {
if (S==2^P2*3^P3*5^P5*79^P79)
return new multiset{S};
else
return null;
}
// otherwise, brute force your way through the factors of 2^P2 * 3^P3 * 5^P5 * 79^P79
for (Q79=P79; Q79>=0; Q79)
for (Q5=P5; Q5>=0; Q5)
for (Q3=P3; Q3>=0; Q3)
for (Q2=P2; Q2>=0; Q2) {
A = 2^Q2 * 3^Q3 * 5^Q5 * 79^Q79; // A is one of the factors
if (A>S) continue;
R = findem(N1,SA,P2Q2,P3Q3,P5Q5,P79Q79); // if A is in the solution, can we solve the subproblem?
if (R!=null) {
R.add(A);
return R;
}
}
}
print findem(4,711,6,2,6,1);
The output is the number of cents in each product. It runs through the (n==1) case 2,707 times. If it were to run through all the rounds without finding anything it would take 49,512 rounds.
Saltine
Re: Math(s)
Saltine wrote:The output is the number of cents in each product. It runs through the (n==1) case 2,707 times. If it were to run through all the rounds without finding anything it would take 49,512 rounds.
You can do somewhat better than that by noting that the item with the factor of 79 has only 7 possible options, 3 of which can be further excluded fairly quickly.
Here's a fairly complete writeup on various solution approaches.
The 711 Problem
Last edited by efnar on Sat Nov 06, 2010 5:55 pm, edited 2 times in total.
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 sum yun gai
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Re: Math(s)
this is a chem problem, and since i never took this part of chemistry i have NO idea what to do.
find the ph:
a) 0.155M HNO2
b) 0.240M CH3NH3I
c) 0.324M KC6H5O
find the ph:
a) 0.155M HNO2
b) 0.240M CH3NH3I
c) 0.324M KC6H5O
ntw3001 wrote:Sass has to come from the heart, not from the shirt.
traubster wrote:I find it irritating whenever I walk through a cemetery and there's not one gravestone that reads something like, "We're all grateful that he's dead. Sorry if he owed you money."
 Saltine
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Re: Math(s)
Wow, not the kind of thing I've done either. I think you need some external information... these are probably solutions in water, and you need to figure out what the concentration of H+ (or OH) is of such a solution in equilibrium, and then you take the base 10 log of it and get your answer or something. I think you need to know how each of those chemicals ionizes to proceed.
Saltine
 loofah
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Re: Math(s)
sum yun gai wrote:this is a chem problem, and since i never took this part of chemistry i have NO idea what to do.
find the ph:
a) 0.155M HNO2
b) 0.240M CH3NH3I
c) 0.324M KC6H5O
It's been a while since I took chemistry, but I think, as Saltine pointed out, you'd have to know the Ka value. You can generally look up Ka (or pKa) values for chemicals in books (or the intertubes).
All of these would be monoprotic, I think, breaking down into:
a) H+ NO2
b) CH3NH3+ I
c) K+ C6H5O
I am so beyond braindead right now, but this looks vaguely like what I remember: http://wiki.answers.com/Q/How_do_you_fi ... d_solution
I can look at it tomorrow (after I've slept) to make sure it seems reasonable.
Saltine wrote:This is all logically consistent, but the artist does not go on to explain that you love Hitler. See, this is why logicians don't write popular music.
 loofah
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Re: Math(s)
I've slept (yay! still tired, but...)
The explanation on the link above is kind of wordy. Here's some pictures that might help (sorry about the poor quality photos).
The other two should go similarly, but the equations (I think) would be:
CH3NH3+ + H2O <> CH3NH2 + H3O+ (The I is just a spectator ion)
C6H5O+ + H2O <> OH + HC6H5O (The K+ is a spectator ion)
For the last one, you would use a Kb instead of a Ka value, and instead of
Ka=[H3O+][A]/[HA], it would be
Kb=[BH+][OH]/[B]
Let me know if you have any questions.
EDIT: is it possible to install some sort of superscript/subscript support for BBcode?
The explanation on the link above is kind of wordy. Here's some pictures that might help (sorry about the poor quality photos).
The other two should go similarly, but the equations (I think) would be:
CH3NH3+ + H2O <> CH3NH2 + H3O+ (The I is just a spectator ion)
C6H5O+ + H2O <> OH + HC6H5O (The K+ is a spectator ion)
For the last one, you would use a Kb instead of a Ka value, and instead of
Ka=[H3O+][A]/[HA], it would be
Kb=[BH+][OH]/[B]
Let me know if you have any questions.
EDIT: is it possible to install some sort of superscript/subscript support for BBcode?
Last edited by loofah on Fri Oct 29, 2010 2:25 am, edited 2 times in total.
Saltine wrote:This is all logically consistent, but the artist does not go on to explain that you love Hitler. See, this is why logicians don't write popular music.
Re: Math(s)
This is roughly my reaction to this thread.
ntw3001 wrote:you can't get raped if you always say yes
Re: Math(s)
As a noncatoriented aside, I revised my writeup of the 711 problem so that I didn't have to wave my hands part way through. I didn't need to worry about 79 dividing the sum of the prospective values, it was enough just to note that the sum was too large.
"Keep in mind that it's hard to rob someone who has taken the precautionary measure of setting him or herself ablaze."
 sum yun gai
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Re: Math(s)
loofah wrote:I've slept (yay! still tired, but...)
The explanation on the link above is kind of wordy. Here's some pictures that might help (sorry about the poor quality photos).
i've got the explanations of how to do the problems, but the Ka and Kb values were not given. all that was given was the above. my girlfriend said she asked her prof about it, and the prof admitted that the Ka or Kb values should have been given in order to do the problems properly
ntw3001 wrote:Sass has to come from the heart, not from the shirt.
traubster wrote:I find it irritating whenever I walk through a cemetery and there's not one gravestone that reads something like, "We're all grateful that he's dead. Sorry if he owed you money."
 Nyperold
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Re: Math(s)
Saltine wrote:Here's a good one you may have heard before: A mother is 21 years older than her son. In six years, she will be five times as old as her son. What is she doing right now?
Let's see:
x = age of son now
y = age of mother now
If x+6=6, y+6=27.
If x+6=5, y+6=26.
On a hunch, let me try this:
x+6=5.25, y+6=26.25.
i.e., x=.75, or 9 months before he was born.
Here's my answer:
I'm gonna go with sum yun gai's answer, there. If you want an artistic rendition, however, that's someone else's purview.
TICK
TICK
TICK
DundunDUNdunDUN!
Critical thinking is the key to a success!
 MysticalDescent
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Re: Math(s)
Feels a little embarrassing to ask such a daft question when I know that the answer is fairly simple, but there's a question from an exam I had last week where I just can't make the answer drop out and I'd appreciate it if someone could show me where I go wrong. I was given a wavefunction and was asked to normalise it to find the normalisation constant, A. The particle could only occur in between 0 < x < infinity. I've written down the important bits of the question here, I'm assuming that I keep going wrong in the integral.
They gave that standard integral at the end of the question. Instead of getting the value of A that they want, I end up with 16a^2 rather than 32a^3.
They gave that standard integral at the end of the question. Instead of getting the value of A that they want, I end up with 16a^2 rather than 32a^3.
 Saltine
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Re: Math(s)
Oookay, limits of all integrals shown are from 0 to ∞:
∫ Ψ*Ψ dx = 1
∫ A ∜x exp(ax) exp (+ihE/2πt) A ∜x exp(ax) exp (ihE/2πt) dx = 1
∫ A² √x exp(2ax) dx = 1
I'm going to change the variable from x to 2ax, so I'm going to multiply both sides by 2a and √(2a)
∫ A² √(2a) √x exp(2ax) 2a dx = 2a√(2a)
∫ A² √(2ax) exp(2ax) d(2ax) = 2a√(2a)
Note that if x=0, 2ax=0, and if x=∞, 2ax=∞, so this is just
∫ A² √u exp(u) du = 2a√(2a)
A² ∫ √u exp(u) du = 2a√(2a)
There's your standard integral
A² √π / 2 = 2a√(2a)
Rearrange
A² = 4a√(2a) / √π
A² = √(16a²)√(2a) / √π
A² = √(32a³/π)
A = ∜(32a³/π)
So, my guess is that you left out the factor of √(2a) necessary to convert √x into √(2ax) for the variable change to work. That is, you ended up with ∫ √x exp(2ax) d(2ax) instead of ∫√(2ax) exp(2ax) d(2ax).
∫ Ψ*Ψ dx = 1
∫ A ∜x exp(ax) exp (+ihE/2πt) A ∜x exp(ax) exp (ihE/2πt) dx = 1
∫ A² √x exp(2ax) dx = 1
I'm going to change the variable from x to 2ax, so I'm going to multiply both sides by 2a and √(2a)
∫ A² √(2a) √x exp(2ax) 2a dx = 2a√(2a)
∫ A² √(2ax) exp(2ax) d(2ax) = 2a√(2a)
Note that if x=0, 2ax=0, and if x=∞, 2ax=∞, so this is just
∫ A² √u exp(u) du = 2a√(2a)
A² ∫ √u exp(u) du = 2a√(2a)
There's your standard integral
A² √π / 2 = 2a√(2a)
Rearrange
A² = 4a√(2a) / √π
A² = √(16a²)√(2a) / √π
A² = √(32a³/π)
A = ∜(32a³/π)
So, my guess is that you left out the factor of √(2a) necessary to convert √x into √(2ax) for the variable change to work. That is, you ended up with ∫ √x exp(2ax) d(2ax) instead of ∫√(2ax) exp(2ax) d(2ax).
Saltine
 MysticalDescent
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Re: Math(s)
That's magnificent. Many, many, many thanks.
I think you're right when you say I missed the factor of √(2a) out. I just didn't substitute in properly and ended up messing it up.
To be honest, my maths isn't up to the same standard as some of the students who I'm working with whose second subject is maths. I just don't see the little tricks I need to do with integration sometimes.
I think you're right when you say I missed the factor of √(2a) out. I just didn't substitute in properly and ended up messing it up.
To be honest, my maths isn't up to the same standard as some of the students who I'm working with whose second subject is maths. I just don't see the little tricks I need to do with integration sometimes.
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