Tremulous Forum
Community => Off Topic => Topic started by: NiTRoX on December 05, 2007, 05:47:11 pm
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Same as Name the game but instead you have to give an answer to the math and/or quiz question.
Graphs
You are given 2 coordinates (1,1) and (5,5). What is the gradient?
Are the lines parallel?
Give reasons for your second answer answer.
1 point for each correct answer, total of 2 points
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the gradiant is 1.
You have 2 lines cut by a line. The angles are for point a 30,30,150,150 for pont b 20, 20, 160, 160.
Are the lines parallel?
hopw I understood the rules right.
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Very good altough it was a bit over detailed.
As a reason you should of told me yes, because they have the same gradient and because the values are bot positive and if divided by themselves they all give 1.
Its your turn demenator
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Very good altough it was a bit over detailed.
As a reason you should of told me yes, because they have the same gradient and because the values are bot positive and if divided by themselves they all give 1.
Its your turn demenator
........
[/quote]the gradiant is 1.
Edit: I said its one.
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Very good altough it was a bit over detailed.
As a reason you should of told me yes, because they have the same gradient and because the values are bot positive and if divided by themselves they all give 1.
Its your turn demenator
........
the gradiant is 1.
[/quote]
I was referring to "are the lines parallel?" question.
Post your math/quiz question please.
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ok
You have a right angle triangular with sides 4 and 3 coulculate 3rd side.
...
Now back to the question you asked about gradiant.
sry I misread that I had to reason my anser. Can I have another question :P ?
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Its a 345 triangle so the 3rd side is 5.
You can also use Pythagoras' Theorem.
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your anser is totaly corect
NiTRoX +1
can I have another question???
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Which man in history was known as the "Iron Chancellor"?. The name has to be in full, not just the surname.
Points so far:
NiTRoX - 1
Demenator - 1
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Otto Eduard Leopold von Bismarck-Schönhausen
wikipedia knows all
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o.0
Correct
Ok the name didn't have to be THAT full.
Points so far:
NiTRoX - 1
Demenator - 2
Your turn.
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how hot will an iron bar with weight of 2.5 kilos get if it absorbes 780 kj (kilo joules). I have to coulculate for unser muself :p.
Edit: and I will not be peresent on forums till tomorow so you can ask next question.
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You have to give me two more values: the change in temperature or the specific heat capacity of the iron bar.
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sry just the heat change.
And heat compasity if I remember right marked with "C" is 480. I mean I am not sure what it is marked with but I remember the heat compasity. It is surely 480. Remmber it from science book. Just in case I will check by tomorow.
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you still have to give me one more value xD
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you still have to give me one more value xD
Why cant you find it yourself :P
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form wikipedia free encyclopedia
Heat capacity of iron "(25 °C) 25.10 J·mol−1·K−1 "
Dont get this but in my scince book it was 480 joules per kelvin.
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30,000?
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30, 000 k?
Anyway its wrong as far as I know.
Sry I din't put much structure let me try again. C is heat compasity.
C= 480j/k
weight - 2.5 kilos
energy - 780 kj
Coulculate tempreture change in Kelvin.
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1.5384615384615384615384615384615 degrees celsius :)
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ask me for formula when you are tired of guesing. Anyway other people should try.
And you should of posted pie whould of been way funnier ;D
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q = MC/\t
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q = MC/\t
this is not the formula. And if "Q" stand for energy it should be capital.
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694.17 degrees Kelvin
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C= 480j/k
weight - 2.5 kilos
energy - 780 kj
(480*2.5)=1200 j for 1 degree kelvin of difference for the entire mass
780000/1200=650 degrees kelvin difference
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do your own homework :D
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C= 480j/k
weight - 2.5 kilos
energy - 780 kj
(480*2.5)=1200 j for 1 degree kelvin of difference for the entire mass
780000/1200=650 degrees kelvin difference
You are right
Survivor +1
ask the next question ;D
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Ohh noes :(
Demenator - 2
NiTRoX - 1
Survivor - 1
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I suck at stating quiz questions. Give me a while.
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I suck at stating quiz questions. Give me a while.
They don't neccesarily have to be math questions. They can be from history,geography, language etc ..
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Topic: Literature history
A certain author has written a story consisting of 6 different parts, dealing with reanimation in the genre of horror. Name both the author and the title of all 6 parts when combined together. As a hint I'll say that the author is most famous for one of his squidlike creations.
Once you get the answer I suggest you read that story. It's a nice one :)
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HP Lovecraft's "Herbert West: Re-Animator"
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Point goes to kozak6.
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What is the first derivative of the hyperbolic cosecant of x with respect to x?
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Is it this?
(http://upload.wikimedia.org/math/d/4/e/d4e96e3c1d1784e567bf497f0c180834.png)
Althrough I dont understand anything about this formula I googlesd it up. So I could as well be wrong.
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Its definately wrong, you need to mention x.
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d/dx(cosh(x)) = sinh(x)
Dur.
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Mmmm....
Close.
I asked for d/dx(csch(x))
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d/dx(csch(x))
1/x(csch(x))
1/x(1/sinh(x))
that's as far as i got but i am only 12 so what can you expect?
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Not quite.
d/dx refers to taking the derivative of a function. The d's don't actually cancel.
Taking the derivative is a rather fascinating mathematical operation you can read all about here: http://en.wikipedia.org/wiki/Derivative .
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Derivatives are a very annoying mathematical operation i have to learn all about in university. You do not want to see what they mess us up with >:E
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(d/dx) csch x = -csch x coth x
csch x = 1/sinhx
d/dx 1/sinhx = - coshx/sinhx^2 = - 1/sinhx * coshx/sinhx = - cschx * cothx
=> (d/dx) csch x = -csch x coth x
Used :
(cosh x)' = sinhx
(U/V)' = (U'V - UV')/V^2
(1)' = 0
nothing really hard here.
EDIT :
Pretty sure that the answer is right, if not then Ill delete this edit :)
Problem :
An infinitely extensible rubber band is attached to a wall at one side and to a car at another one. the initial length of a rubber band is 2 m. There is an ant on the rubber band right near the wall, which moves along the band into the direction of the car.
Simultaneously ant and a car starting to move, speed of ant is 1 m/s speed of the car is 2 m/s. Car is moving away from the wall. When will ant reach the car? (and can it do so? Can it reach it if car moves at 0.5 m/s?)
Take the ant as a dimensionless object. Rubber can be extended infinitely. Car and ant moves at constant speed.
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Whups. Misread it. I thought it was derivative of the hyperbolic cosine of x. :P
d/dx(csch(x)) = -csch(x)coth(x) Troy got it.
Soo...you're saying that the ant is moving in the direction of the car at 1 m/s while the car is moving 2 m/s, and the car is 2 meters in front? So the diagram would look like:
|>..........>>
| <-- wall
> <-- ant
.......... <-- represents 2 meters
>> <-- car
Am I correct? Also, since the rubber band is infinitely extensible, there is no force acting on the ant that would make it move faster or accelerate (Assuming this because you said speeds are constant)?
If that is the case, the ant cannot catch the car and the car will constantly outrun the ant. If the car moves at .5 m/s, then the problem is changed. In that case, the ant, with position equation: x = t, and the car, with position x = 2 + .5t, would be at the same position when t = 4 seconds.
Correct?
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W|
A|
L|Ant -> (1 m/s)
L|----------------------Car -> (2 m/s)
=|rubber band 2m
=|
=|
(Ill get normal drawing soon)
Ant's speed is relative to the surface he moves on. So if he is on top of the car his speed is 1 m/s relative to the car and 3 m/s relative to the floor. There is a sufficient friction force between ant and band to allow ant not to slid on the band.
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If that is the case, the ant cannot catch the car and the car will constantly outrun the ant. If the car moves at .5 m/s, then the problem is changed. In that case, the ant, with position equation: x = t, and the car, with position x = 2 + .5t, would be at the same position when t = 4 seconds.
^^Is that the correct answer?
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no :) Band stretches => ant has to crawl longer distance each time, but there is a trick there.
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The behavior of the ant with respect to the expanding point on which it stands hasn't been clearly defined. Am I right in saying this infinitely stretching band has a velocity at distance x from the wall of C(x/L), where C is the speed of the car and L is the current length of the band and that the ant has velocity Vr + C(x/L), where Vr is the ants velocity relative to the point on the band on which it stands?
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Good, that is the main idea of the problem. And well, ant is a mathematical point + as I said, there is enough friction for it to not slid = it will grip in to a stretching point, gaining the speed of it + its own speed relative to that point. Good start Nux.
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In which case (and please forgive me if I get this terribly wrong) it can be shown that for Vr = 1, C = 2, L = 2 and time t:
V = Vr + C(x/L)
V = 1 + 2(x/2) = 1 + x
x (= Vt) = (1 + x)t
t = x/(1+x)
Where the Cars distance Cx = 2t + 2, for the distances to be equal,
t = 2t+2/(1+2t+2) = 2t+2/2t+3
2t2 + 3t = 2t + 2
2t2 + t - 2 = 0
t = (-1 + sqrt(1+16))/4 = (-1 + sqrt(17))/4 [or t = (-1 - sqrt(17))/4 but this has no meaning here)]
If the C = 0.5 then,
V = 1 + 0.5(x/2) = 1 + x/4
x = (1 + (x/4))t
t = x/(1+(x/4))
Cx = 0.5t + 2
t = 0.5t + 2/(1+0.5t + 2) = 0.5t+2/0.5t+3
0.5t2 + 3t = 0.5t + 2
0.5t2 + 1.5t - 2 = 0
t = (-1.5 + sqrt((1.5)^2+4)) = -1.5 + sqrt(25/4) [or -1.5 - sqrt(25/4) but not meaningful here]
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~6.3891s
~2.5949s
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In which case (and please forgive me if I get this terribly wrong) it can be shown that for Vr = 1, C = 2, L = 2 and time t:
V = Vr + C(x/L)
V = 1 + 2(x/2) = 1 + x
Wrong. You cannot substitute 2m right away. Say if the car moved and the distance is 4m from the wall at which ant (for example) reaches the car. from your equation his speed is 5 m/s which is not possible, since his maximum speed can be only 3 m/s (sitting on the car). L is a variable, not a constant.
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Ah yes. I'd noticed that the first time, but managed to forget it to make things easier for myself =)
When trying to integrate this velocity function, I run into a lot of trouble with these inner functions which are defined in terms of each other. I've tried limits, recursive definitions and pretending variables are constants >.>..
All in all I haven't had enough practice with this sort of problem.
Funnily enough, I made up a new variable Cx which meant exactly the same thing as L. xD
Let's start again:
V (= dx/dt) = Vr + C(x/(Ct+L))
Where L is now the initial length.
As a guess, I'd say this ant has a positive but exponentially decaying acceleration.
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As a guess, I'd say this ant has a positive but exponentially decaying acceleration.
Relatively to what? :)
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Relative to the same frame as the car's 2m/s (the ground). If I meant relative to the band then that's already covered by the ant's stated velocity of 1m/s.
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Okay, I'll pick up where Nuxy left off.
V (= dx/dt) = Vr + C(x/(Ct+L)) = Vr + x/(t+L/C)
Writing this out a little differently, we get: dx/dt - 1/(t+L/C)*x = Vr
Now, I'm going to define a cute little function as: M(t) = 1/(t+L/C)
I multiply our previous equation by my new function: M*dx/dt - (M/(t+L/C))*x = M*Vr
Which happens to be (feel free to check this, but I defined it this way on purpose): M*dx/dt + dM/dt*x = M*Vr
Using a well-known law which I don't know the English name for (product rule! thanks, nux!):
d(M*x)/dt = M*Vr = Vr/(t+L/C)
Integrating both sides (according to t, of course), gives: M*x = Vr*ln(t+L/C)
We then need only divide by M to get: x = Vr*(t+L/C)*ln(t+L/C)
In order to find out when the ant will reach the car, let's equate!
Vr*(t+L/C)*ln(t+L/C) = Ct + L = C*(t+L/C)
Vr*ln(t+L/C) = C
ln(t+L/C) = C/Vr
t + L/C = e^(C/Vr)
t = e^(C/Vr) - L/C
Substituting the given Vr = 1, C = 2, L = 2, we get: t = e^2 - 1 seconds
And for the second part (Vr = 0.5, C = 2, L = 2), we get: t = e^4 - 1 seconds
Yes - so long as the ant is moving, no matter how slow, it will eventually reach the car. Pretty neat. ^_^ So, am I right?
(By the way, I'm mildly disappointed in you, Nux. You said you love math! ^_~)
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I do love maths! I really do!
This is good, because it's made me want to practice more.
I understand everything you've put, but only through practice can you do it yourself.
Edit: Oh, and that law you didn't know the egnlish name of is the 'Product Rule'.
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Well, the one thing I didn't explain is how I got to M, and I can explain that if you wish. But this is easy stuff.... this was the very first sort of problem I learned how to solve in my differential equations course. I can give you a harder problem if I won. ^_^
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Okay, I'll pick up where Nuxy left off.
V (= dx/dt) = Vr + C(x/(Ct+L)) = Vr + x/(t+L/C)
Writing this out a little differently, we get: dx/dt - 1/(t+L/C)*x = Vr
Now, I'm going to define a cute little function as: M(t) = 1/(t+L/C)
I multiply our previous equation by my new function: M*dx/dt - (M/(t+L/C))*x = M*Vr
Which happens to be (feel free to check this, but I defined it this way on purpose): M*dx/dt + dM/dt*x = M*Vr
Using a well-known law which I don't know the English name for (product rule! thanks, nux!):
d(M*x)/dt = M*Vr = Vr/(t+L/C)
Integrating both sides (according to t, of course), gives: M*x = Vr*ln(t+L/C)
We then need only divide by M to get: x = Vr*(t+L/C)*ln(t+L/C)
In order to find out when the ant will reach the car, let's equate!
Vr*(t+L/C)*ln(t+L/C) = Ct + L = C*(t+L/C)
Vr*ln(t+L/C) = C
ln(t+L/C) = C/Vr
t + L/C = e^(C/Vr)
t = e^(C/Vr) - L/C
Substituting the given Vr = 1, C = 2, L = 2, we get: t = e^2 - 1 seconds
And for the second part (Vr = 0.5, C = 2, L = 2), we get: t = e^4 - 1 seconds
Yes - so long as the ant is moving, no matter how slow, it will eventually reach the car. Pretty neat. ^_^ So, am I right?
(By the way, I'm mildly disappointed in you, Nux. You said you love math! ^_~)
Bravo! Very well done. Your turn, sir. :)
just a bit of misunderstanding in the end : the speed of ant is still 1ms and car is now moving at 0.5 ms. But this is not important, since you gave the general formula, which is exactly what I was looking for.
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Yay! ^_^ Okay, here's a fun one, not too hard. Knowing that x3- x -1 = 0, determine the exact value of: (3x2 - 4x)1/3 + x(2x2+3x+2)1/4
(This is a really easy problem to solve with a calculator, but it's much cooler/funner to solve algebraically. Don't disappoint me!)
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Yay! ^_^ Okay, here's a fun one, not too hard. Knowing that x3- x -1 = 0, determine the exact value of: (3x2 - 4x)1/3 + x(2x2+3x+2)1/4
(This is a really easy problem to solve with a calculator, but it's much cooler/funner to solve algebraically. Don't disappoint me!)
2x3 + 4x2 - 1.55
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Sorry, but no. The answer I'm looking for is independent of x.
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using an Easy 3rd Degree Equation Solver, I got x = 1.3247.
...then I picked up a calculator and got 2.0
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Like I said, it's a very easy problem to solve with a calculator. But if you guys don't want to delve into the algebra, that's fine by me. Go ahead, your turn.
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I was just kidding xD
Base equations used:
x3-x-1 = 0
x = x3-1
x3 = x+1
1 = x(x2-1)
The path:
( 3x2-4x )1/3 + x*( 2x2+3x+2 )1/4
( 3(x3-1)2-4x )1/3 + x*( 3x2+3x+2x(x2-1) )1/4
( 3x6-63+3-4x )1/3 + x*( x(3x+3+2x2-2) )1/4
( 3x6-63+3-4(x3-1) )1/3 + x*( x(2x2+2x+1) )1/4
( 3x6-10x3+7 )1/3 + x*( x(2x2+2x+x(x2-1)) )1/4
( 3(x3)2-10x3+7 )1/3 + x*( x2(2x+2+x2-1) )1/4
( 3(x+1)2-10x3+7 )1/3 + x*( x2(x2+2x+1) )1/4
( 3x2+6x+3-10x3+7 )1/3 + x*( x2(x2+1)2 )1/4
( -10x3+3x2+6+10 )1/3 + x*( x(x2+1) )1/2
( -x3-9x3+3x2+6+10 )1/3 + x*( x(x2+x(x2-1) )1/2
( -x3-9(x+1)+3x2+6+10 )1/3 + x*( x2(1+x2-1) )1/2
( -x3+3x2-3x+1 )1/3 + x*( x2(x2) )1/2
( 1-3x+3x2-x3 )1/3 + x*( x4 )1/2
( (1-x)3 )1/3 + x*x2
1-x+x3
(x3-x-1)+2
2
PS1: btw do u know how much it took me to get this through algebra? 4 fucking hours
PS2: Troy, I posted correct answers to your question before anyone else. Why the silent ignore? You could have just said "I need details".
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Yaaaay! =D
PS1: btw do u know how much it took me to get this through algebra? 4 fucking hours
PS2: Troy, I posted correct answers to your question before anyone else. Why the silent ignore? You could have just said "I need details".
Sorry! o_o It took me a while too, but I thought it was fun...
And I hadn't noticed that you had reached the correct answer, since you wrote it in decimal form, so maybe Troy didn't either?
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Yaaaay! =D
PS1: btw do u know how much it took me to get this through algebra? 4 fucking hours
Sorry! o_o It took me a while too, but I thought it was fun...
Yeah it was! (Did I sound angry? xD)
Could you plz name the source (book,etc.)?
I'll think of a challenge tomorrow. // ZzZ...
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I am in Newton's 3D space.
I am at location P, and moving in direction&speed S.
My target is at location T, and moving in direction&speed V.
I am to shoot a projectile of a constant speed f in a direction D, so that it hits my target.
The projectile picks up my current speed, as in Newtonian physics.
Nothing is accelerating/decelerating.
Where should I shoot? (Create a formula to get D!)
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I am in Newton's 3D space.
I am at location P, and moving in direction&speed S.
My target is at location T, and moving in direction&speed V.
I am to shoot a projectile of a constant speed f in a direction D, so that it hits my target.
The projectile picks up my current speed, as in Newtonian physics.
Nothing is accelerating/decelerating.
Where should I shoot? (Create a formula to get D!)
Can I just quote the particle prediction algorithm used for the trapper? :D
Working on 3D model, got the 2D equations, now its just a hassle of maths going from 2D to 3D :)
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Sorry! o_o It took me a while too, but I thought it was fun...
Yeah it was! (Did I sound angry? xD)
Well, your use of the word "fucking" gave me that idea. :P
Could you plz name the source (book,etc.)?
Yup - here it is (http://perplexus.info/show.php?pid=5083). (TamTam is me)
Now for the new problem. Lessee....
Since you want to negate your current speed, and since you want to catch up to the second particle, I'll start with a speed and direction of (V-S). Using only this speed, the projectile would end up moving parallel to the target, matching its speed. So now the problem can be simplified to two static objects. The direction from one object to the other is (T-P), so we'll need to add some speed E in that direction. The magnitude of E doesn't matter terribly, as long as it's greater than zero.
So I'd say f = (V-S) + E*(T-P), and the direction is already included in there, since we're dealing with vectors.
I know you asked specifically for D, but can I get away with D = f/|f|? XD
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f is a linear speed value, like 800m/s. The direction is unknown. Combining speed and direction, the raw trajectory would be f*D+S.
For example:
Alert! Enemy starship approaching P(5,4,3)km, S(1,2,3)km/s from T(11,-2,-3)km, V(3,2,4)km/s! We are shooting a linear rocket moving at f=20km/s. But uhm, where to? A moving target is hard to hit... or is it?
OK let's rename!
First, v := f. Second, we are relatively stationary, and are at the relative origin. The target is at location L (= T-P), and heading in direction&speed H (= V-S).
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Okay, now I have absolutely no idea what you're talking about. XD Wait, nevermind. I got it.
I still think you should fire the rocket at H + E*L, where E > 0. You want that in specifics?
H = (2,0,1), L = (6,-6,-6). Vfinal = (2+6E,-6E, 1-6E)
f = |Vfinal| = √((2+6E)2 + (-6E)2 + (1-6E)2) = √(4+24E+36E2+36E2+1-12E+36E2) = √(108E2+12E+5)
D = Vfinal/f = ( (2+6E)/f, -6E/f, (1-6E)/f )
Why do I feel like I'm making a silly mistake? :D
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NitPicking time! >:D
The projectile picks up my current speed, as in Newtonian physics.
Nothing is accelerating/decelerating.
Kudos for launching a projectile without acceleration. :P (OMG a joke! Yes I know it's not important to the problem/accelerated yet)
For example:
Alert! Enemy starship approaching P(5,4,3)km, S(1,2,3)km/s from T(11,-2,-3)km, V(3,2,4)km/s! We are shooting a linear rocket moving at f=20km/s. But uhm, where to? A moving target is hard to hit... or is it?
Where are these values relative to? Surely we only need the velocity and position relative to us.
Also, why are we firing a rocket if we're in space and it has a constant velocity? Are we using Hollywood physics? I want to fire a hammer instead!
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Haha, Nux! XD But you know, in physics we use infinitesimal acceleration all the time....I've gotten used to it already. ^_^
And who said we're in space, huh? What if we're in starships in a non-vacuum, and our rocket is fired with continuing acceleration that exactly negates the friction? :P
Addition to my solution - the projectile will take 1/E seconds to reach the target.
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And who said we're in space, huh?
Alert! Enemy starship approaching...
Perhaps the starship is approaching on wheels? Maybe it's lost in deep-sea and we're an angry submarine?
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Many sci-fi starships can handle themselves just fine in an atmosphere, and even underwater. ^_^
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I would -have -said -that -be-fore -you -if -it -weren-'t -for t-his -damn -min-us -key sub-tracting- -everything- -I -say ->:(
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Aww. *pat pat* Try shifting that evil minus into an underscore! Maybe that'll help.
Anyway, I want to know if I'm right so I can post another problem. x_x Come on, humancontroller! I don't have all day!
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Well maybe I don't have all day too... to press the refresh webpage button every minute.
I'm asking for a formula, like D = bla(bla/sqrt(bla+bla)-bla)
Why is your answer/start_of_answer weird? well,
f*D = H + E*L
f*D = H + 1/t*L
t*f*D = H*t + L
L = H*t+L
H*t = 0
wtf, t = 0, or |H| = 0?
TH3 SYST3M: http://drop.io/7ovjgpj
We are not in water, not air, not is vacuum, but the world of pure Newtonian space. That's something that doesn't exist. Just like geometry. Humans invented, in their mind, both.
Troy, the blob algorithm is, well, OMFGWTFNOOBHAX?!
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.....except t*f*D != L. I still think I'm right.
*sigh*
D = (H + E*L)/√((Hx+ELx)2 + (Hy+ELy)2 + (Hz+ELz)2)
I'm almost certain that's not what you want....but I can't find a problem with it, nor can I find a better way to write it. :| I phail.
And by the way, I was kidding about not having all day.... XD
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OK I've slept on this one. Yeah you're right, it was just weird for me at a first glance. Still, I'm telling you: |D| = 1
We still don't know the value of E, but keep going :)
Here's an example:
Target: T(-2,4), V(7,6)
Me: P(-3,-5), S(3,3)
Deltas: L(1,9), H(4,3)
rocket speed: v = 13
results of calculation:
direction: D(5/13,12/13) (|D| = 1)
time to hit: t = 1 // but this isn't the question
rocket's relative trajectory: v*D = (5,12) (|v*D| = 13 = v)
rocket's absolute trajectory: S+v*D = (8,15) (absolute speed = 17 !)
target's relative coordinates after t time: L + t*H = (5,12)
rocket's relative coordinates after t time: t*v*d = (5,12)
target's absolute coordinates after t time: T + t*V = (5,10)
rocket's absolute coordinates after t time: P + t*(S+v*D) = (5,10)
// bullseye in both cases
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My |D| = 1, thankyouverymuch.
Your example uses my formula with E = 1, and therefore t = 1/E = 1. If, for example, you had used E = 2, you would have gotten:
Rocket speed = f = √477
D = (6/√477,21/√477)
|D| = 1
t = 1/E = 1/2
target's relative coordinates after t time: L + t*H = (3,10.5)
rocket's relative coordinates after t time: t*f*d = (3,10.5)
target's absolute coordinates after t time: T + t*V = (1.5,7)
rocket's absolute coordinates after t time: P + t*(S+f*D) = (1.5,7)
Still bullseyeing. ^_^ Any E works, I'm telling ya.
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But where do u get E from, and why do you say "you would have gotten f = √477", when f is a given value?
Listen. Here's the given data:
Dretch: T(5,-1), V(7,6)
Me: P(-13,-3), S(3,3)
Lucifer cannon's speed: v (aka. f) = 13
(and although the lucifer cannon doesn't pick up my current speed, and the luci's speed isn't actually "13", but it doesn't matter, it's only for the sake of example)
gimme D, so that i can hit that dretch
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But wait. Why are you giving examples in 2-space? Isn't this 3-space?
I attempted the problem on my own using vector analysis (Velocity is a vector quantity that can be broken up into 3 subsections, x, y, and z by using trigonometry and two angles. Yes, two angles in 3-space. One to determine how far above the ground and how far in the x-direction and another to determine how far off of the xy plane it is (z).)
I got extremely complicated formulas that did not effectively isolate theta and alpha (my two angles). I had 3 equations with 3 unknowns, though. Here's what I had, maybe someone else can extrapolate a bit more:
T = Target
R = projectile
V = Target velocity
S = projectile velocity
[] = anything within there is a subscript.
t = time
Target directions are given in the problem statement, so the only unknowns are time, and theta and alpha.
T- = T[x[0]] + Vcos(theta[Target direction])t
T[y] = T[y[0]] + Vsin(theta[Target direction])t
T[z] = T[z[0]] + Vcos(alpha[Target direction])t
R
R[y] = 0 + Ssin(theta)t
R[z] = 0 + Scos(alpha)t
In my analysis, since there is no acceleration, basic kinematic equations can be used. The equation I used is x[t] = x[0] + vt + 1/2at^2. a = 0 so it's simply x[t] = x[0] + vt. Since V is a velocity magnitude, in order to find the direction in a certain axis direction basic trig must be used to find exactly how much of V works in that direction.
Also, I "moved" my axis system so that the origin was located at where the projectile was fired. It's a simple calculation, you only subtract the particles initial position from the targets initial position and set the particle at the origin. It simplifies the algebra a bit, yet does not yield a fully simplified answer using the tools that I currently know.
So, after that little note you can set those six equations equal (In their respective axes), solve for t, substitute, and you will have 2 equations with only theta and alpha as unknowns. It's only complicated because simplifying trig functions isn't as easy as factoring out common factors in algebraic functions.
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But wait. Why are you giving examples in 2-space? Isn't this 3-space?
An example is just an example. The perfect solution would be for N dimensions, not just 2 or 3.
I attempted the problem on my own using vector analysis (Velocity is a vector quantity that can be broken up into 3 subsections, x, y, and z by using trigonometry and two angles. Yes, two angles in 3-space. One to determine how far above the ground and how far in the x-direction and another to determine how far off of the xy plane it is (z).)
I got extremely complicated formulas that did not effectively isolate theta and alpha (my two angles). I had 3 equations with 3 unknowns, though. Here's what I had, maybe someone else can extrapolate a bit more:
T = Target
R = projectile
V = Target velocity
S = projectile velocity
[] = anything within there is a subscript.
t = time
Target directions are given in the problem statement, so the only unknowns are time, and theta and alpha.
I would say "not only the time, but also the angles are unknown".
T- = T[x[0]] + Vcos(theta[Target direction])t
T[y] = T[y[0]] + Vsin(theta[Target direction])t
T[z] = T[z[0]] + Vcos(alpha[Target direction])t
R
R[y] = 0 + Ssin(theta)t
R[z] = 0 + Scos(alpha)t
In my analysis, since there is no acceleration, basic kinematic equations can be used. The equation I used is x[t] = x[0] + vt + 1/2at^2. a = 0 so it's simply x[t] = x[0] + vt. Since V is a velocity magnitude, in order to find the direction in a certain axis direction basic trig must be used to find exactly how much of V works in that direction.
Also, I "moved" my axis system so that the origin was located at where the projectile was fired. It's a simple calculation, you only subtract the particles initial position from the targets initial position and set the particle at the origin. It simplifies the algebra a bit, yet does not yield a fully simplified answer using the tools that I currently know.
So, after that little note you can set those six equations equal (In their respective axes), solve for t, substitute, and you will have 2 equations with only theta and alpha as unknowns. It's only complicated because simplifying trig functions isn't as easy as factoring out common factors in algebraic functions.
Here's a reference manual to help you with simplifications:
http://www.alcyone.com/max/reference/maths/trigonometry.html
If you can do this with trigonometry, then that's fine by me.
But for a general, N-dimensional solution, you would have to use vectors.
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But where do u get E from, and why do you say "you would have gotten f = √477", when f is a given value?
Ohhhhh. Now I understand why we weren't understanding each other - I thought I was supposed to find f as well as D.
Well then. Since |H + E*L| = f, we can find E:
√((Hx+ELx)2 + (Hy+ELy)2 + (Hz+ELz)2) = f
Hx2 + 2EHxLx + E2Lx2 + Hy2 + 2EHyLy + E2Ly2 + Hz2 + 2EHzLz + E2Lz2 = f2
E2(Lx2 + Ly2 + Lz2) + E(2HxLx + 2HyLy + 2HzLz) + Hx2 + Hy2 + Hz2 - f2 = 0
Slightly simplified, E2*|L|2 + E(2HxLx + 2HyLy + 2HzLz) + |H|2 - f2 = 0
E = -(2HxLx + 2HyLy + 2HzLz) + √( (2HxLx + 2HyLy + 2HzLz)2 - 4*|L|2*|H|2 ) * 1/(2|L|2)
And once you have E, D = (H + E*L)/√((Hx+ELx)2 + (Hy+ELy)2 + (Hz+ELz)2)
Is that too complicated? ???
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But wait. Why are you giving examples in 2-space? Isn't this 3-space?
An example is just an example. The perfect solution would be for N dimensions, not just 2 or 3.
I attempted the problem on my own using vector analysis (Velocity is a vector quantity that can be broken up into 3 subsections, x, y, and z by using trigonometry and two angles. Yes, two angles in 3-space. One to determine how far above the ground and how far in the x-direction and another to determine how far off of the xy plane it is (z).)
I got extremely complicated formulas that did not effectively isolate theta and alpha (my two angles). I had 3 equations with 3 unknowns, though. Here's what I had, maybe someone else can extrapolate a bit more:
T = Target
R = projectile
V = Target velocity
S = projectile velocity
[] = anything within there is a subscript.
t = time
Target directions are given in the problem statement, so the only unknowns are time, and theta and alpha.
I would say "not only the time, but also the angles are unknown".
T- = T[x[0]] + Vcos(theta[Target direction])t
T[y] = T[y[0]] + Vsin(theta[Target direction])t
T[z] = T[z[0]] + Vcos(alpha[Target direction])t
R
R[y] = 0 + Ssin(theta)t
R[z] = 0 + Scos(alpha)t
In my analysis, since there is no acceleration, basic kinematic equations can be used. The equation I used is x[t] = x[0] + vt + 1/2at^2. a = 0 so it's simply x[t] = x[0] + vt. Since V is a velocity magnitude, in order to find the direction in a certain axis direction basic trig must be used to find exactly how much of V works in that direction.
Also, I "moved" my axis system so that the origin was located at where the projectile was fired. It's a simple calculation, you only subtract the particles initial position from the targets initial position and set the particle at the origin. It simplifies the algebra a bit, yet does not yield a fully simplified answer using the tools that I currently know.
So, after that little note you can set those six equations equal (In their respective axes), solve for t, substitute, and you will have 2 equations with only theta and alpha as unknowns. It's only complicated because simplifying trig functions isn't as easy as factoring out common factors in algebraic functions.
Here's a reference manual to help you with simplifications:
http://www.alcyone.com/max/reference/maths/trigonometry.html
If you can do this with trigonometry, then that's fine by me.
But for a general, N-dimensional solution, you would have to use vectors.
=\. None of the formulas on that site you gave me simplify my trig. Oh well. I've given the problem enough thought for now. Unless I come across a new idea in the future, I'll most likely just wait for someone to solve the problem.
But, just to clarify myself, I am using vectors in my analysis. If I truly wanted, I could express position and velocity both in terms of vectors and do some analysis that way, but I doubt I'll get too far by doing it, so I'm not going to even bother. Kudos to making a fantastic problem.
-Bull
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Well then. Since |H + E*L| = f, we can find E:
√((Hx+ELx)2 + (Hy+ELy)2 + (Hz+ELz)2) = f
Hx2 + 2EHxLx + E2Lx2 + Hy2 + 2EHyLy + E2Ly2 + Hz2 + 2EHzLz + E2Lz2 = f2
E2(Lx2 + Ly2 + Lz2) + E(2HxLx + 2HyLy + 2HzLz) + Hx2 + Hy2 + Hz2 - f2 = 0
Slightly simplified, E2*|L|2 + E(2HxLx + 2HyLy + 2HzLz) + |H|2 - f2 = 0
E = -(2HxLx + 2HyLy + 2HzLz) + √( (2HxLx + 2HyLy + 2HzLz)2 - 4*|L|2*|H|2 ) * 1/(2|L|2)
And once you have E, D = (H + E*L)/√((Hx+ELx)2 + (Hy+ELy)2 + (Hz+ELz)2)
almost Correct! You forgot f out of the equation
To show what I mean about a general vector formula, I'll just rewrite this:
E = (-HoL +- √(HoL2 + (LoL)(f2 - HoH))) / (LoL) // o signifies a dot product
D = (H+E*L)/f // if E > 0
OK, Eternal, your turn.
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*cough cough* Yeah, I fail. I forgot the f and I missed a much simpler way of writing the answer. >_< But I'm out of practice! Haven't done any math in a while.
Mmm, my turn.....I don't know. Oh, y'know what? I'd like to see how you guys would solve this:
an+1 = an + 0.5n4 - 40n3 + n - 1
a1 = 1
a101 = ?
I never learned a formal way to solve this, though the solution is simple enough. I'm just wondering how widespread it is.
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an instant c program shows 5071616, now let's see if i can do this analytically xD
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Yep, that's right. :P If nobody succeeds in doing it analytically, you take the next one.
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a101 = 1 + 0.5*(14+24+...+1004) - 40*(13+23+...+1003) + (1+2+...+100) - 100*(1)
The formula for (1+2+...+n) is (n*(n+1)/2), for 100 that's 5050.
The formula for (13+23+...+1003) is ((n*(n+1)/2)2), for 100 that's 25502500.
a101 = 0.5*(14+24+...+1004) -1020095049
But I can't find the formula for (14+24+...+1004).
I'll google some more, or maybe you can tell me :]
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Why google when you can figure these things out on your own? :D This problem (up to n9, I think) is one I played around with and found an answer to when I was 15 or 16. Of course, I was taking a very different approach, so I dunno.
I'd still like to see if someone can do it analytically. ^_^ Just one part left!
EDIT: Oh wow, I just found the general solution (http://en.wikipedia.org/wiki/Faulhaber%27s_formula) online. <3 Awesomeness. So never mind, I lost my curiosity in seeing how you'd do it. XD
14 + 24 +...+ n4 = (6n5 + 15n4 + 10n3 - n)/30
Your turn again!
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P and T are static points in 3D.
P will throw something at T with a velocity of v.
The rock is affected by the gravity vector, G.
Where (direction: D) should P throw.. uhm, so that it hits T?
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D = left
;)
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So, uh, is 'v' the starting velocity? And there's no friction?
Assuming that similarly to the previous problem, you're not causing any acceleration, then the only acceleration is caused by the force of gravity. I'm defining the Z axis as up.
a(t) = -G
V(t) = v - Gt
X(t) = Px + vxt, Y(t) = Py + vyt, Z(t) = Pz + vzt - Gt2/2
Assuming that we picked the right direction, there will be a time t0 for which our projectile hits the target. In that case,
X(t0) = Px + vxt0 = Tx
Y(t0) = Py + vyt0 = Ty
Z(t0) = Pz + vzt0 - Gt02/2 = Tz
Of course, we also know that vx2 + vy2 + vz2 = v2
That gives us four equations, four unknowns. We'd express t0 using v, and then find the different components of v, and thus the solution. I'd do it now, but I'd got other things to do. I'll probably get to it later.
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G is a vector that may or may not be pointing downwards.
Typically, it would be (0,0,-9.81), but that is not guaranteed (only that |G| > 0).
Note the conformation: the "positive direction" of the axes, the negative value in G.
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G is a vector that may or may not be pointing downwards.
Seeing as you put no limitations on the axes, I can simply define them so that G is indeed pointing downwards. And if that doesn't satisfy you, the four equations can easily be changed to:
X(t0) = Px + vxt0 - Gxt02/2 = Tx
Y(t0) = Py + vyt0 - Gyt02/2 = Ty
Z(t0) = Pz + vzt0 - Gzt02/2 = Tz
vx2 + vy2 + vz2 = v2
...which is the same story. And I'm still too lazy to go through the algebra. Maybe later.
EDIT: Okay, let's do this thing.
vx = Gxt0/2 + (Tx - Px)/t0
vy = Gyt0/2 + (Ty - Py)/t0
vz = Gzt0/2 + (Tz - Pz)/t0
vx2 + vy2 + vz2 = (Gxt0/2 + (Tx - Px)/t0)2 + (Gyt0/2 + (Ty - Py)/t0)2 + (Gzt0/2 + (Tz - Pz)/t0)2 = v2
Gx2t02/4 + Gx(Tx - Px) + (Tx2 - 2TxPx + Px2)/t02 + Gy2t02/4 + Gy(Ty - Py) + (Ty2 - 2TyPy + Py2)/t02 + Gz2t02/4 + Gz(Tz - Pz) + (Tz2 - 2TzPz + Pz2)/t02 = v2
Multiply the equation by t02....
Gx2t04/4 + Gx(Tx - Px)t02 + Tx2 - 2TxPx + Px2 + Gy2t04/4 + Gy(Ty - Py)t02 + Ty2 - 2TyPy + Py2 + Gz2t04/4 + Gz(Tz - Pz)t02 + Tz2 - 2TzPz + Pz2 = v2t02
Some rearranging....
t04(Gx2 + Gy2 + Gz2)/4 + t02(Gx(Tx - Px) + Gy(Ty - Py) + Gz(Tz - Pz) - v2) + Tx2 - 2TxPx + Px2 + Ty2 - 2TyPy + Py2 + Tz2 - 2TzPz + Pz2 = 0
t04(GoG)/4 + t02(Go(T-P) - v2) + (T-P)o(T-P) = 0
t02 = v2 - Go(T-P) + √((Go(T-P) - v2)2 - (GoG)*(T-P)o(T-P)) *2/(GoG)
I'm gonna stick in L = T-P to get a cleaner formula....
t0 = √( v2 - GoL + √((GoL - v2)2 - (GoG)*(LoL)) *2/(GoG) )
Now that we have t0, we also have vx, vy and vz, and therefore
D = (vx/v, vy/v, vz/v) or D = Gt0/2v + L/vt0
Good enough? ^_^
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good.
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Okay, I hope this one isn't too hard. Find a function f(x) (though general solutions are nice) for which:
f''(x)*cos(2x) + 4f(x)*cos(2x) = 1
Where f''(x) is the second derivative of f(x).
If this is too hard, I can switch it for something easier. Again, I just want to see the sort of response I'll get.
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Okay, I hope this one isn't too hard. Find a function f(x) (though general solutions are nice) for which:
f''(x)*cos(2x) + 4f(x)*cos(2x) = 1
Where f''(x) is the second derivative of f(x).
If this is too hard, I can switch it for something easier. Again, I just want to see the sort of response I'll get.
my calculus is a little on the rusty side to be solving second order differential equations right now, but my trusty 89 says (in LaTex to avoid a mess of parentheses): f(x) = \frac{cos(2x)*ln(abs(cos(2x)))}{4} + C_{1}*cos(2x) + (\frac{x}{2} + C_{2})*sin(2x) where C1 and C2 are any constants.
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I dunno why, but reading latex is one of my pet peeves:
(http://aycu20.webshots.com/image/36579/2001794787102490274_rs.jpg)
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That is very correct. o_o What is this trusty 89, eh, precious? What is it, eh?
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ti-89, the motorola 68K calculators from can solve many first and second order differential equations (but you have to be careful because they don't tell you when they are wrong, it's easy to check though, since they always give the correct answer when differentiating)
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Okay, then next time I should give a third order equation. :P Calculators are for the weak, anyway.
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MATLAB > 3rd order differential equations i think
a calculator is useless for this one (this is pretty unreadable but i'll let khalsa fix it up again):
det \left( \left[ \begin{array}{cccccc}
-3 & -3 & -5 & -4 & 6 & -6 \\
7 & -4 & 3 & 5 & 6 & -9 \\
-3 & 6 & -1 & -2 & -1 & -2 \\
9 & 9 & 4 & 2 & -6 & 2 \\
-1 & 2 & 9 & -8 & -6 & -3 \\
1 & -8 & -5 & -7 & 3 & 9
\end{array} \right] - \lambda I_{6} \right)
where lambda is any eigenvalue of that 6x6 matrix.
[edit] bonus question:
list all values of eigenvalues of the matrix, i will accept answers accurate to 32 bits or better[/edit]
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Why must you abuse my pet peeves?
(http://aycu29.webshots.com/image/36468/2000845050386533513_rs.jpg)
Also, matlab solved this in under a second >_>
But I'll leave these ones for the plebs.
Khalsa
/me tries to ignore this thread.
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...this is just raw calculating. How boring. If I was going to solve this, I would also have used matlab, so khalsa might as well post the solution and give us a new problem. ^_^
(Since when can matlab solve differential equations? That's new to me. How do you do it?)
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Just need to break down the equation a bit:
http://www.mathworks.com/products/pde/description3.html
http://web.mit.edu/answers/matlab/matlab_ode.html
Khalsa
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if you think the original problem (not the bonus question) requires a lot of calculating then you aren't thinking hard enough
(also wtf partial differential equations have nothing to do with this)
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if you think the original problem (not the bonus question) requires a lot of calculating then you aren't thinking hard enough
As far as I can currently tell, both problems require the same effort. So sure, I'm missing something. *shrug* Thanks, Khalsa.
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hint: attempting to solve the second problem by hand should reveal a trivial solution to the first.
edit: in case i wasn't completely clear in my original post, i want the value of the whole expression, not what lambda has to be to satisfy the conditions (that's the bonus question)
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Just cause kevlarman was being an annoying brat:
Answer is:
0
Proof:
"the solutions to det(A - \lambda I)=0 are the eigenvalues of A" ;
"the eigenvalues of that matrix btw are roughly 1.355 and 16.982"
Khalsa
Edit: My actual proof was "I'm retarded"
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at least take the effort to copy the program i gave you into your 92 and give me eigenvalues to (almost) 64 bits.
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Actually cause I'm a lazy bastard, let's just call your bonus question the question I pose for the people 8) .
It's easy as pie folks, finish it quick and shut kevlarman up.
Khalsa
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....duh. >_< Way to forget what eigenvalues are, Eternal! For some reason, I thought you wanted the polynomial for an unknown lambda. Bah.
But I'm still not planning on doing that determinant by hand, since there would be 720 products. And I only have matlab at school.
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....duh. >_< Way to forget what eigenvalues are, Eternal! For some reason, I thought you wanted the polynomial for an unknown lambda. Bah.
But I'm still not planning on doing that determinant by hand, since there would be 720 products. And I only have matlab at school.
i chose a large matrix for a reason, most calculators (once again 89/92 are smarter than that) choose to calculate a determinant naively. doing this properly should take roughly 30 additions and 36 multiplications.
(the determinant of a triangular matrix is simply the product of the diagonal, swapping 2 rows negates the determinant, multiplying a row by 1/c will multiply the determinant by c, adding another row times a constant to a row does not change the determinant), the only difficulty is the 6th order polynomial that results from this (ti-89s can't find exact solutions unless the polynomial is trivial to factor).