Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[da_nang]

f'(t) = F/m0 * 1/√(1-(f(t))2/c2)

I'll be assuming all but f(t) and f'(t) are constants.
Let f(t) = y

dy/dt = F/m0 * 1/√(1-y²/c²)

√(1-y²/c²) dy =  (F/m0)dt

The right-hand side evaluates to (F/m0)t + C.

To evaluate the left hand side, let y/c = sin(u) and dy = c cos(u) du

Int[√(1-y²/c²) dy] = c Int[√(1-sin(u)²) cos(u) du] = c Int[cos(u)² du] = (c/2)Int[(1 + cos(2u)) du] = (c/2)(u + sin(2u)/2) = (c/2)(arcsin(y/c) + (y/c)*cos(arcsin(y/c))) = (c/2)(arcsin(y/c) + (y/c)*√(1 - (y/c)²)) = (c/2)*arcsin(y/c) + (y/2)√(1 - (y/c)²)

So, (c/2)*arcsin(f(t)/c) + (f(t)/2)√(1 - (f(t)/c)²) = (F/m0)t + C .

You have the function implicitly. I doubt you can get it explicitly.

[aenri]

Posting to watch
also - r/Theydidthemath for some interesting calculations

[Skyrunner]

Wait, what happened? I thought there was another oage of posts?...

[Pnx]

Ok, so I've got the equation z = x² + y² + 1
Which I'm supposed to convert to cylindrical and spherical forms. Cylindrical is easy, you just r² in for x² + y² getting z = r² + 1.

The spherical coordinates thing is a bit harder. I'm not certain how to do this. Except that I want to somehow algebra away the z.

[MagmaMcFry]

If a point has cartesian coordinates (x, y, z) and spherical coordinates (r,t,p), then x = r*cos(t)*cos(p), y = r*cos(t)*sin(p), and z=r*sin(t), or at least something looking like that. To convert the equation z=x²+y²+1 to spherical coordinates, you just insert the spherical forms of x, y, z into the equation. This gives you r*sin(t) = (r*cos(t)*cos(p))²+(r*cos(t)*sin(p))²+1 = r²*cos(t)²+1, so your spherical form is r*sin(t) = r²*cos(t)²+1 (or something similar, depending on how your teacher believes spherical coordinates are defined, so you better pick the right spherical coordinate system).

[da_nang]

Spherical coordinates are defined by (r,θ,φ) where r is the radius, θ is the longitude and φ is the latitude. (Different institutions may switch symbols for the two angles)
Now, look at a generic point (x,y,z) in a Cartesian coordinate system and try to rewrite that in spherical coordinates.

Note that it is preferred to use a right-hand definition for the angles with zero being aligned to the positive side of an axis. I.e. θ acts like the polar angle would in the xy-plane and φ as a polar angle (limited from 0 to π) with the z-axis and a generic axis is the xy-plane working as a y-axis.

EDIT: Also what MagmaMcFry said.

[PsyberianHusky]

Hi. So i have jumped head first into a calculus book and am getting thrown around by the warm up stuff.
The problem is Solve (x-2)(x+3) > 0
It presents the answer in two cases and I would like to know how it gets to these answers algebraicly

Case 1: x-2>0 and x+3>0. Then x>2 and X> -3 but these are equivalent to x>2 alone since X>2 implies X>-3.
Case 2: x-2<0 and x+3<0 Then X<2 and X<-3, Which are equivalent to x>-3 since X<-3 implies X<2 thus (x-2)(x+3)>0 holds when x>2 or X<-3.

So how algebraicly am I able to compare (parenthetical one) and (Parenthetical two) to > x independently as done in the underlined explanations. It has been many years and I forgot what I am allowed to numbers.

[Vector]

.

[MaximumZero]

If a point has cartesian coordinates (x, y, z) and spherical coordinates (r,t,p), then x = r*cos(t)*cos(p), y = r*cos(t)*sin(p), and z=r*sin(t), or at least something looking like that. To convert the equation z=x²+y²+1 to spherical coordinates, you just insert the spherical forms of x, y, z into the equation. This gives you r*sin(t) = (r*cos(t)*cos(p))²+(r*cos(t)*sin(p))²+1 = r²*cos(t)²+1, so your spherical form is r*sin(t) = r²*cos(t)²+1 (or something similar, depending on how your teacher believes spherical coordinates are defined, so you better pick the right spherical coordinate system).

You know, I read this thread almost every day, and I'm really glad that you guys are so chill about math(s), because I have no fucking clue what most of you are saying most of the time. I'd have a very hard time policing this thread if something were to come up that didn't involve mafia-level swearing and accusations. Some days, like today, I just shake my head and feel stupid.

[frostshotgg]

If a point has cartesian coordinates (x, y, z) and spherical coordinates (r,t,p), then x = r*cos(t)*cos(p), y = r*cos(t)*sin(p), and z=r*sin(t), or at least something looking like that. To convert the equation z=x²+y²+1 to spherical coordinates, you just insert the spherical forms of x, y, z into the equation. This gives you r*sin(t) = (r*cos(t)*cos(p))²+(r*cos(t)*sin(p))²+1 = r²*cos(t)²+1, so your spherical form is r*sin(t) = r²*cos(t)²+1 (or something similar, depending on how your teacher believes spherical coordinates are defined, so you better pick the right spherical coordinate system).

You know, I read this thread almost every day, and I'm really glad that you guys are so chill about math(s), because I have no fucking clue what most of you are saying most of the time. I'd have a very hard time policing this thread if something were to come up that didn't involve mafia-level swearing and accusations. Some days, like today, I just shake my head and feel stupid.

NO FUCK YOU YOU CANT CALL HYPERREAL NUMBERS REAL YOU'RE LITERALLY NUMBER HITLER

Like that?

[MaximumZero]

If a point has cartesian coordinates (x, y, z) and spherical coordinates (r,t,p), then x = r*cos(t)*cos(p), y = r*cos(t)*sin(p), and z=r*sin(t), or at least something looking like that. To convert the equation z=x²+y²+1 to spherical coordinates, you just insert the spherical forms of x, y, z into the equation. This gives you r*sin(t) = (r*cos(t)*cos(p))²+(r*cos(t)*sin(p))²+1 = r²*cos(t)²+1, so your spherical form is r*sin(t) = r²*cos(t)²+1 (or something similar, depending on how your teacher believes spherical coordinates are defined, so you better pick the right spherical coordinate system).

You know, I read this thread almost every day, and I'm really glad that you guys are so chill about math(s), because I have no fucking clue what most of you are saying most of the time. I'd have a very hard time policing this thread if something were to come up that didn't involve mafia-level swearing and accusations. Some days, like today, I just shake my head and feel stupid.

NO FUCK YOU YOU CANT CALL HYPERREAL NUMBERS REAL YOU'RE LITERALLY NUMBER HITLER

Like that?

That might be sufficient. I think.

[Jopax]

So I'm not sure if this is strictly math but they've lumped it under applied math. And I'm about to stab someone, well, would stab someone if there was anyone nearby.

Anyways, it's statistics for the most parts. Namely calculating odds of shit happening or not happening. Now, the problem I'm having is that they've only shown us how to do half of it (as in, I can get a starting problem and calculate the odds and probabilities out of that just fine, but doing the reverse, however straightforward it may seems is a whole different beast apparently), and they seem to give both types of problems on exams.

To finally get to the point with a specific problem I'm having. So we have this bag you see, it has 13 metal balls and 7 wooden ones. The odds of taking out a metal one are three times smaller than those of taking out a wooden one (however little sense that makes). It asks what are the odds of pulling out two balls and both of them being metal.
Now, I've figured out the actual odds of pulling out a metal one (1/4) but that's where my knowledge stops, I have no idea how to proceed, I've tried a few things but I have no idea if they're correct approaches since there's no solved problem that I can compare to.

In short, the exam is tommorrow, HALP!

[MagmaMcFry]

Okay, the only way that problem could make sense (EDIT: It's not, see below) is if you painted any single metal ball red and any single wooden ball blue, you'd be thrice as likely to get the blue ball than the red ball.

Now let's assume you draw exactly one ball. Let p be the probability of picking the red ball. Since all metal balls are equally likely, the probability to pick any metal ball is 13p. Now since the blue ball is thrice as likely to be picked as the red ball, the probability of picking the blue ball is 3p, so the probability to pick any wooden ball is 7*3p=21p. Now the probability of picking any ball is equal to 1, but also equal to 13p+21p, which is equal to 34p, so 1=34p which means p=1/34, and the probability of picking a metal ball on your first draw is 13/34 (not 1/4).

Now let's assume you managed to pick a metal ball on your first draw. Now there are 12 metal balls and 7 wooden ones left. Let's examine the probabilities of the second draw. For this, let's paint another metal ball green, and let q be the probability of drawing the green ball from this new set of balls (it's slightly larger than p since there are now less balls to draw from). Now the probability of drawing any metal ball from the bag is 12q, and the probability of drawing the blue ball is 3q, and the probability of drawing any wooden ball is 7*3q = 21q. The probability of drawing any ball is now equal to 1 and also equal to 12q+21q (which is 33q), so q = 1/33 and the probability of drawing a metal ball on your second draw is 12q = 12/33=4/11.

Now let's examine the original problem: drawing two metal balls. This probability is equal to the probability of drawing a metal ball first, multiplied by the probability of drawing a metal ball second after drawing a metal ball first. We know the probability of drawing a metal ball first is 13/34, and the probability of drawing a metal ball second after drawing a metal ball first is 4/11, so your total probability is 13/34*4/11=26/187.

[Bauglir]

Hm, well, you've got 20 balls total, and 13 are metal. So when you choose one at random, you've got a 13/20 chance of choosing metal. Compared to the 7/20 chance of choosing wood, it's about twice as likely (13 is about 14, which is 2*7) that you'll pull a metal ball than you would a wooden one, which disagrees with what you're being told in the setup. You're 3 times likelier to choose wood than metal, which means that there's some hidden factor that makes metal much less likely than it otherwise would be. You can't actually come to a good conclusion about how this is going to influence the rest of the problem, but the simplest assumption is that there's a single factor skewing your odds of getting metal downward, I think. To figure this out, you can compare your actual chance of drawing metal to your expected chance - in this case, 13/20 vs 5/20. So you're 5/13 times as likely to draw metal as you'd expect to be from the numbers alone.

Now, you've correctly figured out that you've got a 1/4 chance of drawing metal the first time, based on what you've been told about the actual odds. If you put the ball back before drawing a second one, your odds of drawing metal the second time are the same, so your odds of drawing metal both times are (1/4)*(1/4) = 1/16. In this case, telling you how many balls of each kind there were was a red herring meant to confuse you, which is something many teachers are fond of.

If you don't put the ball back, or you draw two balls at the same time (I believe these are equivalent, a better statistics person than I could confirm it), you have one fewer metal ball that you could draw. So, now, your expected odds are 12/19, but there's that mysterious voodoo factor that you have to account for. Since you've got no reason to assume it should change, I believe that your chance on the second draw of drawing metal is (12/19)*(5/13) = 60/247. Multiplied by your 1/4 chance of drawing metal the first time, your final chance is 15/247.

EDIT: We have a ninja, I'm gonna read his post. If we disagree, trust him over me.

EDIT: And I'm completely lost as to that explanation. I defer to a wiser man, but I cannot see what marking individual balls adds to the problem. It's probably a more sensible strategy than my arbitrary factor, but I'll be damned if I can follow the logic.

[Jopax]

I honestly like your approach better Bauglir, since MMF's is just doubly confusing me. We just haven't had any problems so far that had you use both the real odds of something happening and the odds they gave you.

For example, we had a coin flipping one, where the chance of heads was 1/10 and we were supposed to figure out how many throws it took to get four heads. It was fairly straightforward and gave the expected answer of forty. I think the same should apply here, because if I've learned anything, it's that math shouldn't make real-world sense in most cases but instead works within any ruleset you give it as a sort of mental exercise.