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[MonkeyHead]

Gentlemen, I am in need of a function.



I want it to look like this, but I can't seem to figure out a way to get the result I want.

EDIT: y = ( -((x-2)^2) )/(x^2)+1 seems to be correct.

EDIT2:

Spoiler: What it was for (click to show/hide)

(they are vaguely evenly spaced becaue they are both pulling towards and away from eachother)
Function is (-square(dist-500)/square(dist)+1)*0.000001 currently.

Anyway, I was thinking about n-body simulations and wanted them to bounce into each other. So I thought I could just bake it into the gravity. If they get too close then it turns negative and pushes them away from eachother. Works pretty much perfectly. It's kinda awesome to see single high-speed particles flying around and bouncing into stuff like a billiard ball.

Actually kinda lying about the perfect thing. If some get too close by some fluke, then they bounce away from eachother at a ludicrously high speed. If I try to make a black hole, then the damn thing flies apart.

You might want to look into flocking simulations - it might be your thing.

[Furtuka]

line l = vector->x(t) = (1-t, 1+t, t) 
Find the unit tangent vector, the curvature, and the tangent line to the l at the point where (a) t=2 and (b) at any point on the line.

I think I got the unit tangent vector and curvature, but how do I get the tangent line? For a is it (-2t/√(62), 7t/√(62), 3t/√(62))?

[Reelya]

If you have a vector and a point it passes through, those define a unique line. The most basic form of the line equation is y=ax+b. You already worked out "a" from the tangent vector's direction, and you know the tangent point (call it x₁,y₁), so you substitute in the knowns into this equation:

y₁ = ax₁ + b

And solve for "b", which is the only unknown.

[Furtuka]

I think I might have been putting the tangent line where it should be vector then. Would the non unit tangent vector for the first one be (-2, 7, 3)?

[Furtuka]

Spoiler (click to show/hide)

I've been staring at this problem for a very long time and have no idea what I'm doing. Please help

[crazysheep]

@Caroline: If I understand your spoilered notes correctly, you have 3 variables but only 2 equations, which makes it harder to get to the answer without some other criteria to eliminate nonsensical answers.

Spoiler: my working (click to show/hide)

Disclaimer: This may not make sense from a mathematical perspective, so I hope it doesn't confuse you too much.

1. Let the amount of coffee and milk Sandra takes be represented by c and m respectively.
From the cup volume, we get eqn (1): c + m = 8
We can rearrange this to express c in terms of m: c = 8 - m.

2. Let the total amount of coffee and milk consumed be represented by c_T and m_T respectively.
Given that we know how much Sandra's beverage contributed to the overall liquid mix, we get the following:
eqn (2): m = (1/4)m_T
eqn (3): c = 8 - m = (1/6)c_T
Rearranging (2) and (3) gives:
eqn (2): m_T = 4m
eqn (3): c_T = 48 - 6m

3. Summing (2) and (3) together gives:
m_T + c_T = 48 - 2m

Using your assertion that:

since the total amount of coffee and milk gives us the ounces, which divided by eight gives us the amount of cups

We can write a new equation relating those variables:
eqn (4): m_T + c_T = 8n, where n is the number of cups/family members.

Equating (3) and (4), we get:
8n = 48 - 2m, which simplifies to:
eqn (5) n = 6 - (1/4)m

4. This is where

no cups have less than 0 ounces of any of the two.

comes into play: it means that m cannot be 0, thus n must be less than 6. m cannot be 8 either, as this would give c = 0, thus n must be greater than 4.

We also know that n must be an integer, as it would not make sense for the total number of family members to be non-integer values. Thus any value of m must be divisible by 4... which leaves us with the only possible value for m: 4.

That gives n = 5.

[Furtuka]

I'm still confused. In the first one how do I use the drawing to show that it has the same length as v?

[crazysheep]

Thank you.
 :DIt didn't confuse me very much at all, but I'll reread it when I'm not extremely sleepy to make sure I caught everything.

No worries then

[Furtuka]

I'm still confused. In the first one how do I use the drawing to show that it has the same length as v?

Well, you know what v is supposed to be- that's the arrow with the line pointing to it to the equation is v. You have the line r, which is equal to x. All you need to do is draw in w (should be vertical), draw in the cross product of w and x, and show that that's the same direction and distance.

If x and w are variable values, how am I suppose to know what the distance of their cross product is?

[da_nang]

I'm still confused. In the first one how do I use the drawing to show that it has the same length as v?

Well, you know what v is supposed to be- that's the arrow with the line pointing to it to the equation is v. You have the line r, which is equal to x. All you need to do is draw in w (should be vertical), draw in the cross product of w and x, and show that that's the same direction and distance.

If x and w are variable values, how am I suppose to know what the distance of their cross product is?

Since they're perpendicular to one another, the magnitude of the cross product is the product of the magnitudes of the cross product operands.

[Furtuka]

By factor do you mean each of the vectors? And I still don't understand how I can assume it to be the same distance as v

[da_nang]

Yes, each vector. You can also prove it's the same magnitude if you recall that |v| = |w||r| for circular motion.

[Bauglir]

I'm looking for a function that behaves similarly to the exponential function, but whose inversion is a bit friendlier to algebra than a logarithm. To be more specific, I want an always-increasing, smooth y = f(x) such that y approaches 0 as x approaches -∞, y approaches ∞ as x approaches +∞, there are no inflection points, and f^-1(g(y)/x + 1) = x can be separated to give x as some function in terms of only y. It's that last bit that's ruling out an exponential function.

To give more information, and because I'm sure I've worded that last bit in a way that's not technically correct, right now I'm working with y = arctan(x*(1+e^x)), and I need to invert it. Having an explicit inversion is an important goal for the thing I'm working on, which is a number generation scheme (that formula is its CDF). Getting to x = ln(tan(y)/x -1) is straightforward enough, but can't separate that logarithm and so I can't separate the variables and get a result that gives x in terms of y. What I need is a function that I can replace the exponential with that will let me proceed, but I can't for the life of me think of one.

Is there such a function?

Note, I'm not actually terribly attached to the x*(1+e^x) thing - what I really need is a function of x that's close to y = x below 0, and rapidly diverges from it above 0 (in any direction, actually - as long as the difference between the two is always increasing, and the no inflection points thing holds, it'll do what I need). What I've asked for above is framing the question in what seems the easiest way to me, but it might well not be, and I'll happily take better ideas. I gave a logarithm a shot as the obvious next guess, and it's not much better (you wind up having (ex)^x in terms of y, unless I've missed something obvious).

[MagmaMcFry]

Note, I'm not actually terribly attached to the x*(1+e^x) thing - what I really need is a function of x that's close to y = x below 0, and rapidly diverges from it above 0 (in any direction, actually - as long as the difference between the two is always increasing, and the no inflection points thing holds, it'll do what I need).

How about y=-sqrt(x²+1)? It doesn't really diverge rapidly, but it has no inflection and its inversion is given by x=+-sqrt(y²-1) for any y < -1.

If you want a completely invertible function, then I propose y = (3x-sqrt(5x²+4))/4, its inversion is given by x = 3y+sqrt(5y²+1).

[Bauglir]

Glancing at that second one, it looks like it ought to do quite nicely. I'll see what happens. Thank you for being amazing.