Wikipedia:Reference desk/Archives/Mathematics/2013 October 31

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

October 31

Calculus homework - related rates

Hi, we are learning how to solve "related rates" word problems in calculus. So far I know how to do problems involving the changing area of a circle, Pythagorean distance problems involving right triangles and the classic problem involving the changing water level in a conical tank.

Generally, the solutions involve taking a geometrical equation, then taking the derivative of that equation using implicit differentiation then plugging the given values into the results in order to solve the problem (usually there is one missing derivative, which we solve for algebraically).

But we were given a different kind of related rates problem that I don't know how to solve.

An object is moving along a circle of radius 1 . When the object is at the point (4.5,−3.5) , the x value of its position is changing at the rate of 9 units per second. How fast is the y value of its position changing at that moment?

I looked up the answer and it is 12, but I don't know the method of solving this problem. I'm guessing it has something to do either with the equation of the circumference of the circle or the (x-a)^2 + (y-b)^2 = r^2 equation. But I really don't know where to go from here (i.e., how to set up the implicit differentiation, so I can substitute the radius 1 and the rate of change of x, which is 9).

How do you go about solving this type of problem?--Jerk of Thrones (talk) 09:47, 31 October 2013 (UTC)

There is insufficient information when we don't know anything about the center of the circle. For example, if the center is (4.5,−4.5) then the object is moving horizontally right above the center and the answer would have been 0. PrimeHunter (talk) 10:38, 31 October 2013 (UTC)

Indeed. Implicit differentiation gives us:

$2{\dot {x}}(x-a)+2{\dot {y}}(y-b)=0$

which we can re-arrange to get

${\dot {y}}=-{\dot {x}}\left({\frac {x-a}{y-b}}\right)$

We know ${\dot {x}}$, x and y, but unless we are also given one of a or b (from which we could infer the other, as we know r) then we are stuck. Gandalf61 (talk) 10:57, 31 October 2013 (UTC)

............Seems reasonable to assume the coordinates were meant to be (4/5,–3/5) instead of (4.5,–3.5) and the circle's center is (0,0) by default. --CiaPan (talk) 14:19, 31 October 2013 (UTC)

Ah, yes ! Good catch. Gandalf61 (talk) 14:36, 31 October 2013 (UTC)

Then the velocity vector is orthogonal to the position vector, so ${\frac {\dot {y}}{\dot {x}}}={\frac {v_{y}}{v_{x}}}=-{\frac {r_{x}}{r_{y}}},$ which implies $v_{y}=-v_{x}\cdot {\frac {r_{x}}{r_{y}}}=-9\;\mathrm {unit/s} \cdot {\frac {\frac {4}{5}}{-{\frac {3}{5}}}}=9\;\mathrm {unit/s} \cdot {\frac {4}{3}}=(3\cdot 4)\;\mathrm {unit/s} =12\;\mathrm {unit/s}$ --CiaPan (talk) 17:58, 31 October 2013 (UTC)

Dot-notation corrected. --CiaPan (talk) 08:15, 4 November 2013 (UTC)

Yes, sorry, It was (4/5, -3/5). I think when I copied and pasted the problem, the fractions didn't copy correctly, so I accidentally changed it to 4.5 and -3.5. Thanks for your help!--Jerk of Thrones (talk) 06:10, 6 November 2013 (UTC)

Distribution functions

Suppose I have two functions f() and g(), where f() returns either 0 or 1, with probability 0.5 for each, and g() always returns 0.5. The question is, how would I construct a function h(x) where x can be at least any real between 0 and 1, and returns h(0) -> f() and h(1) -> g()? Obviously you can now say "well, apply coefficients to f and g, or return f or g with a certain probability", but that is not what I mean, I mean is there some sort of abstract mathematical concept I should read more about, if I want to know how to go smoothly from a fair coin flip to a 100% chance of the coin landing on its side? Ginsuloft (talk) 20:24, 31 October 2013 (UTC)

Hm, I'm a bit unclear what you mean. I think you are looking for random variables, rather than functions (although a random variable are functions from the sample space into the reals. If f, g are random variables with the properties as above, then you can simply set $h=(1-x)f+xg$ where $0\leq x\leq 1$. Such a random variable would return f for x=0 and g for x=1. For any other x it would be the RV that returns 0.5+x/2 and 0.5-x/2 with probability 0.5 each.

A different way to achieve a smooth transition would be to consider 3 events, namely 1 (heads), 0 (tails) and 0.5(coin on side) and define h as $P(h=1)=(1-x)/2,P(h=0)=(1-x)/2,P(h=0.5)=x$. Don't know if any of this is close to what you are after. 86.179.30.226 (talk) 22:26, 31 October 2013 (UTC)

No, I'm afraid that's not it. To clarify, the points h(0) and h(1) should be extreme cases; usually the function should return a pretty uniform distribution, meaning that normally you shouldn't be able to guess what the returned function returns. This is for the purposes of experimenting with connectionism. Anyway, it probably isn't possible to get an answer here, as I can't really put my wants into words without giving implementation details (and therefore solving the problem myself). Thanks anyway to you and anyone else who tries to help. Ginsuloft (talk) 22:39, 31 October 2013 (UTC)

I think I basically understand what you are asking for. The most common approach is to feed the output of h into a sigmoid function, with a parameter setting the slope of the sigmoid. A commonly used choice of sigmoid is

${\frac {1}{1+e^{\alpha (1/2-x)}}}$

If α is zero, the output is always 1/2. If α is infinite, the result is either 0 or 1 with equal probability. Intermediate values of α give you a graded distribution. Looie496 (talk) 23:04, 1 November 2013 (UTC)

Yes, thank you for making sense of my nonsense. That was exactly what I was looking for. I never realized the answer was just a sigmoid function. I knew that sigmoids are widely used in neural nets but never would have thought that even some improvised probabilistic model I came up by myself would have made use of them. Ginsuloft (talk) 17:11, 2 November 2013 (UTC)