Wikipedia:Reference desk/Archives/Mathematics/2007 September 15

= September 15 =

Blackboard Bold
Does anyone know of a source (freeware or cheap anyway) of a good quality truetype Blackboard bold font? -- SGBailey 08:32, 15 September 2007 (UTC)


 * You can download it and other math-related TrueType fonts for free at this site from the math department at Union College:
 * — Michael J  21:15, 15 September 2007 (UTC)
 * — Michael J  21:15, 15 September 2007 (UTC)

rainfall
1 inch of rain equals how many gallons of water per acre? jodyjer0801 —Preceding unsigned comment added by Jodyjer0801 (talk • contribs) 13:19, 15 September 2007 (UTC)
 * http://www.google.com/search?q=1+inch+in+gallons+per+acre --Spoon! 13:38, 15 September 2007 (UTC)


 * (potentially unnecessary) warning: there are various gallons in the world, so make sure you get the right one. In answer to the question, why take the Google way out? An acre is a furlong by a chain; a furlong is ten chains; a chain is four rods, poles or perch; a rod, pole or perch is five and a half yards; a yard is three feet; a foot is twelve inches; a (US liquid) gallon is 231 cubic inches. Where's the problem? --Algebraist 17:24, 15 September 2007 (UTC)


 * That way of calculating sounds really funny, because I live in a metric country. I'd just say that an inch is 0.0254 meters, an acre is 4.047*10^3 square meters so an inch times an acre is 102.8 cubic meters; a US gallon is 3.785*10^(-3) cubic meters, so the volume is 27160 gallons.  &#x2013; b_jonas 21:28, 17 September 2007 (UTC)
 * I suspect that Algebraist's post might be intended to subtly mock the imperial system. -- Meni Rosenfeld (talk) 21:37, 17 September 2007 (UTC)
 * You do not need to mock the Imperial System, it's quite capable of mocking itself. 210.49.155.132 13:13, 19 September 2007 (UTC)

Catenary
With reference to a catenary, what calculations can be made if you know its length and the horizontal distance between the points at each end that secure it in place? asyndeton 15:57, 15 September 2007 (UTC)
 * Are the points at the end at the same height? If so, this information completely determines the catenary, as far as I know. Tesseran 18:13, 15 September 2007 (UTC)
 * Yes, they are at the same height. What do you mean by 'determines the catenary'? asyndeton 18:26, 15 September 2007 (UTC)
 * It means there is only one catenary with those parameters. --Taejo|대조 12:07, 18 September 2007 (UTC)

See Catenary - you can get the shape of the curve it hangs in, and the 'force/tension' at a given point along, as well as related things such as the amount of droop/angle at a given point.87.102.43.253 18:38, 15 September 2007 (UTC)


 * The equation of a hanging catenary in (x,y) terms is usually expressed wrt an origin a certain distance directly below the lowest point, this distance (c, say) being a parameter of a particular curve. If the suspension points are 2a apart, then the distance of the lowest point below them (b, say) is b=c[cosh(a/c)-1]. If the total length of the curve is 2s, then s=c[sinh(a/c)]. c cannot be determined explicitly, but is the solution of the equation 2c[sinh(a/2c)]=(s^2-b^2)^0.5 As far as I can see, there is no straightforward answer to such questions as "A heavy flexible inextensible string hangs from two points 10 units apart, with its lowest point 5 units below them. What is its length?"…86.132.239.70 19:19, 15 September 2007 (UTC)


 * The equation of a catenary is $$y = a \cosh \left( \frac{x}{a} \right)$$ You want to find a, which will tell you the exact equation of the catenary, and everything. Suppose that you know the horizontal distance between the ends is w, and you know the length of the catenary is s. The length is given by (from arc length) $$s = \int_{-w/2}^{w/2} \sqrt { 1 + [f'(x)]^2 }\, dx $$.
 * $$f'(x) = \sinh \left( \frac{x}{a} \right)$$, so $$\sqrt { 1 + [f'(x)]^2 } = \cosh \left( \frac{x}{a} \right)$$
 * And the length is $$s = \int_{-w/2}^{w/2} \cosh \left( \frac{x}{a} \right) dx = a [ \sinh \left( \frac{x}{a} \right) ]_{-w/2}^{w/2} = 2a \sinh \left( \frac{w}{2a} \right) = s $$
 * You know s and w, and you need to numerically solve for a (with the restriction that a is positive, of course, so that gravity points downward). a will have units of length.
 * As for the vertical position of the catenary, the height of the place where the ends are hung is $$a \cosh \left( \frac{w}{2a} \right)$$, and everything else is relative to that. You can add a constant to adjust the height to be relative to some "ground" if necessary.
 * --Spoon! 20:35, 15 September 2007 (UTC)

need some help.
If you started with a population of 2 wolves and ended up with a population of 150,000 wolves after 2000 years (with a generation every 5 years), considering the growth rate was relativly the same how many cumulative individuals would have lived in the population? Can someone solve this for me or at least tell me the equation is a way some one who has only finished algebra II would be able understand and solve with a TI-89. Thanks for any help I get! -Icewedge 19:54, 15 September 2007 (UTC)
 * Depends on what you mean by "generation"... Under some simplified assumptions, we can model this as a geometric progression; Let $$a_0 = 2$$ be the starting population, and $$a_n$$ the population after n generations. We assume that no wolf survives more than a generation, and that $$a_n = a_0q^n$$ for some q. Then $$150000=2q^{400}$$ so $$q=75000^{1/400}$$. Now we are looking for $$\sum_{k=0}^{400}a_k = \frac{a_{401}-a_0}{q-1} \approx 5 \cdot 10^6$$ (give or take a million or two, due to sensitivity to starting population and general model inaccuracies). -- Meni Rosenfeld (talk) 21:44, 15 September 2007 (UTC)

(ec: I wrote this before Meni's answer above, but forgot to submit it. :/) I'm assuming that by a constant growth rate you mean that
 * $$\frac{dp}{dt} = cp\,$$,

where $$t$$ is time, $$p$$ is the population size and $$c$$ is the annual growth rate the population. The solution to that differential equation, giving the population size as a function of time, is
 * $$p(t) = p_0\,e^{ct}\,$$,

where $$p_0 = p(0)$$ is the initial size of the population and $$e$$ is the base of the natural logarithm. To obtain the total number of "wolf-years" elapsing over the 2000 years, $$P$$, you need to integrate this over time to get
 * $$\int_0^{2000} p(t)\,dt = p_0 \int_0^{2000} e^{ct}\,dt = p_0 \frac{e^{2000c} - e^0}{c} = \frac{p(2000)-p(0)}{c}$$.

From the constraints $$p(0) = 2$$ and $$p(2000) = 150,000$$ you should be able to solve for $$c$$ and plug it into the expression above. Note that, to get the (approximate) total number of wolves that have ever lived, you still need to divide the result by the average lifespan of a wolf — which I'd suppose to be the 5 years you've given above, though your terminology is rather unclear. (Bonus exercise: Why is the result only approximate? Consider in particular the limit as the lifespan of a wolf tends to infinity.)

In any case, though my method is slightly different from Meni's, I also get a result a little over 5 or so. Which is about as much as can be expected from such a simplistic model. —Ilmari Karonen (talk) 22:37, 15 September 2007 (UTC)