Wikipedia:Reference desk/Archives/Mathematics/2007 February 16

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February 16

Pull tabs

What's the average mass and volume of a pull tab from an aluminum can?

^π⁄₆ (units unspecified). Go find a trivia forum and ask there. This question has no mathematical content. --KSmrq^T 10:57, 16 February 2007 (UTC)

I don't know, but here's how to find out. First, start a donation drive and get all your friends to donate their pull tabs to you. Collect maybe 1000 pull tabs. Then, weight all the pull tabs on a scale. Divide by the number of pull tabs, and that will be the average mass of one pull tab. Finding the average volume is trickier, since you haven't really defined what volume you're looking for - there's a lot of empty space in a pull tab! So are you looking for the volume of just the aluminum particles (which means I would estimate based on the calculated mass and the given density of aluminum)? -sthomson06 (Talk) 16:00, 16 February 2007 (UTC)

To find average volume fill a bucket to the brim with water, add your 1000 pull tabs to the bucket, collect the water displaced, measure its volume and divide by 1000. Running around naked crying out "Eureka" is an optional part of the procedure. Gandalf61 16:58, 16 February 2007 (UTC)

That part's optional? Shoot, if I had known it was optional it would have saved me a TON of time. :/ Dugwiki 17:16, 16 February 2007 (UTC)

Although it's an optional step that's not to say it wont affect your results. Provider uk 18:04, 20 February 2007 (UTC)

Are you sure water displacement would work? Most pull tabs I've seen are curled up on themselves at the edge, and I'm pretty sure they'd trap air. Black Carrot 07:50, 20 February 2007 (UTC)

How about if you use the average mass of a pull-tab and the density of aluminum to calculate the volume ? StuRat 00:37, 27 February 2007 (UTC)

Ellipses and hyperbolas

How can I find the equation of an ellipse, centered on a certain point, that passes through two given points?

eg, center at (0,0), passing through (1,(-10*2^(1/2))/3) and (-2, (5*5^(1/2))/3).

And how can I find the equation of a parabola, with a given vertex, that passes through two given points?

(I know this must be an very easy question but I am pretty dumb and I can't figure it out).

Thank you kind people!

It's fairly simple - I'll give an example for the parabola - the equation of the parabola I assume is y = a(x-b)²+c

Now you have three points - the vertex (the turning point at the top or bottom of the parabola), and two others.

The x value of the turning point gives b eg if the peak is at (4,7) then b=4

Now place the other two pairs of values into the equation - this gives you two linear eqautions in terms of the unknowns 'a' and 'c' - you should be able to easily solve these equations by eliminating one of the unknowns first and finding the other.

For the ellipse the method is similar - the centre of the ellipse is (0,0) therefor the equation for the ellipse (x-x1)²+b(y-y1)²=f² simplifies to (x)²+b(y)²=f² - replace values for x and y to give two linear equations with f² and b as the unknowns and solve as linear equations.87.102.20.186 11:26, 16 February 2007 (UTC)

If the ellipse center is at (x_c,y_c) translate to the origin. Further assume (as you seem to) that the ellipse is axis-aligned. Then its equation is

${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1.\,\!$

Let c = ¹⁄_a² and d = ¹⁄_b². This gives a linear equation in c and d:

$(x^{2})c+(y^{2})d=1.\,\!$

When we substitute the coordinates of the first point, (x₁,y₁), we get an equation that c and d must satisfy; likewise when we substitute the coordinates of the second point. We know how to solve a pair of simultaneous linear equations. Do so, then recover a and b using the positive square roots. Be careful; some pairs of points allow no solution.

The parabola raises a question. As before, place the vertex at the origin; but now we cannot expect incidence with most pairs of points if we insist on axis alignment and a fixed vertex. (The equation would be y = ax².) Therefore we must be free to rotate the parabola, and need a suitable formula. However, I see no way to remain with linear equations; the "aiming" of the parabola would seem to require at least a quadratic variable. Perhaps, as in the title you gave this section, you wanted a hyperbola, not a parabola? --KSmrq^T 13:09, 16 February 2007 (UTC)

Negative exponents and differential calculus help

I am currently working my way through Calculus Made Easy by Silvanus P. Thompson. I have just read the section on the power rule and am stuck on the example showing that it applies to negative exponents. In particular, the equation is set out like this.

Let $y=x^{-2}$. Then proceed as before:

$y+dy=(x+dx)^{-2}$

$y+dy=x^{-2}\left(1+{\frac {dx}{x}}\right)^{-2}$

I do not understand how he gets the $y+dy=x^{-2}\left(1+{\frac {dx}{x}}\right)^{-2}$ part from the part above. Can someone please explain it to me? Thanks. --80.229.152.246 18:47, 16 February 2007 (UTC)

I'll try. It's just an algebraic manipulation. Here:

$(x+dx)^{-2}={\frac {1}{(x+dx)^{2}}}={\frac {1}{x^{2}+2xdx+(dx)^{2}}}$ $={\frac {1}{x^{2}(1+2/x\cdot dx+(dx)^{2}/x^{2})}}$ $={\frac {1}{x^{2}(1+dx/x)^{2}}}=x^{-2}\left(1+{\frac {dx}{x}}\right)^{-2}$

Hope that makes sense. –King Bee ^{(T • C)} 19:20, 16 February 2007 (UTC)

$(x+dx)=x\left(1+{\frac {dx}{x}}\right)$ --Spoon! 21:28, 16 February 2007 (UTC)

Thanks for the help guys. I still don't understand why he would want it written like that though. Oh well, I think I should leave the book for a few years. Thanks very much. --80.229.152.246 09:49, 19 February 2007 (UTC)

Take a look at some precalculus topics, if you get stuck with something. Mr.K. (talk) 14:06, 19 February 2007 (UTC)

You did not ask why the author did this, but how he got there. As to the why, it should be because it is a step towards the final desired result. In general, if n is positive,

$(1+\epsilon )^{-n}=1-n\epsilon +\epsilon ^{2}U(\epsilon )\,$,

in which U(ε) goes to a definite limit as ε tends to zero. For example, if n = 2, U(ε) = (3+2ε)(1+ε)^–2. This can be used to see how (A+δ)^–n differs from A^–n for very small δ, where A ≠ 0, since

$(A+\delta )^{-n}=(A(1+{\frac {\delta }{A}}))^{-n}=A^{-n}(1+{\frac {\delta }{A}})^{-n}\,$

$=A^{-n}(1-n{\frac {\delta }{A}}+({\frac {\delta }{A}})^{2}U({\frac {\delta }{A}}))\,$

$=A^{-n}-n\delta A^{-(n+1)}+\delta ^{2}A^{-(n+2)}U({\frac {\delta }{A}}))\,$.

If δ is sufficiently small, the third term is negligeably small compared to the second term. Putting A = x and δ = dx, you get the formulas from your textbook, at least for the first line.  --Lambiam^Talk 16:46, 19 February 2007 (UTC)