Testwiki:Reference desk/Archives/Mathematics/2008 August 23

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August 23

Pyramid Volume section contains wrong derivation of volume: a negative sign in the integral...

Hi,

Just noticing in the Pyramid_(geometry) following "The volume is given by the integral"

There's a negative sign (-A / 3h^2) and there shouldn't be! Right? link title

How would I fix this? I don't know how to generate the math images... Oh wait, it's

$\frac{A}{h^2}\underset{0}{\overset{h}{\int }}(h-y{)}^{2}dy=\frac{A}{3h^2}(h-y{)}^3{|}_0^h=\frac{1}{3}Ah.$

Right? Thanks! InverseSubstance (talk) 18:12, 23 August 2008 (UTC)

No, the sign is correct. Algebraist 18:16, 23 August 2008 (UTC)

More explicitly, the sign gets flipped once by the (implied) substitution for $-y$, and it gets flipped again because you evaluate at $y=h$ where the antiderivative is 0 and subtract the positive evaluation at $y=0$. Hope this helps. --Tardis (talk) 18:26, 23 August 2008 (UTC)

Wait. I understand the evaluation part. But not the integration part. So it seems to me what your saying is, fex:

$\int (1-x)dx=-\frac{1}{2}(1-x{)}^2$ Right? What rule of integration does this fall under? Thanks... InverseSubstance (talk) 19:25, 23 August 2008 (UTC)

There is no rule for that, but it's useful to note that $\int (a+bx{)}^{n}dx=\frac{(a+bx{)}^{n+1}}{(n+1)b}+c$ for complex a and b. x42bn6 Talk Mess 20:15, 23 August 2008 (UTC)

It's called the substitution rule. You write $u=1-x\Rightarrow du=-dx,dx=-du$ so $\int (1-x)dx=\int u(-du)=-\int udu=-\frac{u^2}{2}+C=-\frac{(1-x{)}^2}{2}+C$. For definite integrals, you have to transform the limits too, unless you immediately (as I did here, and as the original example does) switch back to the original variable before evaluating. --Tardis (talk) 04:34, 24 August 2008 (UTC)

You can see it's true by differentiating, the key rule there being the chain rule, the derivative of -x (and, therefore, 1-x) is -1. --Tango (talk) 20:50, 23 August 2008 (UTC)

How to find the shorter path?

What algorithms can you use to find the shorter path in a net (like a very complex subway net with thousands of nodes)? —Preceding unsigned comment added by Mr.K. (talk • contribs) 19:16, 23 August 2008 (UTC)

Is Shortest path problem#Algorithms what you're looking for? -- BenRG (talk) 19:44, 23 August 2008 (UTC)

Geometric figure for xⁿ

If x is a segment, x² is a square, x³ a cube, what geometric figure is x⁴, and is there a figure for every xⁿ? —Preceding unsigned comment added by Mr.K. (talk • contribs) 19:17, 23 August 2008 (UTC)

x⁴ is the hypervolume of a tesseract, and x⁵ a five-dimensional hypercube, and so on. There can't be a three-dimensional figure with a volume of x⁴, because the units come out wrong. For example if x is 2 meters, then x⁴ is 16 meters⁴, but a volume has to have units of meters³. -- BenRG (talk) 19:49, 23 August 2008 (UTC)

It is interesting to note that the first solution to the quartic equation was published together with the first solution to the cubic equation (by Cardano, I think), but Cardano downplayed the solution of the quartic equation, because at that time raising things to the power of 4 was commonly regarded as meaningless, as there is no corresponding geometrical object in three dimensions. Eric. 213.173.233.179 (talk) 09:48, 24 August 2008 (UTC)