Oh yeah, that too. I think I ran across that once, it looks like a few calculus problems I may have seen... I guess I've been wrapping too many gifts recently. Kinda got stuck on the physical notion of cube-roots and areas (-: And of course, I should have said "double the volume of..." rather than "double the size of...". Size matters, but volume is more specific. - Robert On Wed, Dec 28, 2011 at 04:59, David Makin <makinmagic@tiscali.co.uk> wrote:
Hi,
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If you double the size of a cube, its edge length goes up by the cube root of 2, and therefore its surface area (and thus, the amount of paper required to wrap it) increases by the cube root of 4. The same thing applies to other shapes so long as the doubled shape is similar (in the Euclidean-geometrical sense of the word) to the original, and isn't one of those monsters that all the kids (like Helge and Georg) keep going on about.
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Now that's also pretty interesting :)
The answer I was looking for was the value of x such that the area under y=x^2 is equal to the area under y=sqrt(x) between the origin and x (area=4/3).
Because I'm very paranoid and a little rusty when it comes to actually performing such calculations I checked it online via this rather nice tool:
Using maxima I get three answers, two of them complex: (%i4) solve(integrate(x^2, x, 0, y) = integrate(sqrt(x), x, 0, y), y); Is y positive, negative, or zero? positive; (sqrt(3) %i - 1) sqrt(y) (sqrt(3) %i + 1) sqrt(y) (%o4) [y = ------------------------, y = - ------------------------, 2/3 2/3 2 2 1/3 y = 2 sqrt(y)] All of them are in the form y=K sqrt(y), which is easy to solve manually but annoying. I cannot afford Mathematica. -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com