Wikipedia:Reference desk/Archives/Mathematics/2013 August 23

= August 23 =

the shape of heart
What is the geometric shape of the heart? Can I say that the heart is a truncated cone? (I do not think it is a truncated cone that is not really like that)46.210.149.99 (talk) 00:14, 23 August 2013 (UTC)


 * Do you mean a physical heart or the traditional heart shape? If the latter, see Limaçon.  Bubba73 You talkin' to me? 00:39, 23 August 2013 (UTC)


 * See . Other than the Cardioid there isn't a recognized heart curve in mathematics. I think most graphic designers would use splines to get the desired shape. --RDBury (talk) 01:49, 23 August 2013 (UTC)


 * I mean about a physical heart (no about traditional heart shape), Is there any geometric shape for the real heart? 95.35.80.206 (talk) 09:04, 23 August 2013 (UTC)


 * Well, it's kind of a blob, not any regular geometric shape. If we wanted to model it as a single geometric primitive, maybe an ellipsoid or a closed surface of revolution would be close enough. StuRat (talk) 09:19, 23 August 2013 (UTC)

is truncated cone not close more than surface of revolution? 95.35.80.206 (talk) 11:30, 23 August 2013 (UTC)


 * Looking at images of the human heart here on Wikipedia and elsewhere, I struggle to see any symmetry in its shape at all. I doubt whether a truncated cone or any more general surface of revolution would be a good model of the shape of the heart. It might help if you explained why you want to model the shape of the heart - a rough or approximate model might be good enough for some purposes, but not close enough for others. Gandalf61 (talk) 12:22, 23 August 2013 (UTC)

I need to model the shape of heart because one of my people asked me about, and I didn't know to answer, I promised him that I cheak it out latter. I think that it's good question. In the beginning I'd thought to answer that the heart doesn't have any defined shape because it's not so symetry. afterward I understood that the heart has defined shape. It's like triple or cone without apex.95.35.239.134 (talk) 13:31, 23 August 2013 (UTC)


 * I'm not seeing a truncated cone at all. If I were to try to model it as a single primitive, I'd start with this image: .  I'd then draw a line down the central vertical muscle band, dead center at the top and on the right side of the muscle band at the bottom, and use that as the axis of revolution, with the left half of the outer edge used as the generating curve.  This would probably be close enough, say, for a video game, where you want to keep the complexity down to improve performance.  The ellipsoid suggestion would be less accurate, but might be even quicker to model. StuRat (talk) 20:52, 23 August 2013 (UTC)

Intersection of two "curves" question.
Consider the equation x^y=y^x. (with x and y both >1). The solutions fall into two "curves", one of which is the line y=x and the other "looks" hyperbolic (with asymtotes at x=1 and y=1) and includes 2,4 and 4,2. Since it curves so that the intersection of the curves would be "below" (3,3), (and we are working with exponentials) so it would seem natural for the point to be (e,e), but how do I prove it?

— Preceding unsigned comment added by Naraht (talk • contribs) 15:54, 23 August 2013‎


 * xy = yx → ln(xy) = ln(yx) → y ln(x) = x ln(y) → $ln(x)⁄x$ = $ln(y)⁄y$
 * or
 * xy = yx → xy$\sqrt{x^{y}}|undefined$ = xy$\sqrt{y^{x}}|undefined$ → x$\sqrt{x}$ = y$\sqrt{y}$
 * Since the maxima and minima are to be found among the roots of the first derivative, we have
 * [$ln(t)⁄t$]  '  = $1 - ln(t)⁄t^{2}$
 * or
 * [t$\sqrt{t}$]  '  = $^{t}$\sqrt{t}$⁄t^{2}$ . (1 - ln[t])
 * both of which are zero when ln(t) = 1 or t = e. — 79.113.213.6 (talk) 22:11, 23 August 2013 (UTC)


 * (edit conflict) Take logs of both sides and divide through by xy to obtain $$\frac{\log x}{x}=\frac{\log y}{y}.$$ Plot $$f(z)=\frac{\log z}{z}$$ against z -- it starts at (1, 0), rises monotonically to a peak at z=e, and then declines monotonically toward zero as z goes to infinity. As you say x=y gives a set of solutions, represented by x and y both being at the same point on the z axis. All other solutions are found by drawing horizontal lines at various heights and letting the horizontal locations of the intersections with the curve f(z) be x and y or y and x. Start infinitesimally close to x=1 with the horizontal line near the horizontal axis, so the corresponding y is arbitrarily large. Now raise the horizontal line -- x goes up, and y goes down. Eventually x and y come together at the peak of the curve, which is at x=y=e. After that, the solutions with x>y are found by continuously lowering the horizontal line again. So you're right -- your two curves coincide at x=y=e. Duoduoduo (talk) 22:13, 23 August 2013 (UTC)


 * Alternatively, x and y are the values of u where a straight line of positive slope through the origin cuts the curve log u. x and y are equal when they coincide at the value of u where the line is tangential to the curve. It's easy to show that this is e, for any logarithmic base.→31.53.1.113 (talk) 22:29, 23 August 2013 (UTC)


 * The gradient of $$x^y-y^x$$ will be zero at the point of intersection. The gradient is $$(y x^{y-1} - y^x\log y, x^y\log x - x y^{x-1})$$.  Setting to zero gives $$y x^{y-1}=y^x\log x \implies y=x\log x$$ and similarly $$x=y\log y$$.  Let $$f(x)=x\log x$$.  The x coordinate of the solution point must be a fixed point of $$f(f(x))$$.  (In particular, it must be &ge; 1, for otherwise $$f(x)$$ is not in the domain of $$f$$.)  The function $$f(f(x))$$ is a convex function on $$[1,\infty)$$ with $$f(f(1))=0<1$$, so it can have at most one fixed point.  This fixed point is at $$x=e$$.   Sławomir Biały  (talk) 23:54, 23 August 2013 (UTC)
 * Just for fun, a related problem: For given b, how may times to the graphs of y=bx and y=logbx intersect? --RDBury (talk) 07:13, 24 August 2013 (UTC)