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Cubes, spheres, and derivatives.

  1. Jul 22, 2013 #1
    The volume of a sphere with radius r is

    [itex]v = \frac{4}{3}\pi r ^{3}[/itex]

    It makes sense that its derivative with respect to radius is the surface area of the sphere.

    [itex]\frac{dv}{dr} = a = 4\pi r ^{2}[/itex]


    The volume of a cube with side length n is
    [itex]n^{3}[/itex]

    The derivative of this is just 3n^(2), which is not the surface area of the cube.

    But, if instead of writing the volume of the cube in terms of its side length, I can write it in terms of half of its side length (call it a, so that a = (1/2)n) and then the volume is

    [itex]v = (2a)^{3}[/itex]
    [itex]v = 8a^{3}[/itex]

    While the derivative with respect to a is

    [itex]\frac{dv}{da} = 24a^{2}[/itex]

    Which is the surface area of that cube.


    So, why is expressing these formulas in terms of "half lengths" (radius, and half-side, vs diameter and side) "special"?
     
  2. jcsd
  3. Jul 23, 2013 #2
    C'mon guys, this is interesting!
     
  4. Jul 23, 2013 #3

    verty

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    Here is another interesting tidbit. The volume of a pyramid is 1/3 base * height, this is also true for a cone (1/3 * πr^2 * h). But now, this is interesting. The volume of a sphere is 4/3 πr^3 = 1/3 * 4πr^2 * r. This is just 1/3 of the radius * the surface area. Does this mean a sphere is related to a pyramid or cone with height r and base 4πr^2? Can we transform a sphere into a cone by slicing it and flattening the surface onto the plane? If not, why do we have this similarity?
     
  5. Jul 23, 2013 #4

    LCKurtz

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    Here's one I found interesting when I learned it. Consider a sphere of radius ##r## inscribed in a cylinder. The lateral surface area of the cylinder and the area of the sphere are equal. Furthermore, the projection that maps areas of the sphere radially along the cylinder's radius onto the cylinder is area preserving. I would guess that mapmakers know that although I don't know if I have ever seen a map of the world projected like that onto a cylinder and rolled flat. Probably everyone but me knows about that kind of map...
     
  6. Jul 23, 2013 #5

    lurflurf

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    We could say this results from Stokes' theorem
    $$\int_{\partial \Omega} \omega=\int_{ \Omega} \mathop{d} \omega$$

    and that your formulation introduces irrelevant parameterization. In fact we can chose the parameterization to make the result true. That is often done in differential geometry in which a parameterization is chosen to preserve a desired relation. To illustrate say
    Volume=f(r(s))
    Surface area=g(r(s))
    [f(r(s))]'=g(r(s))
    implies (chain rule)
    f'(r(s))r'(s)=g(r(s))
    so all we need do is chose r such that r'(s)=1.

    Often the family we want to consider is homogeneous then we have
    Volume=V=A t^3
    Surface area=S=B t^2
    then we have r'(t)=1 when
    r(t)=3V/S=(3A/B)t

    we might then like to write
    Volume=C (t/3)^3
    Surface area=S=C (t/3)^2
    where C=A*27=B*9

    Here is a nice related article
    http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.rmjm/1181068766
     
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