The volume of a sphere with radius r is(adsbygoogle = window.adsbygoogle || []).push({});

[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]

Whichisthe 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"?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Cubes, spheres, and derivatives.

Loading...

Similar Threads - Cubes spheres derivatives | Date |
---|---|

I Deriving the volume of a sphere using semi-circles | Oct 11, 2016 |

Expanding a simple cube root | Oct 12, 2014 |

Viewing Inside Faces of a Cube | Sep 23, 2011 |

How cubed root of x is not differentiable at 0 | Sep 20, 2011 |

The Boundary of a Countable Union of Almost Disjoint Cubes | Sep 26, 2010 |

**Physics Forums - The Fusion of Science and Community**