- #1
1MileCrash
- 1,342
- 41
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"?
[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"?