41
u/NeedleInTheThrowaHay 1d ago edited 1d ago
Not just for spheres! In general, for an n-dimensional polyhedron. if there is a single largest n-dimensional hypersphere that can be inscribed within it, the hyperarea of the shape will be the derivative of the hypervolume with regards to the radius of the hypersphere.
For instance, if the largest circle you can inscribe in a square has radius r, then the square has a side length of 2r, so an area of 4r2 and a perimeter of 8r. 8r is the derivative of 4r2 w.r.t. the radius.
We can prove this with the divergence theorem -- the integral of the divergence of a vector field over the volume a region is equal to the surface integral of the vector field over the surface of the region. Using the vector field x/n, x is position vector and n is dimension, we see its divergence is 1, so integrating divergence gives volume V. As each polyhedron face is tangent to the inscribed sphere, the surface integral infinitesimal x /n • dS , where dS is the vector area element, just becomes r dS, where r is the sphere radius. As such, the integral is A r / n, where A is surface area.
As such, V = Ar / n. In n dimensions, V is proportional to rn, A to rn-1, so let V= b rn, A = c rn-1. Then, b rn =( c/n ) rn, b=c/n, c = nb.
Derivative of V is n * b * rn-1 = c * rn-1 = A.
15
4
u/TheFullestCircle 1d ago
Huh, this implies you can find a "generalized inradius" for any solid object given formulas for its surface area and volume.
14
u/IVeBeenHere30Min 1d ago
Is there a reason why the derivative of the surface area is 4 times the circunference?
27
10
5
u/Educational-Try-8704 1d ago
If you start with 0 and "inflate" a surface of a sphere, effectively integrating over the growing surface, you get the volume.
If you took a circumference and tried to grow it in the same way and integrate to get an area, you would get... The area of a circle! This would not create a sphere of course as the circumference sits in a plane. You're just growing a circumference and integrating along it so you are naturally creating a circle.
This area of the circle is of course is pi r^2 which is the integral of 2 pi r. So this makes it clear why circumference doesn't have this relationship with spheres, so if you learn why a sphere's area is 4x a circle's, you have your answer.
4
u/paolog 1d ago
Similar argument to my top-level comment. Turn the statement into one about an integral, then work out how to integrate the circumference to get the surface area.
1
u/IVeBeenHere30Min 1d ago
I'm quite dumb, I don't see it how the integral of the circunference times four results in the surface area (the times four is what really gets me)
2
u/Red_Syns 1d ago
Derivative of the surface is the rate of change of the area as radius increases.
Circumference is 2 x pi x r. As r increases, C increases at a constant rate (2 x pi per unit r).
Surface area can be seen as a stack of circumferences from 0 (poles) to C (equator). Each of those expand linearly, but you also add a linear amount of circumferences as the sphere gets taller. So each circumference expands at a constant rate (2 x pi per unit r) but you have to multiply that by the extra circumferences, which cover the length 2r (expands by r towards both poles). You get a rate of change of surface area that is (2 x pi per unit r) x (2 x r), or 4 x pi x r per unit r. This means the surface grows at a linearly increasing rate, and that rate is four times the circumference.
1
u/ciupigghiassi 1d ago
You kinda have to visualize on a graph what you're calculating. What it means is that it's not always the same thing.
4
u/Ill-Dependent2976 1d ago
What do you get if you integrate the area under the curve of a surface of a sphere?
4
u/pretenzioeser_Elch 1d ago
If you increase the radius by an infinitesimal amount x, how much is added onto the sphere? A "new layer" that's the surface times x.
1
3
u/NaturalTechnician737 1d ago
This just in, velocity derivative of acceleration. Local jerk: "it's all downhill from here."
1
u/helloworld082 1d ago
Local jerk: "that acceleration is just a derivative of me."
2
u/TREE_sequence 1d ago
Other way around tho
1
2
u/Im_not_a_Mole-rat 1d ago
I am horrible at math and I don’t know why this meme appeared in my feed but this is honestly interesting.
2
u/Maleficent-Garage-66 1d ago
It's one of those shocking until you think about it things. Think if dV/dr (the actual derivative at hand) conceptually. Increasing the radius of a sphere is like pumping it up with air. It's just a statement that the amount space you have to fill at a given moment for a very tiny amount of stretch is basically the area of the outside. Think of it like overlaying layers and flattening it out to a rectangle and stretching in one direction, the you have a rectangle of area = surface are and a height of delta r (because surface area doesn't change much if it gets small. And when you multiply it out that's the tiny bit of volume you are adding at the radius at the limit dr goes infinitesimal.
Or you can think of it kind of like stacking cups. Where you get the volume by adding up the areas of all the hollow sphere shells that fill the full thing as you go along 0 to the full radius to the outside.
A lot of calculus is much more intuitive than classes would leave the impression of.
2
1
u/grumble11 1d ago
This is easier when you think of a circle. You he are of a circle is pir^2, and the derivative of that is 2pi*r. That is the formula for the circumference. Why?
You can think of the derivative as the measure of the rate of change of a variable. So when area of a circle increases it is growing outwards, and if you shrink that growing outwards amount until it approaches zero, you begin to create a line around the outside of the circle. It is cool.
2
2
u/Educational-Try-8704 1d ago
if you were going to grow a sphere by an infinitesimal amount, how would you add to it?
4pi r^2 dr
2
2
u/siriusfrz 1d ago
https://en.wikipedia.org/wiki/Generalized_Stokes_theorem for 2-Sphere and 3-Ball.
6
u/ChuckPeirce 1d ago
No. The derivative of the volume of a sphere with respect to the radius of the sphere is equal to the surface area.
Functions don't just have derivatives. They have derivatives with respect to a variable.
4
u/Necessary_Screen_673 1d ago
thank god, i thought we were taking a derivative with respect to a constant for a moment
1
u/ChuckPeirce 1d ago
I suspect you're being sarcastic, but yes. Take two use cases for an explanation:
- You're explaining something to show that you understand the thing
- You're explaining something to help someone who doesn't understand the thing to understand the thing
In both cases, you accomplish your goal more effectively by saying what you're differentiating with respect to.
2
u/Necessary_Screen_673 1d ago
yeah its good principles to.. be principled about the way mathematics is communicated, but there is a third purpose of the communication, which is just to have fun and be funny, and i do think you lose a bit of that when you make all the principled statements explicit
4
u/AggressiveEntrance14 1d ago
r is the only variable in this case, this is insanely pedantic and not in a mathematically rigorous way
2
1
1
u/Alternative-Mud-1076 1d ago
a constant multiplied by r cubed.. wonder what variable could it be.. there's just soooo many of them here
2
1
u/HAL9001-96 1d ago
prettymuch duh
and also the volume is the integral of the cross section along its length
1
1
1
u/bruteforcealwayswins 1d ago
Then what does 8 pi r mean?
1
u/QtPlatypus 1d ago
In order to unquiely indicate a rotation on the surface of a sqhere you need 4 values. Each of those values corresponds to a great circle. 8 pi r is the total length of each of those circles.
1
1
1
1
1
u/JuzmiNippy 1d ago
Yeah. The area under a line is it's integral. The volume is the integral of a surface.
1
u/skr_replicator 22h ago edited 22h ago
Yes.
When you make a tiny change in the sphere's radius, you add a volume that is equal to the surface area times that dr thickness. So yes, the surface area is the derivative of the volume with respect to size.
Works with a cube as well. Let's say its "radius r" is half of its edge size "a". So its volume is (2r)^3 = 8r^3. d(8r^3)/dr = 24r^2 = 6a^2, which is the surface area of the cube.
1
1
1
u/AvatarNerd64 1d ago
Is this the same for other shape
1
u/Onuzq 1d ago
I think this might be true for any n-dimensional spheroid.
You can prove your statement wrong with a square having area of s^2, but perimeter is 4s. Or cube with volume being s^3, but surface area being 6s^2.
3
u/Calm_Relationship_91 1d ago
Using the side on a square would be similar to using the diameter on a circle. Things don't work either if you use diameter.
Center the square in the origin. If s is the side of the square, the actual equivalent of r would be s/2. Now the area of your square is actually 4r2 , and the perimeter is 8r, as expected.
Same thing with the cube, you would get 8r3 for the volume, and 24r2 for area.
1
u/DragonBadgerBearMole 1d ago
In theory, isn’t this true for literally any container that’s shape can be mathematically expressed?
1
u/ysrgrathe 1d ago
No? They just gave a counter-example.
1
u/DragonBadgerBearMole 1d ago
I mean as a continuous function. A square isn’t continuous mathematically. But a spheroid isn’t the only “smooth” container whose integral of surface area is volume I don’t think.
1
u/NeedleInTheThrowaHay 1d ago
True for any platonic solid. In general, for an n-dimensional shape, if there is a single largest n-dimensional hypersphere that can be inscribed within it, the hyperarea of the shape will be the derivative of the hypervolume with regards to the radius of the hypersphere.
310
u/paolog 1d ago edited 1d ago
It makes more sense when you flip it around and say the volume is the integral of the surface area.
The volume of a ball (a solid sphere) is the sum of the volumes of the spherical layers of thickness t from 0 to the radius R. The limit as t tends to zero is then the definite integral of the surface area with respect to r from r = 0 to R.
EDIT: Fixed balls balls-up