question

[vivid spade]

my teacher was saying smth ab how the derivative of sphere's volume ( 4/3pir^3) is equal to surface area(4pir^2)and it also works with area of a circle (pir^2) and the circumference (2pir). I don't get why this works tho cuz I thought differentiation was just how to find the gradient of a curve?

[cloud turtle]

I don’t even get this icl

V = 4/3 pir^3

With this u can see how the volume changes when u change the radius

[inland heath]

dr/dx of 4/3pir^3

= 4pir²

which is the SA

[cloud turtle]

Dv/dr = 4pir^2

[vivid spade]

ya i get that

[cloud turtle]

Change in volume over change in radius

[vivid spade]

but like how does it come together

[inland heath]

?

[vivid spade]

like how does it relate to gradient

[inland heath]

you could insert a random point

and get a value

[cloud turtle]

Describes area I guess

Idk

[inland heath]

that would be the gradient at that specific point, which would be area

A=4pir²

[cloud turtle]

Nah he gets that part bones

[vivid spade]

she^

[cloud turtle]

But he doesn’t get why

She

Mb mb

[vivid spade]

dwdw

i just dont get how it applied to 3d

applies

[inland heath]

you can differeniate in 3d

it's just MUCH harder

like uni level maths

[cloud turtle]

Think about it like this

U have a football

[vivid spade]

wthh

[cloud turtle]

U wrap it perfectly

What’s the area of the wrapping?

[vivid spade]

ermm

[cloud turtle]

Same as the surface area

[vivid spade]

ya

[cloud turtle]

Because it wraps it perfectly

[inland heath]

I mean realistically you should be able to

[cloud turtle]

If u keep wrapping

U continue increasing the volume of ball

[inland heath]

since diff is the rate of change of something

[cloud turtle]

But the surface area each time

[vivid spade]

ohh stays the same