Why-is-circumference-of-a-circle-derivative-of-its-area-d-dr-pir-2-2pir-circumference-

Question Number 3228 by Rasheed Soomro last updated on 08/Dec/15

W h y i s c i r c u m f e r e n c e o f a c i r c l e d e r i v a t i v e o f

i t s a r e a ?

\frac{d}{d r} (π r^{2}) = 2 π r = c i r c u m f e r e n c e .

Commented by 123456 last updated on 08/Dec/15

it also work to sphere too (? ?)

V = \frac{4 π r^{3}}{3}

S = 4 π r^{2}

Commented by prakash jain last updated on 08/Dec/15

May be it has to with formula related to

miximum area / volume given perimeter / area .

Commented by 123456 last updated on 08/Dec/15

isoperimetric inequality (or something) ?

Commented by Filup last updated on 08/Dec/15

Reversing the question, you get :

Why is the area the integral of its

circumferance ?

The circumferance, o r l o c u s, follows :

x^{2} + y^{2} = r^{2}

The area of a full circle is then given by :

A = 2 \int_{- r}^{r} \sqrt{r^{2} - x^{2}} d x = π r^{2}

Commented by Rasheed Soomro last updated on 08/Dec/15

C i r c u m f e r e n c e h e r e i s m e a s u r e o f c i r c l e (l o c u s) n o t

c i r c l e i t s e l f .

Y o u h a v e m e a n t b y c i r c u m f e r e n c e l o c u s (c i r c l e)

i t s e l f .

Commented by Filup last updated on 08/Dec/15

i s e e

Commented by Yozzi last updated on 08/Dec/15

I t w o u l d a p p e a r t o m e t h a t t h e r e e x i s t s

a d i m e n s i o n a l i n t e r p r e t a t i o n o f t h e

d e r i v a t i v e / a n t i - d e r i v a t i v e o f

c e r t a i n p h y s i c a l e q u a t i o n s .

S o t h a t, i n d e f i n i n g a c i r c l e b y

a q u a n t i t y i n i t s h i g h e s t p o s s i b l e

d i m e n s i o n i n 2 D - s p a c e (a r e a), t h e

d e r i v a t i v e y i e l d s a 1 D - s p a c e p r o p e r t y

o f t h e c i r c l e - t h e l e n g t h o f t h e l i n e

w h i c h c o n s t i t u t e s t h e c i r c l e . I f w e

g o f u r t h e r, 2 π t h e n r e p r e s e n t s a

p o i n t i f w e p e r h a p s d e c i d e d t o p l o t

p o i n t s P (r, \frac{d^{n} A}{{d r}^{n}}) i n t h e p l a n e, k n o w i n g

t h e l o c u s o f a c i r c l e h a v i n g a r e a A

a n d r a d i u s r . S o, P (r, \frac{d^{0} A}{{d r}^{0}}) g i v e s a

s h a d e d c i r c l e o f r a d i u s r; P (r, \frac{d A}{d r}) g i v e s a p l a i n

c i r c l e o f r a d i u s r; P (r, \frac{d^{2} A}{{d r}^{2}}) g i v e s

a p o i n t a n d P (r, \frac{d^{3} A}{{d r}^{3}}) d e g e n e r a t e s

t h e c i r c l e c o m p l e t e l y .

(T h i s i s j u s t m y v i e w : {i t}^{'} s p r o b a b l y

w r o n g .)

Answered by prakash jain last updated on 08/Dec/15

Let us say we try to find out how

much such curve exists . By the problem

definition y = f (r, x) . w h e r e r i s a n

i n d e p e n d e n t p a r a m e t e r .

Length of a curve L (r) = \int_{0}^{r} \sqrt{1 + f_{x} {(r, x)}^{2}} d x

A rea of curve A (r) = \int_{0}^{r} f (r, x) d x

The problem statement is find all

f (r, x) s u c h t h a t L (r) = \frac{d}{d r} A (r)

\int_{0}^{r} \sqrt{1 + f_{x} {(r, x)}^{2}} d x = \frac{d}{d r} \int_{0}^{r} f (r, x) d x

\int_{0}^{r} \sqrt{1 + f_{x} {(r, x)}^{2}} d x = f (r, r) + \int_{0}^{r} f_{r} (r, x) d x \dots (A)

So we need to solve the above equation to

find f (r, x) .

f (r, x) = \sqrt{r^{2} - x^{2}} s a t i s f i e d

\sqrt{1 + f_{x} {(r, x)}^{2}} = \sqrt{1 + \frac{x^{2}}{r^{2} - x^{2}}} = \sqrt{\frac{r^{2}}{r^{2} - x^{2}}} = \frac{r}{\sqrt{r^{2} - x^{2}}}

f_{r} (r, x) = \frac{r}{\sqrt{r^{2} - x^{2}}}, f (r, r) = 0

I will try to find more solution for A .