Member • Mar 1, 2014
Circle is special
And vice versa while integrating ??
Member • Mar 1, 2014
Member • Mar 1, 2014
Member • Mar 1, 2014
Member • Mar 1, 2014
That is some logic.A.V.RamaniIf f(x) = a x^b , then df(x)/dx = abX^(b-1)
It just happens that the area of the circle has a differential equal to the circumference as a consequence of the above.
See what happens if you use the area of the circle as a function of the diameter.
Area = (pi/4)D^2
The differential wrt d = (pi/2)D, which is not the circumference.
Member • Mar 4, 2014
A.V.RamaniIf f(x) = a x^b , then df(x)/dx = abX^(b-1)
It just happens that the area of the circle has a differential equal to the circumference as a consequence of the above.
See what happens if you use the area of the circle as a function of the diameter.
Area = (pi/4)D^2
The differential wrt d = (pi/2)D, which is not the circumference.
OH!!!!! that not working for diameter....A.V.RamaniIf f(x) = a x^b , then df(x)/dx = abX^(b-1)
It just happens that the area of the circle has a differential equal to the circumference as a consequence of the above.
See what happens if you use the area of the circle as a function of the diameter.
Area = (pi/4)D^2
The differential wrt d = (pi/2)D, which is not the circumference.
Member • Sep 15, 2014
Shah SharathWhy differentiation of area of circle is its circumference
And vice versa while integrating ??
Member • Sep 16, 2014
Member • Sep 17, 2014
Member • Sep 17, 2014
Please study the problem in detail and refelect on the actions performed.shiwa436As from the definitions of integration and differentiation from the Wikipedia,
Integration of perimeter 2pir with respect to its radius r gives its signed area in both the planes i.e x, y planes..
Differential gives sensitivity of position of y axis equivalent w.r.t x axis one... i.e position of a dot on the edge along the diameter