- #1
- 3,802
- 95
Given the volume of a sphere V is given by [tex]V=\frac{4}{3}\pi r^3[/tex]
Taking the derivative of the V wrt r gives [tex]\frac{dV}{dr}=4\pi r^2[/tex]
which is the Surface Area of the sphere: [tex]\frac{dV}{dr}=A_s[/tex]
Taking the derivative of As wrt r gives [tex]\frac{dA_s}{dr}=8\pi r[/tex]
which I don't recognise as anything too significant.
It is not far off the area of a circle ([tex]2\pi r[/tex]) but it's not quite.
My question is: Is there any physical significance for the rate of change of the surface area of a sphere? Since the rate of change of the volume of the sphere describes the 3-d surface area of the sphere. Then the rate of change of this should describe...? Maybe the 2-d surface area (the shape could be truncated in some odd way).
Taking the derivative of the V wrt r gives [tex]\frac{dV}{dr}=4\pi r^2[/tex]
which is the Surface Area of the sphere: [tex]\frac{dV}{dr}=A_s[/tex]
Taking the derivative of As wrt r gives [tex]\frac{dA_s}{dr}=8\pi r[/tex]
which I don't recognise as anything too significant.
It is not far off the area of a circle ([tex]2\pi r[/tex]) but it's not quite.
My question is: Is there any physical significance for the rate of change of the surface area of a sphere? Since the rate of change of the volume of the sphere describes the 3-d surface area of the sphere. Then the rate of change of this should describe...? Maybe the 2-d surface area (the shape could be truncated in some odd way).