Is the derivative of the volume of a sphere equal to its surface area?

The rate of change of the volume of the sphere is equal to the surface area of the sphere. The outside of the paint is the new boundary of the sphere, and the inside of the paint is added to the volume. This explains why the derivative (rate of change) of the volume is the surface area (SA).

Why is the derivative of a circle the circumference?

If you increase the radius of a circle by a tiny amount, dR, then the area increases by (2πR)(dR). . That is, the derivative of the area is just the circumference. This makes the “differential nature” of the circumference a little more obvious.

What is the circumference of a sphere?

The Circumference of a circle or a sphere is equal to 3.1416 times the Diameter.

What is the derivative of the area of a circle?

The change in area, dA, is dA = (2πR)dR. So, . That is, the derivative of the area is just the circumference. This, by the way, is one of the arguments for using “τ” instead of “π” . τ = 2π, so the area of a circle is . This makes the “differential nature” of the circumference a little more obvious.

What is the area of a circle in terms of volume?

The area of a circle is, and the circumference is, which is the derivative. The volume of a sphere is, and the surface area is, which is again the derivative. This, it turns out, is no coincidence! If you describe volume, V, in terms of the radius, R, then increasing R will result in an increase in V that’s proportional to the surface area.

How do you find the circumference of a sphere?

If you were to obtain circumference from a sphere based on surface area, for instance, the formula would be (after much simplification) C=8pi*r, which is 4 times bigger than 2pi*r . Taking the perimeter of a cube based on its surface area, would give us P=48r, which is 6 times bigger than the usual 8r.

How do you find the perimeter of a circle?

The circle (and sphere) is not really that special. It also works for the square if you measure it using not the side length $s$, but half that, $h=s/2$. Then its area is $A=(2h)^2=4h^2$ with derivative $dA/dh=8h$ which is its perimeter. $\\begingroup$ It works for the cube as well.