Relation of Radius, Surface Area, and Volume of a Sphere

time

Take the derivative of

V =

4π

3

r

3

to get

∂V

∂t

= 4π

2

r

∂r

∂t

.

Using

∂V

∂t

= 1 and solving,

∂r

∂t

=

1

4π

2

r

.

Also, A = 4π

2

r

,

so

∂A

∂t

= 8πr

∂r

∂t

=

8πr

4π

2

r

=

2

r

.

Imagine that you are blowing up a spherical balloon at the rate of

1

3

cm

/s

. How do the radius and surface area of the balloon change with its volume? We can find the answer using the formulas for the surface area and volume for a sphere in terms of its radius.