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WOLFRAM|DEMONSTRATIONS PROJECT

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.
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