time
Take
the
derivative
of
Global
`
V
=
4
π
r
3
3
to
get
∂
Global
`
V
∂
t
=
4
π
r
2
∂
r
∂
t
.
Using
∂
Global
`
V
∂
t
=
1
and
solving
,
∂
r
∂
t
=
1
4
π
r
2
.
Also
,
A
=
4
π
r
2
,
so
∂
A
∂
t
=
8
π
r
∂
r
∂
t
=
8
π
r
4
π
r
2
=
2
r
.
0
2
4
6
8
10
∂
Global
`
V
∂
t
,
∂
r
∂
t
,
∂
A
∂
t
0
.
0
0
.
2
0
.
4
0
.
6
0
.
8
1
.
0
1
.
2
radius
0
2
4
6
8
10
0
.
0
0
.
5
1
.
0
1
.
5
surface
area
