T
at position (
x,y,z
) in space is given by the formula
T
= 50 +
xyz
. Suppose a bird is at position (2
,
3
,
10), if the bird ﬂies oﬀ in direction
h
1
,

1
,
2
i
what is the rate of change of the temperature with respect to distance? In
what direction should the bird ﬂy in order to get warmer the most quickly?
Solution:
The gradient of
T
is
5
T
(
x,y,z
) =
h
yz,xz,xy
i
, and at the point (2
,
3
,
10),
5
T
(2
,
3
,
10) =
h
30
,
20
,
6
i
and the length is
√
900 + 400 + 36 =
√
1336. The bird
should ﬂy in the direction of
h
30
,
20
,
6
i
or
1
√
1336
h
30
,
20
,
6
i
(which is just the unit
vector in that direction) in order to get warm the fastest.
So for the rate of change in the direction of the vector
h
1
,

1
,
2
i
is
u
=
1
√
6
h
1
,

1
,
2
i
and the rate of change is
D
u
T
(2
,
3
,
10) =
h
30
,
20
,
6
i ·