Page 378 - Advanced engineering mathematics

358 CHAPTER 11 Vector Differential Calculus

The directional derivative of ϕ at P 0 in the direction of u is
D u ϕ(2,−1,π)
1
π
π
= ((−4 − e )i + 4j − 2e k) · √ (i − 2j + k)
6
1 π π
= √ (−4 − e − 8 − 2e )
6
−3
= √ (4 + e ).
π
6
If a direction is specified by a vector that is not of length 1, divide it by its length before
computing the directional derivative.
Now imagine standing at P 0 and observing ϕ(x, y, z) as (x, y, z) moves away from P 0 .In
what direction will ϕ(x, y, z) increase at the greatest rate? We claim that this is the direction of
the gradient of ϕ at P 0 .

THEOREM 11.1

Let ϕ and its first partial derivatives be continuous in some sphere about P 0 , and suppose that
∇ϕ(P 0 ) = O. Then
1. At P 0 , ϕ(x, y, z) has its maximum rate of change in the direction of ∇ϕ(P 0 ).This
maximum rate of change is ∇ϕ(P 0 ) .
2. At P 0 , ϕ(x, y, z) has its minimum rate of change in the direction of −∇ϕ(P 0 ).This
minimum rate of change is − ∇ϕ(P 0 ) .

For condition (1), let u be any unit vector from P 0 and consider
D u ϕ(P 0 ) =∇ϕ(P 0 ) · u
= ∇ϕ(P 0 ) u cos(θ)

= ∇ϕ(P 0 ) cos(θ)
where θ is the angle between u and ∇ϕ(P 0 ). Clearly D u ϕ(P 0 ) has its maximum when cos(θ)=1,
which occurs when θ = 0, hence when u is in the same direction as ∇ϕ(P 0 ).
For condition (2), D u ϕ(P 0 ) has its minimum when cos(θ) =−1, hence when θ = π and
∇ϕ(P 0 ) is opposite u.

EXAMPLE 11.7
2 y
Let ϕ(x, y, z) = 2xz + z e and P 0 : (2,1,1). The gradient of ϕ is
y
2 y
∇ϕ(x, y, z) = 2zi + z e j + (2x + 2ze )k
so
∇ϕ(2,1,1) = 2i + ej + (4 + 2e)k.
The maximum rate of increase of ϕ(x, y, z) at (2,1,1) is in the direction of 2i + ej + (4 + 2e)k,
and this maximum rate of change is

2
2
4 + e + (4 + 2e) ,
√
or 20 + 16e + 5e .
2

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October 15, 2010 16:11 THM/NEIL Page-358 27410_11_ch11_p343-366