Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Directional
Derivatives

To this point we’ve only looked at the two partial derivatives f x ( x, y ) and f y ( x, y )
...
We now need to discuss how to find the rate of change of f
if we allow both x and y to change simultaneously
...
For instance one could be changing faster than the other and
then there is also the issue of whether or not each is increasing or decreasing
...

The main idea that we need to look at is just how are we going to define the changing of x and/or
y
...
Let’s also suppose that both x and y are increasing and that, in this case, x is increasing
twice as fast as y is increasing
...

To help us see how we’re going to define this change let’s suppose that a particle is sitting at
( x0 , y0 ) and the particle will move in the direction given by the changing x and y
...
Now that we’re thinking of this changing x and y as a
direction of movement we can get a way of defining the change
...
For our example we will say that we want the rate
of change of f in the direction of v = 2,1
...
There is still a small problem with this however
...
For instance all of the following vectors point in the same direction
as v = 2,1
...
We will
do this by insisting that the vector that defines the direction of change be a unit vector
...
This means that for the example that
we started off thinking about we would want to use

v=

2 1
,
5 5

since this is the unit vector that points in the direction of change
...

Sometimes we will give the direction of changing x and y as an angle
...
The unit vector that points in this
direction is given by,

u = cos θ ,sin θ
Okay, now that we know how to define the direction of changing x and y its time to start talking
about finding the rate of change of f in this direction
...

Definition
The rate of change of f ( x, y ) in the direction of the unit vector u = a,b is called the
directional derivative and is denoted by Du f ( x, y )
...

However, in practice this can be a very difficult limit to compute so we need an easier way of
taking directional derivatives
...

To see how we can do this let’s define a new function of a single variable,

g ( z ) = f ( x0 + az, y0 + bz )

where x0 , y0 , a, and b are some fixed numbers
...

Then by the definition of the derivative for functions of a single variable we have,

g ( z + h) − g ( z )
g ′ ( z ) = lim
h →0
h

and the derivative at z = 0 is given by,

g ( h) − g ( 0)
g ′ ( 0 ) = lim
h →0
h

If we now substitute in for g ( z ) we get,

g ( h ) − g ( 0)
f ( x0 + ah, y0 + bh ) − f ( x0 , y0 )
g ′ ( 0 ) = lim
= lim
= Du f ( x0 , y0 )
h →0
h →0
h
h
So, it looks like we have the following relationship
...
Let’s rewrite g ( z ) as follows,

g ( z ) = f ( x, y ) where x = x0 + az and y = y0 + bz
We can now use the chain rule from the previous section to compute,

dg ∂f dx ∂f dy
g′ ( z ) =
=
+
= f x ( x , y ) a + f y ( x, y ) b
dz ∂x dz ∂y dz

So, from the chain rule we get the following relationship
...


Du f ( x, y ) = f x ( x, y ) a + f y ( x, y ) b
This is much simpler than the limit definition
...
There are similar formulas that can be derived by
the same type of argument for functions with more than two variables
...


Example 1 Find each of the directional derivatives
...
[Solution]
of θ =
3
(b) Du f ( x, y, z ) where f ( x, y, z ) = x 2 z + y 3 z 2 − xyz in the direction of
v = −1, 0,3
...

3
We’ll first find Du f ( x, y ) and then use this a formula for finding Du f ( 2, 0 )
...

In this case let’s first check to see if the direction vector is a unit vector or not and if it isn’t
convert it into one
...


v = 1+ 0 + 9 = 10 ≠ 1
So, it’s not a unit vector
...
So, the unit vector that we need is,

1
1
3
u=
−1, 0,3 = −
, 0,
10
10
10
The directional derivative is then,

⎛ 1 ⎞
⎛ 3 ⎞ 2
2
3
Du f ( x, y, z ) = ⎜ −
⎟ ( 2 xz − yz ) + ( 0 ) ( 3 y z − xz ) + ⎜
⎟ ( x + 2 y z − xy )
⎝ 10 ⎠
⎝ 10 ⎠
1
2
3
=
x
+
y
z − 3xy − 2 xz + yz )
3
6
(
10
[Return to Problems]

There is another form of the formula that we used to get the directional derivative that is a little
nicer and somewhat more compact
...

Let’s start with the second one and notice that we can write it as follows,

Du f ( x, y, z ) = f x ( x, y, z ) a + f y ( x, y, z ) b + f z ( x, y, z ) c
= f x , f y , f z i a, b, c
In other words we can write the directional derivative as a dot product and notice that the second
vector is nothing more than the unit vector u that gives the direction of change
...

Now let’s give a name and notation to the first vector in the dot product since this vector will
show up fairly regularly throughout this course (and in other courses)
...

With the definition of the gradient we can now say that the directional derivative is given by,

Du f = ∇f iu
where we will no longer show the variable and use this formula for any number of variables
...
This notation will be used when we want to note
the variables in some way, but don’t really want to restrict ourselves to a particular number of
variables
...

Let’s work a couple of examples using this formula of the directional derivative
...

(a) Du f ( x ) for f ( x, y ) = x cos ( y ) in the direction of v = 2,1
...
[Solution]

Solution
(a) Du f ( x ) for f ( x, y ) = x cos ( y ) in the direction of v = 2,1
...


∇f = cos ( y ) , −x sin ( y )

Also, as we saw earlier in this section the unit vector for this direction is,

u=

2 1
,
5 5

The directional derivative is then,

2 1
Du f ( x ) = cos ( y ) , − x sin ( y ) i
,
5 5
1
=
( 2 cos ( y ) − x sin ( y ) )
5
[Return to Problems]

( )

(b) Du f ( x ) for f ( x, y, z ) = sin ( yz ) + ln x 2 at (1,1, π ) in the direction of v = 1,1, −1
...
To do this we will first
compute the gradient, evaluate it at the point in question and then do the dot product
...


2
∇f ( x, y, z ) = , z cos ( yz ) , y cos ( yz )
x
2
∇f (1,1, π ) = , π cos (π ) , cos (π ) = 2, −π , −1
1
Next, we need the unit vector for the direction,

v = 3

u=

1 1
1
,
,−
3 3
3

Finally, the directional derivative at the point in question is,

1 1
1
Du f (1,1, π ) = 2, −π , −1 i
,
,−
3 3
3
1
=
( 2 − π + 1)
3
3−π
=
3
[Return to Problems]

Before proceeding let’s note that the first order partial derivatives that we were looking at in the
majority of the section can be thought of as special cases of the directional derivatives
...
The same can be
done for f y and f z
We will close out this section with a couple of nice facts about the gradient vector
...

Theorem
The maximum value of Du f ( x ) (and hence then the maximum rate of change of the function

f ( x ) ) is given by ∇f ( x ) and will occur in the direction given by ∇f ( x )
...
First, if we start with the dot product form Du f ( x ) and use a nice
fact about dot products as well as the fact that u is a unit vector we get,

Du f = ∇f iu = ∇f

u cos θ = ∇f cos θ

where θ is the angle between the gradient and u
...
Therefore the maximum

value of Du f ( x ) is ∇f ( x )

Also, the maximum value occurs when the angle between the

gradient and u is zero, or in other words when u is pointing in the same direction as the
gradient, ∇f ( x )
...


Example 3 Suppose that the height of a hill above sea level is given by
2
2
z = 1000 − 0
...
02 y
...
So even though most hills aren’t this symmetrical it will at least be
vaguely hill shaped and so the question makes at least a little sense
...
There are a couple of questions to answer here, but using the theorem
makes answering them very simple
...


∇f ( x ) = −0
...
04 y

The maximum rate of change of the elevation will then occur in the direction of

∇f ( 60,100 ) = −1
...
2 ) + ( 4 )
2

2

= 17
...
176

Before leaving this example let’s note that we’re at the point ( 60,100 ) and the direction greatest
rate of change of the elevation at this point is given by the vector −1
...
Since both of the
components are negative it looks like the direction of maximum rate of change is points up the
hill towards the center rather than away from the hill
...

Fact
The gradient vector ∇f ( x0 , y0 ) is orthogonal (or perpendicular) to the level curve f ( x, y ) = k
at the point ( x0 , y0 )
...


Proof
We’re going to do the proof for the 3 case
...
We’ll also
need some notation out of the way to make life easier for us let’s let S be the level surface given
by f ( x, y, z ) = k and let P = ( x0 , y0 , z0 )
...

Now, let C be any curve on S that contains P
...
In other
words, t0 be the value of t that gives P
...
Or,

f ( x (t ) , y (t ) , z (t )) = k

Next, let’s use the Chain Rule on this to get,

∂f dx ∂f dy ∂f dz
+
+
=0
∂x dt ∂y dt ∂z dt

Notice that ∇f = f x , f y , f z

and r′ ( t ) = x′ ( t ) , y′ ( t ) , z′ ( t ) so (4) becomes,

∇f i r′ ( t ) = 0
At, t = t0 this is,

∇f ( x0 , y0 , z0 ) i r′ ( t0 ) = 0

This then tells us that the gradient vector at P , ∇f ( x0 , y0 , z0 ) , is orthogonal to the tangent

(4)

vector, r′ ( t0 ) , to any curve C that passes through P and on the surface S and so must also be
orthogonal to the surface S
...
We will see the first application of this
in the next chapter

---