1.2. THE GRADIENT AND DIRECTIONAL DERIVATIVES 15
2.5
-2.5
-2.5 2.5
Notice that all the gradient vectors point directly away from the origin. is corresponds to our
notion that the gradient is the direction of fastest growth.
˙
e function
2x C y
x
2
C y
2
C 1
from Example 1.30, the rolling ball question, will probably have a
more interesting gradient than the simple paraboloid in Example 1.31.
Example 1.32 Let H.x; y/ D
2x C y
x
2
C y
2
C 1
. For all points .x; y/ with coordinates in the set
f˙3; ˙2:5; ˙2; ˙1:5; ˙1; ˙0:5; 0g, plot the point and the vector starting at that point in the
direction rH.x; y/.
16 1. ADVANCED DERIVATIVES
Solution:
We are plotting the vectors drawn from the gradient
rH.x; y/ D
2y
2
2x
2
2xy C2
.x
2
C y
2
C 1/
2
;
x
2
y
2
4xy C 1
.x
2
C y
2
C 1/
2
which yields the picture:
3.5
-3.5
-3.5 3.5
e gradient vectors point in many directions. A ball rolling on this surface could potentially
have a very complex path.
˙
1.2. THE GRADIENT AND DIRECTIONAL DERIVATIVES 17
Example 1.33 Find a reasonable sketch of the vector field associated with the gradient
of g.x; y/ D sin.x/ C cos.y/ from Example 1.28. Use grid points with coordinates ˙n for
n D 0; 1; : : : 8.
Solution:
We are plotting the vectors drawn from the gradient
rg.x; y/ D
.
cos.x/; sin.y/
/
which yields the picture:
9.0
-9.0
-9.0 9.0
e periodicity of the vector field is easy to see. e different cells of the periodicity are slightly
different because the periods are multiples of , and the grid used to display the vectors is sized
in multiples of one.
˙
18 1. ADVANCED DERIVATIVES
Now that we have the gradient, it is possible to define the derivative in a particular direction.
Knowledge Box 1.4
If z D f .x; y/ is a differentiable function of two variables, and Eu D .a; b/ is a
unit vector, then the derivative of f .x; y/ in the direction of Eu is:
r
Eu
f .x; y/ D Eu rf .x; y/ D af
x
C bf
y
If Ev D .r; s/ is any vector, then the derivative of f .x; y/ in the direction of Ev
is:
r
Ev
f .x; y/ D
Ev
jEvj
rf .x; y/
Notice that we are continuing the practice of using unit vectors to designate directions – even
when computing the derivative of a function in the direction of a general vector, we first coerce
it to be a unit vector.
It is worth mentioning that, when computing directional derivatives, we start with a scalar
quantity – the function f .x; y/. When we compute the gradient, we get the vector quantity
rf .x; y/ D .f
x
; f
y
/ but then return to a scalar function of two variables r
Eu
f .x; y/. It is im-
portant to keep track of the type of object – scalar or vector – that you are working with.
Example 1.34 Find the derivative of
f .x; y/ D x
2
C y
2
in the direction of Eu D .1=2;
p
3=2/.
Solution:
e vector Eu is a unit vector so, starting with f
x
D 2x, f
y
D 2y, we get:
r
Eu
f .x; y/ D
1
2
2x C
p
3
2
2y D x C
p
3y
˙
e directional derivative occasionally comes up in the natural course of trying to solve a prob-
lem. ere is one very natural application: finding level curves. First let’s define level curves.
1.2. THE GRADIENT AND DIRECTIONAL DERIVATIVES 19
Knowledge Box 1.5
Level Curves
If z D f .x; y/ defines a surface, then the level curve of height c of f .x; y/ is
the set of points that solve the equation
f .x; y/ D c:
Example 1.35 Plot the level curves for c 2 f1; 2; 3; 4; 5; 6g for
f .x; y/ D x
2
C 2y
2
Solution:
e equation x
2
C 2y
2
D c is an ellipse that is
p
2 times as far across in the x direction
as the y directions. Solving for the points where x D 0 or y D 0 gives us the extreme points of
the ellipse for each value of c, and we get the following picture.
2.0
-2.0
-3.0 3.0
c=1
c=2
c=6
Level curves for f .x; y/ D x
2
C 2y
2
.
Since the values of c used to derive the level curves are equally spaced, the fact that the curves are
getting closer together gives a sense of how steeply the graph of f .x; y/ D x
2
C 2y
2
is sloped.
˙
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