ME 241/ BME 270
Worksheet on the  operator (called “del” or “nabla”)
Adapted from “Introduction to Electrodynamics” by David J. Griffiths, Prentice Hall, 1981
Part One: The definition of the  operator
 All of differential vector calculus flows from the  operator.
 You must memorize the definition of . You will then be able to reconstruct everything else.



  iˆ  ˆj  kˆ
x
y
z
Notes:
1.  is not a “vector” in the usual sense. It doesn’t really have a meaning until we provide it
with something to act on.
2.  is a “vector operator” that can act either on a scalar function, or on a vector field
3. Note that the unit vectors are to the left. This is just so that no one will think that the
partial derivatives are acting on the unit vectors
Problem 1. Rewrite the definition of  with the unit vectors on the right.
Part Two: Vector notation and multiplication
There are two ways to multiply vectors: the dot product and the cross product
Problem 2: Rewrite this vector in a clearer way: ( x 2  y 2 z 2 )(iˆ  ˆj kˆ)
Problem 3: Find the dot product of
Problem 4: Find the cross product of


A  uiˆvˆj wkˆ and V  VxiˆVy ˆj Vz kˆ


A  uiˆvˆj wkˆ and V  VxiˆVy ˆj Vz kˆ
Fluid Mechanics: Worksheet on the  operator
page 2 of 8
Part Three: The derivative in one dimension
Suppose I have a function of one variable: f(x). The derivative df/dx tells me how rapidly the
function f(x) varies when I change x by a tiny amount:
 df 
df   dx
 dx 
Note: This is exactly the same thing as saying that the slope of a curve is its “rise over run.”
Problem 5. Indicate some values of df, dx, and df/dx in the following figure:
f
x
Part Four: The  operator acts on a scalar function: Gradient
Suppose we have a function of three variables, such as the temperature in a room: T = T(x,y,z).
The derivative of a function is supposed to tell us how fast a function varies, if I move a small
distance. But for a function of three variables, things seem hopeless, because how fast the
temperature varies depends on which direction I move.
Fortunately, we can use partial
derivatives to solve this problem:
dT 
T
T
T
dx 
dy 
dz
x
y
z
This can be rewritten using the dot product as:
 T ˆ T ˆ T
dT  
i
j
y
z
 x


kˆ   dx iˆ  dy ˆj  dz kˆ

This is the gradient of T,
written as T

This is the differential
length element dl
Note: The gradient is a vector and thus it has both a magnitude and a direction
 The gradient T points in the direction of maximum increase of the function T
 The magnitude |T| gives the slope (rate of increase) along that maximal direction
Adapted from “Introduction to Electrodynamics” by David J. Griffiths, Prentice Hall, 1981
Fluid Mechanics: Worksheet on the  operator
page 3 of 8
Problem 6 (homework): The height of a certain hill (in feet) is given by:
h (x, y) = 10(2xy – 3x2 - 4y2 – 18x + 28y + 12)
where y is the distance in miles north, and x is the distance in miles, east, of the Tech building
How steep is the slope (in feet per mile) at a point 1 mile north and 1 mile east of the Tech
building? In what direction is the slope steepest, at that point?
Adapted from “Introduction to Electrodynamics” by David J. Griffiths, Prentice Hall, 1981
Fluid Mechanics: Worksheet on the  operator
page 4 of 8
Part Five: The  operator dotted with a vector field: Divergence
Remember that vectors can multiply in two different ways: by the dot product and by the cross
product. Similarly, the  operator can act on a vector field either via the dot product or via the
cross product. The divergence is what you get when the  operator is dotted with a vector
field.


Problem 7: Find   V , where V  uiˆ vˆj wkˆ
Notes:
1. The divergence yields a scalar, just like any dot product
2. You can’t take the divergence of a scalar: that is meaningless
3. The divergence is a measure of how much a vector field “spreads out” from a given
point.
4. A common mistake is to write:
V V V
 V 


This is WRONG.
x y z
Problem 8: Sketch a vector function with divergence = r, the radial distance from the origin
Problem 9: Sketch a vector function that has zero divergence
Problem 10 (homework): Find the divergence of v = x2 i + 3xz2 j -2xz k
Adapted from “Introduction to Electrodynamics” by David J. Griffiths, Prentice Hall, 1981
Fluid Mechanics: Worksheet on the  operator
page 5 of 8
Part Six: The  operator crossed with a vector field: Curl
As we saw above, the divergence resulted from the dot product of the  operator and a vector
field. The cross product of the  operator and a vector field gives the curl.
Problem 11: Find


  V , where V  uiˆ vˆj wkˆ
Notes:
1. The curl of a vector field is a vector, just like any cross product
2. You can’t take the curl of a scalar: it is meaningless
3. The curl is indeed a measure of how much a vector field “curls around” a given point.
Problem 12: Sketch a vector function with a large curl
Problem 13: (homework):
Find the curl of v = x2 i + 3xz2 j -2xz k
Part Seven: Gaining some intuition for divergence and curl
Problem 14: Suppose that you are standing on an edge of a pond.
If you sprinkle some sawdust into a point of positive divergence, what will happen?
If you sprinkle some sawdust into a point of negative divergence, what will happen?
If you drop a paddlewheel into a point of large curl, what will happen?
Adapted from “Introduction to Electrodynamics” by David J. Griffiths, Prentice Hall, 1981
Fluid Mechanics: Worksheet on the  operator
page 6 of 8
Part Eight: Product Rules
You do NOT need to memorize these. You will not use them in this course. They are here so
that you have a single, complete worksheet on the  operator that you can go back and consult in
the future:
Adapted from “Introduction to Electrodynamics” by David J. Griffiths, Prentice Hall, 1981
Fluid Mechanics: Worksheet on the  operator
page 7 of 8
Part Nine: Second Derivatives
There are only five ways we can combine the gradient, the divergence, and the curl to get second
derivatives. Here are all five of them. You do not need to memorize these, but you do need to
know how to take the Laplacian of a scalar and the Laplacian of a vector.
(1)
(2)
(3)
(4)
The curl of a gradient is always zero
The divergence of the curl is always zero
The gradient of the divergence does not have a special name or notation
 For some reason, it very rarely occurs in physical applications – we won’t use it.
 It is not the same as the divergence of the gradient
The divergence of the gradient is the Laplacian, and has the special notation: 2
Problem 15: Write out 2T explicitly, assuming T is a scalar function. Indicate whether
2T is a vector or a scalar.
Very Important Note: The Laplacian of a vector V is a vector quantity whose x-component is
the Laplacian of Vx, and so forth. This is merely a convenient extension of the meaning of 2
If V = ui + vj + wk then
(5)
2V  (2u) i + (2v) j + (2w) k
The curl of the curl doesn’t really give us anything new:
  (  V) = (•V) - 2V
This is just
(3) above
This is just the
Laplacian of vector V
but it is sometimes used as a way to rigorously define the Laplacian of a vector, instead of just
the “extension” method we saw above:
[(•V)] - [  (  V)] = 2V
Adapted from “Introduction to Electrodynamics” by David J. Griffiths, Prentice Hall, 1981
Fluid Mechanics: Worksheet on the  operator
page 8 of 8
Problem 16 (homework): Calculate the Laplacian of the following functions:
T = x2 + 2xy + 3z + 4
V  x 2 iˆ  3xz 2 ˆj  2 xz kˆ
Adapted from “Introduction to Electrodynamics” by David J. Griffiths, Prentice Hall, 1981