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.

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 written 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:

Problem 3: Find the dot product of

Problem 4: Find the cross product of

Fluid Mechanics: Worksheet on the  operatorpage 1 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:

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:

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:

This can be rewritten using the dot product as:

This is the gradient of T, This is the differential

written as T length elementdl

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

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?

Part Five: The  operator dotted with a vector: 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 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

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 mistakeis to write:

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 = x2i + 3xz2j -2xz k

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

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 function with a large curl

Problem 13: (homework): Find the curl of v = x2i + 3xz2j -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?

The sawdust will spread out

If you sprinkle some sawdust into a point of negative divergence, what will happen?

The sawdust will be drawn into a “sink”

If you drop a paddlewheel into a point of large curl, what will happen?

It will spin

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:

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) The curl of a gradientis always zero

(2) The divergence of the curl is always zero

(3) 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

(4) The divergence of the gradient is the Laplacian, and has the special notation: 2

Problem 15: Write2T out completely, assuming T is a scalar function. Indicate whether 2T is a vector or a scalar.

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 2V (2u) i + (2v) j + (2w) k

(5)The curl of the curl doesn’t really give us anything new:

 ( V) = (•V) - 2V

This is justThis is just the

(3) aboveLaplacian 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

Problem 16 (homework): Calculate the Laplacian of the following functions:

T = x2 + 2xy + 3z + 4

Adapted from “Introduction to Electrodynamics” by David J. Griffiths, Prentice Hall, 1981