**Chapter 9**

**Vector differential calculus. Grad, Div, Curl**

**1 Vectors in 2-Space and 3-Space**

A scalar is a quantity that is determined by its magnitude; A vector is a quantity that is determined by both its magnitude and its direction, thus it is an arrow of a directed line segment. Sometimes, we denote vectors by lowercase boldface letters, , etc. In handwriting, you may use arrows, for instance (in place of **a**).

**Note.** A vector of length 1 is called a unit vector.

**1.1 Equality of vectors**

Two vectors and are equal, written , if they have the same length and same direction. Hence a vector can be arbitrarily translated; that is, its initial point can be chosen arbitrarily.

**1.2 Components of a vector**

We choose an Cartesian coordinate system in space. Let be a given vector with initial point and terminate point . Then

are called the components of the vector with respect to that coordinate system. and we simply write

**Note:** The zero vector **0** has length and no director.

Also if it is given , that means for the components.

**Example 1.2.** The vector with the initial point and terminal point has the components

Hence

**Exercise 1.2.** Find the component and magnitude of with the initial point (-1,3) and terminal point (-4, 2).

**1.3 Addition of Vectors**

The sum of two vectors and is obtained by adding the corresponding components.

**1.3.1 Basic Properties of Vector Addition**

**1.4 Scalar Multiplication (Multipulication by a Number)**

The product of any vector and any scalar (real number ) is .

**1.4.1 Basic Properties of Scalar Multiplication**

**Example 1.4.** Given and , find .

**Solution.**

**Exercise 1.4.** Given and , find , , .

**1.5 Unit Vectors** , ,

Besides , another popular way of writing vector is

where , , are the unit vectors in the positive directions of the axes of a Cartesian coordinate system.

**2 Inner Product (Dot Product) of Vector**

The inner product of two vectors and is the product of their lengths times the cosine of their angle.

(1)

**Note:** if or .

The angle , , between and is measured when the initial points of the vectors coincide.

For and ,

(2)

**2.1 Orthogonality**

A vector is called orthogonal to a vector if . Then is also orthogonal to , and we call and orthogonal vectors.

The inner product of two nonzero vectors is 0 if and only if these vectors are perpendicular.

**2.2 Length and Angle**

From (1) with gives . Hence

(3)

From (1) and (3) we obtain for the angle between two nonzero vectors.

(4)

**Example 2.2.** Find the inner product and the lengths of and as well as the angle between these vectors.

**Solution.**

and (4) gives the angle

**Exercise 2.2**. Let and . Find , and .

**2.3 Some properties of the vectors**

From the definition we see that the inner product has the following properties,

**Note.** Gemetrically, (g) with says that one side of a triangle must be shorter than the sum of the other two sides.

**3 Application of the inner product**

**3.1 Work done by a force expressed as an Inner product**

Let the body be given a displacement . Then the work done by a constant force in the displacement is defined as

If and are orthogonal, then the work done is zero.

**3.2 Projection (Component) of a vector in a given direction**

The projection of a vector in a given direction of a vector , is defined by

(5)

where p is the length of the projection of on the straight line l parallel to .

Multiplying (5) by , we have

(6)

If is a unit vector, then (6) simply gives

(7)

The figure below shows the projection p of in the direction of and the projection of in the direction of .

**3.3 Orthogonal Straight Lines in the Plane**

**Example 3.3.**

Find the straight line through the point P: (1, 3) in the xy-plane and perpendicular to the straight line .

**Solution.**

First we express a general straight line

as with . By (2), now the line through the origin and parallel to .

By orthogonality, the vector is perpendicular to . That means is perpendicular to and because and are parallel. is called the normal vector of .

Now a normal vector of the given line is . Thus is perpendicular to if , for instance, if .

Hence is given by . It passes through P: (1,3) when . Therefore, the required line is .

**Exercise 3.3. **Find the straight line through the point P: (-3, -6) in the xy-plane and perpendicular to the straight line .

**3.4 Normal Vector to a Plane**

**Example 3.4.** Find a unit vector perpendicular to the plane

**Solution.**

By (2), we may write any plane in space as

where .

The unit vector in the direction of is

Dividing by , we have

From (7), we see that p is the projection of in the direction of . This projection has the same constant value for the position vector of any point in the plane. This holds if and only if is perpendicular to the plane. is called a unit normal vector of the plane (the other being ).

Furthermore, from this and the definition of projection it follows that |p| is the distance of the plane from the origin. Representation is called Hesse's normal form of a plane. In our case, , , , , and the plane has the distance 7/6 from the origin.

**Exercise 3.4.** Find a unit vector perpendicular to the plane .

**4 Vector Product (Cross Product)**

The vector product (also called **cross product** or **outer produc**t) . (read ) of two vectors is the vector

as follows. If have the same or opposite direction, or if then . In any other case has the length

(8)

is the angle between . The direction of is perpendicular to both and such that , in this order, form a right-hand triple.

In components, let and .Then has the components

Since

(10)

**Example 4.1. Vector Product**

For the vector product of and we obtain,

Then,

**Note.** Vector products of the standard basis vectors

(11)

**4.1 General Properties of Vector Products**

**Exercise 4.1.** Try to prove (b), (c) and (d).

**5 Typical Applications of vector products**

**5.1 Moment of a force**

**Example 5.1.**

In mechanics the moment $m$ of a force about a point Q is defined as the product , where d is the distance between Q and the line of action L of . If is vector from Q to any point A on L, then and

Since is the angle between , we see from (8) that . The vector

is called the **moment vector** or **vector moment** of about Q. Its magnitude is m. If , its direction is that of the axis of the rotation about Q that has the tendency to produce. This axis is perpendicular to both .

**5.2 Velocity of a rotating body**

**Example 5.2.**

A rotation of a rigid body B in space can be simply and uniquely described by a vector as follows. The direction of is that of the axis of rotation and such that the rotation appears clockwise if one looks from the initial point of to its terminal point. The length of is equal to the angular speed of the rotation, that is, the linear (or tangential) speed of a point of $B$ divided by its distance from the axis of rotation.

Let $P$ be any point of B and d its distance from the axis. Then P has the speed . Let be the position vector of P referred to a coordinate system with origin 0 on the axis of rotation. Then , where is the angle between . Therefore,

From this and the definition of vector product we see that the velocity vector of P can be represented in the from

This simple formula is useful for determining at any point B.

**6 Scalar triple product**

The most important product of vectors with more than two factors is the scalar triple product or mixed triple product of three vectors , , . It is denoted by and defined by

(12)

**Note.** Because of the dot product it is a scalar.

6.1 Properties and applications of scalar triple products

**Exercise 6.1.** Try to use Figure 13 to prove (b).

**Example 6.1**. Tetrahedron

A tetrahedron is determined by three edge vectors , as indicated in Fig.14. Find its volume when .

**Solution.** The volume V of the parallelepiped with these vectors as edge vectors is the absolute value of

the scalar triple product

**Ex 6.1.1. **Try to obtain the scalar triple product

Hence V=72. The minus sign indicates that if the coordinates are rght-handed, the triple is left-handed. The volume of a tetrahedron is of that of the parallelepiped, hence 12.

**7 Scalar functions and vector functions**

**7.1 Scalar function**

Scalar function f maps a point P=(x,y,z) in space to a scalar value

and its value is dependent on P.

We said f defines a ''scalar field" in the domain space.

**Example 7.1.** The distance f(P) of any point P(x,y,z) from a fixed point in space is a scalar function whose domain of definition is the whole space.

f(P) defines a scalar field in space.

**7.2 Vector function**

Vector function maps each point P=(x,y,z) to a vector, i.e.

and its value is dependent on the point P in space.

We say that defines a ''vector field" in the domain space.

**Example 7.2.** Let a particle A of mass M be mixed at a point ans let a particle B of mass m be free to take up various positions P in space. Then A attracts B.

Consider the gravitational force between A and B,

when and .

This vector function describes the gravitational force action on B.

if we consider the gravitational potential energy,

which is a scalar function.

**8 Vector calculus**

**8.1 Derivative of a vector function**

A vector function is said to be differentiable at a point t if the following limit exists:

(13)

This vector is called the derivative of .

In components with respect to a given Cartesian coordinate system,

(14)

Hence is obtained by **differentiating each component separately**.

**Example 8.1.** Find the derivative of .

**Solution**

The familiar differentiation rules continue to hold for differentiating vector functions, for instance,

and in particular,

**Note.** Chain rule can also apply in differentiating vector functions.

Let and , , , Then

The first partial derivatives with respect to u and v are

For example, if and we define polar coordinates , by , , then the first derivatives with respect to and are

**Exercise 8.1.** Find when .

The sides of a triangle are functions of time as , . Computer the rate of change of the area of the triangle at t=1.

8.2 Partial derivatives of a vector function

Suppose that the components of a vector function

are differentiable functions of n variables . Then the partial derivative of with respect to is denoted by and is defined as the vector function

**Example 8.2.** Let . Then

**Exercise 8.2.** Find the first partial derivatives of .

The amount of water in a river

Find the velocity vector of the water.

**9 Parametric Representation of curves**

We usually represent a curve C in space in terms of its projections into the xy-plane and xz-plane, that is

Also it can be expressed in the following parametric form with parameter t:

One can also interpret (18) as a vector function

To each value there corresponds a point of C with position vector with coordinates .

**Example 9.1 Circle**

The unit circle in the xy-plane can be represented parametrically by

where . Indeed, .

**Example 9.2 Straight line**

A straight line L with direction equal to that of a vector , which passes through a point A with position vector can be parametrically represented as

For instance, the straight line in the xy-plane through A:(3,2) having slope 1 is

**Example 9.3 Circular Helix**

A Helix C can be represented by the vector function

It lies on the cylinder . If c>0, the helix is shaped like a right-handed screw. If c<0, it looks a left-handed screw. If c=0, then it becomes a circle.

**Exercise 9.1.** Consider a function , prove that represents the surface of a unit sphere .

**10 Tangent to a curve**

The tangent to a curve C at a point P of C is the limiting position of a straight line L that passes through P and a point Q of C, as Q approaches P along C.

If C is given by , and P and Q correspond to and , then a vector in the direction of L is

In the limit this vector becomes the derivative

If is a tangent vector of C at P and the corresponding unit vector is the unit tangent vector

Then the tangent to C at P is given by

(19)

where is the position vector of P and w is the parameter in (19).

**Example 10.1** Find the tangent to the ellipse at .

**Solution.** The parametric representation of the ellipse is

The derivative is And $P$ corresponds to since .

From (19), the tangent should be

**Exercise 10.1** An object is rotating around the point at a distance r and angular velocity . At time t=0, its location is (a+r,b).

**11 Length of a curve**

Let , , represent a curve C, its length is given by

If we replace the fixed b in (20) with a variable t, the integral becomes a function of t, denoted by and called the arc length function or simply the arc length of C.

(21)

**Example 11.1** Consider the helix , find the length of segment between t=0 and t=.

**Solution.** By (20),

**Exercise 11.1** A space craft file in a helix as .

**12 Gradient of a scalar field. Directional derivative**

** **

**12.1 Gradient**

The gradient of a given scalar function f(x,y,z) is denoted by grad f or and is the vector function defined by

(22)

Here x, y, z are Cartesian coordinates in a domain in 3-space in which f is defined and differentiable.

**Example **12.1 Find the gradient of .

**Solution.** .

**Exercise 12.1** What is ?

**12.2 Directional derivative**

The directional derivative or of a function f(x,y,z) at a point P in the direction of a vector is defined by

Here Q is a variable point on the straight line L in the direction of , and |s| is the distance between P and Q. Also, s>0 if Q lies in the direction of , s<0 if Q lies in the direction of , and s=0 if Q=P.

Consider is a unit vector, i.e. . Then L is given by

(24)

where the position vector of P.

By (23) and apply the chain rule, we obtain

(25)

By differentiating (24), it gives . Hence (25) is simply the inner product of and ; that is,

(26)

**Note.** If the direction is given by a vector of any length , then

**Example 12.2** Find the directional derivative of at P:(2,1,3) in the direction of .

**Solution**. gives at P the vector . Since ,

The minus sign indicates that at P the function f is decreasing in the direction of .

**Exercise 12.2** Find the directional derivative of $f=xyz$ at $P:(-1,1,3)$, $\vec{a}=[1,-2,2]$.

**12.3 Gradient as the direction of maximum change**

Since the directional derivative along direction of is given by

where is the angle between and .

Observe that, for a given f, is maximized when , i.e. when is parallel to . Thus, at any point (x,y,z), the direction of the gradient, i.e. the direction of the vector , is indeed the direction of maximum change of the function f at that point.

**12.3.1 Computing Maxima/Minima via the Gradient**

Since the gradient vector of a function points in the direction of maximum increase, hence the maximum or minimum value of an unconstrained function occurs when the gradient vector equals zero.

**Unconstrained optimization**: The maximum/minimum of a function occurs when =0.

**Example 12.3.1**: Consider the function

Its gradient equals .

Setting this vector to zero gives the equations , , and . Solving these equations together gives us that (x,y,z) = (1,0,0) is a maximum/minimum of the function. To check whether in fact it is a maximum or minimum requires more work...

**Constrained optimization**: The maximum/minimum of a function occurs when =0.

Suppose, further, one is given constraints on the variables of f, then one needs to check the boundary as well, as in the example below.

**Example 12.3.2: **Consider the function in Example 12.3.1, find its maximum/minimum in the region .

The figure above shows the region. First, we need to check whether the solution to the unconstrained problem is feasible or not. As it can be seen, (1,0,0) is not in the region A, and hence it is not feasible.

Next, we need to consider all four edge line segments E1~E4.

E1), then . Let the gradient equal zero, we get y=0. It is not in the line E1, so it is not feasible.

E2), then . Let the gradient equal zero, x=1. Since (1,2,0) is on the line, it is a possible solution.

E3), then . Let the gradient equal zero, y=0. It is not in the line E3, so it is not feasible.

E4), then . Let the gradient equal zero, x=1. Since (1,1,0) is on the line, it is a possible solution.

Finally, we need to consider all four corner points P1, P2, P3 and P4. That is, (1,2,0), (1,1,0), (0,1,0), (0,2,0). After comparing all possible solutions, we find the minimum is (1,1,0).

**Checking for maxima or minima**: Just as in the scalar case, the double derivative of the function tells us whether the function attains a maximum or a minimum at a point, similarly for functions of many variables, a similar concept called the Hessian helps.

**(Advanced) Hessian:** is the square matrix of second-order partial derivatives of a function. Consider the function . Its Hessian is

Note that, if the second derivatives of *f* are all continuous in a neighborhood D, then the Hessian of *f *is a symmetric matrix throughout D. Let the gradients equal zero, we get x=1, y=0. If the **determinant** of Hessian at (1,0) is positive, then *f* attains a **local minimum** at (1,0). If the determinant is negative, then (1,0) is a local maximum. If the determinant equals zero, then (1,0) is a saddle point.* *In this case, the determinant is 24, so it is a local minimum.* *

For higher dimensional Hessian, if the Hessian is **positive definite** at point A , then *f* attains a **local minimum** at A. If the Hessian is **negative definite** at A, then *f* attains a **local maximum** at A. If the Hessian has both positive and negative eigenvalues, then A is a saddle point* * for *f.*

**Positive definite**: An *n* × *n* real symmetric matrix *M* is *positive definite* if * ** * for all non-zero vectors *z* with real entries (), where *z*^{T}denotes the transpose of *z*.

Try to find out Example 12.3.1, is it a maximum or minimum?

* *

**12.4 Gradient as surface normal vector**

Let f be a differentiable scalar function in space and f(x,y,z)=c=const represent a surface S. Now let C be a curve on S through a point P of S. Then C has a representation . For C to lie on the surface S, the components of must satisfy f(x,y,z)=c that is,

(27)

Recall a tangent vector of C is

And the tangent vectors of all curves on S passing through P will generally form a plane, called the *tangent plane* of S at P. The normal of this plane is called t*he surface normal* to S at P. A vector in the direction of the surface normal is called a *surface normal vector* of S at P. We can obtain such a vector by differentiating (27) w.r.t t,

Hence if at P of S is not the zero vector, then is perpendicular to all the vector in the tangent plane, so that it is a normal vector of S at P.

**Example 12.4 Gradient as surface normal vector. Cone**

Find a unit normal vector of the cone of revolution at the point P:(1,0,2).

**Solution. **We have .

**Ex 12.4.1** Find and .

Thus,

**Ex 12.4.2** Find .

points downward since it has a negative $z$-component. The other unit normal vector the cone at P is .

**Exercise 12.3.** Find a normal vector of the surface , P:(4,3,8).

**13 Divergence of a vector field**

Let be a differentiable vector function, where x, y, z are Cartesian coordinates, and let be the components of . Then the function

(28)

is called the* divergence* of or the divergence of the vector field defined by .

**Example 13.1** Find the div when .

**Solution.** div .

Another common motion for the divergence is

**Note. **The divergence div is a scalar function, i.e. its values depend only on the points in space and on but not on the choice of the coordinates in (28).

The physical meaning of the Divergence is a important relation

(29)

*(: a velocity vector of a the motion of a fluid in some region.)*

which is called the condition for the conservation of mass or the continuity equation of a compressible fluid flow.

If the flow is steady, that is, independent of time, then and the continuity equation is

(30)

If the density is constant, so that the fluid is incompressibility, then equation (30) becomes

(31)

This relation is known as the condition of incompressibility. It expresses the fact that the balance of outflow and inflow for a given volume element is zero at any time. Roughly speaking, the divergence measures "outflow minus inflow" for some volume of object.

**Exercise 13.1** If the velocity vector of water in a river is , write u as a function of x,y,z.

**14 Curl of a vector field**

Let be a differentiable vector function of the Cartesian coordinates x,y,z. Then the *curl* of the vector function or of the vector field given by is

(32)

This is the formula when x,y,z are *right-handed*. If they are *left-handed*, the determinant has a minus sign in front.

The curl also has an important physical meaning and it is related to the rotation of an object or fluid.

**Note.** Curl is a vector. That is, it has a length and direction that are independent of the particular choice of a Cartesian coordinate system in space.

**Example 14.1** Let with right-handed x,y,z. Then (32) gives

**Exercise 14.1** Let rotation vector of a rotating body be (0,0,),

**15 Laplacian of a scalar field**

Let f(x,y,z) be a scalar function, the laplacian of f denoted by is defined as

(33)

**Note.** One application is the **Laplace's equation**

It is the most important partial differential equation in physics.

**Example 15.1** Find where .

**Solution. **.

**Exercise 15.1** Prove that gravitational/electric potentials, both of which have the form satisfy . *(Hint: and use chain rule.)*

**16 Some basic formulas for grad, div, curl**

Let be twice differentiable scalar functions and be twice differentiable vector functions.

**Exercise 16.1** Prove (19) - (28).

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