DEFINITION Equality of Vectors

Two vectors a and b are equal, written a = b, if they have the same length and the same direction [as explained in Fig. 166; in particular, note (B)]. Hence a vector can be arbitrarily translated; that is, its initial point can be chosen arbitrarily.

EXAMPLE 1 Components and Length of a Vector

The vector a with initial point P: (4, 0, 2) and terminal point Q: (6, −1, 2) has the components

Hence a = [2, −1, 0]. (Can you sketch a, as in Fig. 168?) Equation (2) gives the length

If we choose (−1, 5, 8) as the initial point of a, the corresponding terminal point is (1, 4, 8).

If we choose the origin (0, 0, 0) as the initial point of a, the corresponding terminal point is (2, −1, 0); its coordinates equal the components of a. This suggests that we can determine each point in space by a vector, called the position vector of the point, as follows.

THEOREM 1 Vectors as Ordered Triples of Real Numbers

A fixed Cartesian coordinate system being given, each vector is uniquely determined by its ordered triple of corresponding components. Conversely, to each ordered triple of real numbers (a₁, a₂, a₃) there corresponds precisely one vector a = [a₁, a₂, a₃], with (0, 0, 0) corresponding to the zero vector 0, which has length 0 and no direction.

Hence a vector equation a = b is equivalent to the three equations a₁ = b₁, a₂ = b₂, a₃ = b₃ for the components.

DEFINITION Addition of Vectors

The sum a + b of two vectors a = [a₁, a₂, a₃] and b = [b₁, b₂, b₃] is obtained by adding the corresponding components,

Geometrically, place the vectors as in Fig. 170 (the initial point of b at the terminal point of a); then a + b is the vector drawn from the initial point of a to the terminal point of b.

DEFINITION Scalar Multiplication (Multiplication by a Number)

The product ca of any vector a = [a₁, a₂, a₃] and any scalar c (real number c) is the vector obtained by multiplying each component of a by c,

Geometrically, if a ≠ 0, then ca with c > 0 has the direction of a and with c < 0 the direction opposite to a. In any case, the length of ca is |ca| = |c||a|, and ca = 0 if a = 0 or c = 0 (or both). (See Fig. 175.)

EXAMPLE 2 Vector Addition. Multiplication by Scalars

With respect to a given coordinate system, let

Then −a = [−4, 0, −1], 7a = [28, 0, 7], and

EXAMPLE 3 ijk Notation for Vectors

In Example 2 we have , and so on.

PROBLEM SET 9.1

1–5 COMPONENTS AND LENGTH

Find the components of the vector v with initial point P and terminal point Q. Find |v|. Sketch |v|. Find the unit vector u in the direction of v.

1. P: (1, 1, 0), Q: (6, 2, 0)
2. P: (1, 1, 1), Q: (2, 2, 0)
3. P: (−3.0, 4, 0, −0.5), Q: (5.5, 0, 1.2)
4. P: (1, 4, 2), Q: (−1, −4, −2)
5. P: (0, 0, 0), Q: (2, 1, −2)

6–10 Find the terminal point Q of the vector v with components as given and initial point P. Find |v|.

• 6. 4, 0, 0; P (0, 2, 13)
• 7.
• 8. 13.1, 0.8, −2.0; P: (0, 0, 0)
• 9. 6, 1, −4; P: (−6, −1, −4)
• 10. 0, −3, 3; P: (0, 3, −3)

11–18 ADDITION, SCALAR MULTIPLICATION

Let a = [3, 2, 0] = 3i + 2j; b = [−4, 6, 0] = 4i + 6j, c = [5, −1, 8] = 5i − j + 8k, d = [0, 0, 4] = 4k.

Find:

• 11.
• 12. (a + b) + c, a + (b + c)
• 13. b + c, c + b
• 14. 3c − 6d, 3(c − 2d)
• 15. 7(c − b), 7c − 7b
• 16.
• 17. (7 − 3)a, 7a − 3a
• 18. 4a + 3b, −4a − 3b
• 19. What laws do Probs. 12–16 illustrate?
• 20. Prove Eqs. (4) and (6).

21–25 FORCES, RESULTANT

Find the resultant in terms of components and its magnitude.

• 21. p = [2, 3, 0], q = [0, 6, 1], u = [2, 0, −4]
• 22. p = [1, −2, 3], q = [3, 21, −16], u = [−4, −19, 13]
• 23.
• 24. p = [−1, 2, −3], q = [1, 1, 1], u = [1, −2, 2]
• 25. u = [3, 1, −6], v = [0, 2, 5], w = [3, −1, −13]

26–37 FORCES, VELOCITIES

• 26. Equilibrium. Find v such that p, q, u in Prob. 21 and v are in equilibrium.
• 27. Find p such that u, v, w in Prob. 23 and p are in equilibrium.
• 28. Unit vector. Find the unit vector in the direction of the resultant in Prob. 24.
• 29. Restricted resultant. Find all v such that the resultant of v, p, q, u with p, q, u as in Prob. 21 is parallel to the xy-plane.
• 30. Find v such that the resultant of p, q, u, v with p, q, u as in Prob. 24 has no components in x- and y-directions.
• 31. For what k is the resultant of [2, 0, −7], [1, 2, −3], and [0, 3, k] parallel to the xy-plane?
• 32. If |p| = 6 and |q| = 4, what can you say about the magnitude and direction of the resultant? Can you think of an application to robotics?
• 33. Same question as in Prob. 32 if |p| = 9, |q| = 6, |u| = 3.
• 34. Relative velocity. If airplanes A and B are moving southwest with speed |v_A| = 550 mph, and northwest with speed |v_B| = 450 mph, respectively, what is the relative velocity v = v_B − v_A of B with respect to A?
• 35. Same question as in Prob. 34 for two ships moving northeast with speed |v_A| = 22 knots and west with speed |v_B| = 19 knots.
• 36. Reflection. If a ray of light is reflected once in each of two mutually perpendicular mirrors, what can you say about the reflected ray?
• 37. Force polygon. Truss. Find the forces in the system of two rods (truss) in the figure, where |p| = 1000 nt. Hint. Forces in equilibrium form a polygon, the force polygon.

  

  Problem 37

• 38. TEAM PROJECT. Geometric Applications. To increase your skill in dealing with vectors, use vectors to prove the following (see the figures).

  (a) The diagonals of a parallelogram bisect each other.

  (b) The line through the midpoints of adjacent sides of a parallelogram bisects one of the diagonals in the ratio 1 : 3.

  (c) Obtain (b) from (a).

  (d) The three medians of a triangle (the segments from a vertex to the midpoint of the opposite side) meet at a single point, which divides the medians in the ratio 2 : 1.

  (e) The quadrilateral whose vertices are the midpoints of the sides of an arbitrary quadrilateral is a parallelogram.

  (f) The four space diagonals of a parallelepiped meet and bisect each other.

  (g) The sum of the vectors drawn from the center of a regular polygon to its vertices is the zero vector.

  

  Team Project 38(a)

  

  Team Project 38(d)

  

  Team Project 38(e)

DEFINITION Inner Product (Dot Product) of Vectors

The inner product or dot product a • b (read “a dot b”) of two vectors a and b is the product of their lengths times the cosine of their angle (see Fig. 178),

The angle γ, 0 γ π, between a and b is measured when the initial points of the vectors coincide, as in Fig. 178. In components, a = [a₁, a₂, a₃], b = [b₁, b₂, b₃], and

THEOREM 1 Orthogonality Criterion

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

EXAMPLE 1 Inner Product. Angle Between Vectors

Find the inner product and the lengths of a = [1, 2, 0] and b = [3, −2, 1] as well as the angle between these vectors.

Solution. , and (4) gives the angle

EXAMPLE 2 Work Done by a Force Expressed as an Inner Product

This is a major application. It concerns a body on which a constant force p acts. (For a variable force, see Sec. 10.1.) Let the body be given a displacement d. Then the work done by p in the displacement is defined as

that is, magnitude |p| of the force times length |d| of the displacement times the cosine of the angle α between p and d (Fig. 179). If α < 90°, as in Fig. 179, then W > 0. If p and d are orthogonal, then the work is zero (why?). If α > 90°, then W < 0, which means that in the displacement one has to do work against the force. For example, think of swimming across a river at some angle α against the current.

EXAMPLE 3 Component of a Force in a Given Direction

What force in the rope in Fig. 180 will hold a car of 5000 lb in equilibrium if the ramp makes an angle of 25° with the horizontal?

Solution. Introducing coordinates as shown, the weight is a = [0, −5000] because this force points downward, in the negative y-direction. We have to represent a as a sum (resultant) of two forces, a = c + p, where c is the force the car exerts on the ramp, which is of no interest to us, and p is parallel to the rope. A vector in the direction of the rope is (see Fig. 180)

The direction of the unit vector u is opposite to the direction of the rope so that

Since |u| = 1 and cos γ > 0, we see that we can write our result as

We can also note that γ = 90° − 25° = 65° is the angle between a and p so that

Answer: About 2100 lb.

EXAMPLE 4 Orthonormal Basis

By definition, an orthonormal basis for 3-space is a basis {a, b, c} consisting of orthogonal unit vectors. It has the great advantage that the determination of the coefficients in representations v = l₁a + l₂b + l₃c of a given vector v is very simple. We claim that l₁ = a • v, l₂ = b • v, l₃ = c • v. Indeed, this follows simply by taking the inner products of the representation with a, b, c, respectively, and using the orthonormality of the basis, a • v = l₁a • a + l₂a • b + l₃a • c = l₁, etc.

For example, the unit vectors i, j, k in (8), Sec. 9.1, associated with a Cartesian coordinate system form an orthonormal basis, called the standard basis with respect to the given coordinate system.

EXAMPLE 5 Orthogonal Straight Lines in the Plane

Find the straight line L₁ through the point P: (1, 3) in the xy-plane and perpendicular to the straight line L₂: x − 2y + 2 = 0; see Fig. 183.

Solution. The idea is to write a general straight line L₁: a₁x + a₂y = c as a • r = c with a = [a₁, a₂] ≠ 0 and r = [x, y], according to (2). Now the line through the origin and parallel to L₁ is a • r = 0. Hence, by Theorem 1, the vector a is perpendicular to r. Hence it is perpendicular to and also to L₁ because L₁ and are parallel. a is called a normal vector of L₁ (and of ).

Now a normal vector of the given line x − 2y + 2 = 0 is b = [1, −2]. Thus L₁ is perpendicular to L₂ if b • a = a₁ − 2a₂ = 0, for instance, if a = [2, 1]. Hence L₁ is given by 2x + y = c. It passes through P: (1, 3) when 2 · 1 + 3 = c = 5. Answer: y = −2x + 5. Show that the point of intersection is (x, y) = (1.6, 1.8).

EXAMPLE 6 Normal Vector to a Plane

Find a unit vector perpendicular to the plane 4x + 2y + 4z = −7.

Solution. Using (2), we may write any plane in space as

where a = [a₁, a₂, a₃] ≠ 0 and r = [x, y, z]. The unit vector in the direction of a is (Fig. 184)

Dividing by |a|, we obtain from (13)

From (12) we see that p is the projection of r in the direction of n. This projection has the same constant value c/|a| for the position vector r of any point in the plane. Clearly this holds if and only if n is perpendicular to the plane. n is called a unit normal vector of the plane (the other being −n).

Furthermore, from this and the definition of projection, it follows that |p| is the distance of the plane from the origin. Representation (14) is called Hesse's² normal form of a plane. In our case, a = [4, 2, 4], c = −7, , and the plane has the distance from the origin.

PROBLEM SET 9.2

1–10 INNER PRODUCT

Let a = [1, −3, 5], b = [4, 0, 8], c = [−2, 9, 1]. Find:

1. a • b, b • a, b • c
2. (−3a + 5c) • b, 15(a − c) • b
3. |a|, |2b|, |−c|
4. |a + b|, |a| + |b|
5. |b + c|, |b| + |c|
6. |a + c|² + |a − c|² − 2(|a|² + |c|²)
7. |a • c|, |a||c|
8. 5a • 13b, 65a • b
9. 15a • b + 15a • c, 15a •(b + c)
10. a •(b − c), (a − b) • c

11–16 GENERAL PROBLEMS

• 11. What laws do Probs. 1 and 4–7 illustrate?
• 12. What does u • v = u • w imply if u = 0? If u ≠ 0?
• 13. Prove the Cauchy–Schwarz inequality.
• 14. Verify the Cauchy–Schwarz and triangle inequalities for the above a and b.
• 15. Prove the parallelogram equality. Explain its name.
• 16. Triangle inequality. Prove Eq. (7). Hint. Use Eq. (3) for |a + b| and Eq. (6) to prove the square of Eq. (7), then take roots.

17–20 WORK

Find the work done by a force p acting on a body if the body is displaced along the straight segment from A to B. Sketch and p. Show the details.

• 17. p = [2, 5, 0], A: (1, 3, 3), B: (3, 5, 5)
• 18. p = [−1, −2, 4], A = (0, 0, 0), B: (6, 7, 5)
• 19. p = [0, 4, 3], A = (4, 5, −1), B: (1, 3, 0)
• 20. p = [6, −3, −3], A: (1, 5, 2), B: (3, 4, 1)
• 21. Resultant. Is the work done by the resultant of two forces in a displacement the sum of the work done by each of the forces separately? Give proof or counterexample.

22–30 ANGLE BETWEEN VECTORS

Let a = [1, 1, 0], b = [3, 2, 1], and c = [1, 0, 2]. Find the angle between:

• 22. a, b
• 23. b, c
• 24. a + c, b + c
• 25. What will happen to the angle in Prob. 24 if we replace c by nc with larger and larger n?
• 26. Cosine law. Deduce the law of cosines by using vectors a, b, and a − b.
• 27. Addition law. cos (α − β) = cos α cos β + sin α sin β. Obtain this by using a = [cos α, sin α], b = [cos β, sin β] where 0 α β 2π.
• 28. Triangle. Find the angles of the triangle with vertices A: (0, 0, 2), B: (3, 0, 2), and C: (1, 1, 1). Sketch the triangle.
• 29. Parallelogram. Find the angles if the vertices are (0, 0), (6, 0), (8, 3), and (2, 3).
• 30. Distance. Find the distance of the point A: (1, 0, 2) from the plane P: 3x + y + z = 9. Make a sketch.

31–35 ORTHOGONALITY is particularly important, mainly because of orthogonal coordinates, such as Cartesian coordinates, whose natural basis [Eq. (9), Sec. 9.1], consists of three orthogonal unit vectors.

• 31. For what values of a₁ are [a₁, 4, 3] and [3, −2, 12] orthogonal?
• 32. Planes. For what c are 3x + z = 5 and 8x − y + cz = 9 orthogonal?
• 33. Unit vectors. Find all unit vectors a = [a₁, a₂] in the plane orthogonal to [4, 3].
• 34. Corner reflector. Find the angle between a light ray and its reflection in three orthogonal plane mirrors, known as corner reflector.
• 35. Parallelogram. When will the diagonals be orthogonal? Give a proof.

36–40 COMPONENT IN THE DIRECTION OF A VECTOR

Find the component of a in the direction of b. Make a sketch.

• 36. a = [1, 1, 1], b = [2, 1, 3]
• 37. a = [3, 4, 0], b = [4, −3, 2]
• 38. a = [8, 2, 0], b = [−4, −1, 0]
• 39. When will the component (the projection) of a in the direction of b be equal to the component (the projection) of b in the direction of a? First guess.
• 40. What happens to the component of a in the direction of b if you change the length of b?

DEFINITION Vector Product (Cross Product, Outer Product) of Vectors

The vector product or cross product a × b (read “a cross b”) of two vectors a and b is the vector v denoted by

I. If a = 0 or b = 0, then we define v = a × b = 0.

II. If both vectors are nonzero vectors, then vector v has the length

where γ is the angle between a and b as in Sec. 9.2.

Furthermore, by design, a and b form the sides of a parallelogram on a plane in space. The parallelogram is shaded in blue in Fig. 185. The area of this blue parallelogram is precisely given by Eq. (1), so that the length |v| of the vector v is equal to the area of that parallelogram.

III. If a and b lie in the same straight line, i.e., a and b have the same or opposite directions, then γ is 0° or 180° so that sin γ = 0. In that case |v| = 0 so that v = a × b = 0.

IV. If cases I and III do not occur, then v is a nonzero vector. The direction of v = a × b is perpendicular to both a and b such that a, b, v—precisely in this order (!)—form a right-handed triple as shown in Figs. 185–187 and explained below.

Another term for vector product is outer product.

EXAMPLE 1 Vector Product

For the vector product v = a × b of a = [1, 1, 0] and b = [3, 0, 0] in right-handed coordinates we obtain from (2)

We confirm this by (2**):

To check the result in this simple case, sketch a, b, and v. Can you see that two vectors in the xy-plane must always have their vector product parallel to the z-axis (or equal to the zero vector)?

EXAMPLE 2 Vector Products of the Standard Basis Vectors

We shall use this in the next proof.

THEOREM 1 General Properties of Vector Products

(a) For every scalar l,

(b) Cross multiplication is distributive with respect to vector addition; that is,

(c) Cross multiplication is not commutative but anticommutative; that is,

(d) Cross multiplication is not associative; that is, in general,

so that the parentheses cannot be omitted.

PROOF

Equation (4) follows directly from the definition. In (5α), formula (2*) gives for the first component on the left

By (2*) the sum of the two determinants is the first component of (a × b) + (a × c), the right side of (5α). For the other components in (5α) and in 5(β), equality follows by the same idea.

Anticommutativity (6) follows from (2**) by noting that the interchange of Rows 2 and 3 multiplies the determinant by −1. We can confirm this geometrically if we set a × b = v and b × a = w; then |v| = |w| by (1), and for b, a, w to form a right-handed triple, we must have w = −v.

Finally, i ×(i × j) = i × k = −j, whereas (i × i) × j = 0 × j = 0 (see Example 2). This proves (7).

EXAMPLE 3 Moment of a Force

In mechanics the moment m of a force p about a point Q is defined as the product m = |p|d, where d is the (perpendicular) distance between Q and the line of action L of p (Fig. 190). If r is the vector from Q to any point A on L, then d = |r| sin γ, as shown in Fig. 190, and

Since γ is the angle between r and p, we see from (1) that m = |r × p|. The vector

is called the moment vector or vector moment of p about Q. Its magnitude is m. If m ≠ 0, its direction is that of the axis of the rotation about Q that p has the tendency to produce. This axis is perpendicular to both r and p.

EXAMPLE 4 Moment of a Force

Find the moment of the force p about the center Q of a wheel, as given in Fig. 191.

Solution. Introducing coordinates as shown in Fig. 191, we have

(Note that the center of the wheel is at y = −1.5 on the y-axis.) Hence (8) and (2**) give

This moment vector m is normal, i.e., perpendicular to the plane of the wheel. Hence it has the direction of the axis of rotation about the center Q of the wheel that the force p has the tendency to produce. The moment m points in the negative z-direction, This is, the direction in which a right-handed screw would advance if turned in that way.

EXAMPLE 5 Velocity of a Rotating Body

A rotation of a rigid body B in space can be simply and uniquely described by a vector w as follows. The direction of w is that of the axis of rotation and such that the rotation appears clockwise if one looks from the initial point of w to its terminal point. The length of w is equal to the angular speed ω (> 0) 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 ωd. Let r be the position vector of P referred to a coordinate system with origin 0 on the axis of rotation. Then d = |r| sin γ, where γ is the angle between w and r. Therefore,

From this and the definition of vector product we see that the velocity vector v of P can be represented in the form (Fig. 192)

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

THEOREM 2 Properties and Applications of Scalar Triple Products

(a) In (10) the dot and cross can be interchanged:

(b) Geometric interpretation. The absolute value |(a b c)| of (10) is the volume of the parallelepiped (oblique box) with a, b, c as edge vectors (Fig. 193).

(c) Linear independence. Three vectors in R³ are linearly independent if and only if their scalar triple product is not zero.

PROOF

(a) Dot multiplication is commutative, so that by (10)

EXAMPLE 6 Tetrahedron

A tetrahedron is determined by three edge vectors a, b, c, as indicated in Fig. 194. Find the volume of the tetrahedron in Fig. 194, when a = [2, 0, 3], b = [0, 4, 1], c = [5, 6, 0].

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

Fig. 194. Tetrahedron

Hence V = 72. The minus sign indicates that if the coordinates are right-handed, the triple a, b, c is left-handed. The volume of a tetrahedron is of that of the parallelepiped (can you prove it?), hence 12.

Can you sketch the tetrahedron, choosing the origin as the common initial point of the vectors? What are the coordinates of the four vertices?

PROBLEM SET 9.3

1–10 GENERAL PROBLEMS

1. Give the details of the proofs of Eqs. (4) and (5).
2. What does a × b = a × c with a ≠ 0 imply?
3. Give the details of the proofs of Eqs. (6) and (11).
4. Lagrange's identity for |a × b|. Verify it for a = [3, 4, 2] and b = [1, 0, 2]. Prove it, using sin² γ = 1 − cos² γ. The identity is

   

5. What happens in Example 3 of the text if you replace p by −p?
6. What happens in Example 5 if you choose a P at distance 2d from the axis of rotation?
7. Rotation. A wheel is rotating about the y-axis with angular speed ω = 20 sec⁻¹. The rotation appears clockwise if one looks from the origin in the positive y-direction. Find the velocity and speed at the point [8, 6, 0]. Make a sketch.
8. Rotation. What are the velocity and speed in Prob. 7 at the point (4, 2, −2) if the wheel rotates about the line y = x, z = 0 with ω = 10 sec⁻¹?
9. Scalar triple product. What does (a b c) = 0 imply with respect to these vectors?
10. WRITING REPORT. Summarize the most important applications discussed in this section. Give examples. No proofs.

11–23 VECTOR AND SCALAR TRIPLE PRODUCTS

With respect to right-handed Cartesian coordinates, let, a = [2, 1, 0], b = [−3, 2, 0], c = [1, 4, −2], and d = [5, −1, 3]. Showing details, find:

• 11. a × b, b × a, a • b
• 12. 3c × 5d, 15d × c, 15d • c, 15c • d
• 13. c × (a + b), a × c + b × c
• 14. 4b × 3c + 12c + × b
• 15. (a + d) × (d + a)
• 16. (b × c) • d, b • (c × d)
• 17. (b × c) × d, b × (c × d)
• 18. (a × b) × a, a × (b × a)
• 19. (i j k), (i k j)
• 20. (a × b) × (c × d), (a b d)c − (a b c)d
• 21. 4b × 3c, 12|b × c|, 12|c × b|
• 22. (a − b c − b d − b), (a c d)
• 23. b × b, (b − c) × (c − b), b • b
• 24. TEAM PROJECT. Useful Formulas for Three and Four Vectors. Prove (13)–(16), which are often useful in practical work, and illustrate each formula with two examples. Hint. For (13) choose Cartesian coordinates such that d = [d₁, 0, 0] and c = [c₁, c₂, 0]. Show that each side of (13) then equals [−b₂c₂d₁, b₁c₂d₁, 0], and give reasons why the two sides are then equal in any Cartesian coordinate system. For (14) and (15) use (13).
• (13) b × (c × d) = (b • d)c − (b • c)d
• (14) (a × b) × (c × d) = (a b d)c − (a b c)d
• (15) (a × b) • (c × d) = (a • c)(b • d) − (a • d)(b • c)
• (16)

25–35 APPLICATIONS

• 25. Moment m of a force p. Find the moment vector m and m of p = [2, 3, 0] about Q: (2, 1, 0) acting on a line through A: (0, 3, 0). Make a sketch.
• 26. Moment. Solve Prob. 25 if p = [1, 0, 3], Q: (2, 0, 3), and A: (4, 3, 5).
• 27. Parallelogram. Find the area if the vertices are (4, 2, 0), (10, 4, 0), (5, 4, 0), and (11, 6, 0). Make a sketch.
• 28. A remarkable parallelogram. Find the area of the quadrangle Q whose vertices are the midpoints of the sides of the quadrangle P with vertices A: (2, 1, 0), B: (5, −1. 0), C: (8, 2, 0), and D: (4, 3, 0). Verify that Q is a parallelogram.
• 29. Triangle. Find the area if the vertices are (0, 0, 1), (2, 0, 5), and (2, 3, 4).
• 30. Plane. Find the plane through the points , B: (4, 2, −2), and C: (0, 8, 4).
• 31. Plane. Find the plane through (1, 3, 4), (1, −2, 6), and (4, 0, 7).
• 32. Parallelepiped. Find the volume if the edge vectors are i + j, −2i + 2k, and −2i − 3k. Make a sketch.
• 33. Tetrahedron. Find the volume if the vertices are (1, 1, 1), (5, −7, 3), (7, 4, 8), and (10, 7, 4).
• 34. Tetrahedron. Find the volume if the vertices are (1, 3, 6), (3, 7, 12), (8, 8, 9), and (2, 2, 8).
• 35. WRITING PROJECT. Applications of Cross Products. Summarize the most important applications we have discussed in this section and give a few simple examples. No proofs.

EXAMPLE 1 Scalar Function (Euclidean Distance in Space)

The distance f(P) of any point P from a fixed point P₀ in space is a scalar function whose domain of definition is the whole space. f(P) defines a scalar field in space. If we introduce a Cartesian coordinate system and P₀ has the coordinates x_0, y₀, z₀, then f is given by the well-known formula

where x, y, z are the coordinates of P. If we replace the given Cartesian coordinate system with another such system by translating and rotating the given system, then the values of the coordinates of P and P₀ will in general change, but f(P) will have the same value as before. Hence f(P) is a scalar function. The direction cosines of the straight line through P and P₀ are not scalars because their values depend on the choice of the coordinate system.

EXAMPLE 2 Vector Field (Velocity Field)

At any instant the velocity vectors v(P) of a rotating body B constitute a vector field, called the velocity field of the rotation. If we introduce a Cartesian coordinate system having the origin on the axis of rotation, then (see Example 5 in Sec. 9.3)

where x, y, z are the coordinates of any point P of B at the instant under consideration. If the coordinates are such that the z-axis is the axis of rotation and w points in the positive z-direction, then w = ωk and

An example of a rotating body and the corresponding velocity field are shown in Fig. 197.

EXAMPLE 3 Vector Field (Field of Force, Gravitational Field)

Let a particle A of mass M be fixed at a point P₀ and let a particle B of mass m be free to take up various positions P in space. Then A attracts B. According to Newton's law of gravitation the corresponding gravitational force p is directed from P to P₀, and its magnitude is proportional to 1/r², where r is the distance between P and P₀, say,

Here G = 6.67 · 10⁻⁸ cm³/(g · sec²) is the gravitational constant. Hence p defines a vector field in space. If we introduce Cartesian coordinates such that P₀ has the coordinates x₀, y₀, z₀ and P has the coordinates x, y, z, then by the Pythagorean theorem,

Assuming that r > 0 and introducing the vector

we have |r| = r, and (−1/r)r is a unit vector in the direction of p; the minus sign indicates that p is directed from P to P₀ (Fig. 198). From this and (2) we obtain

This vector function describes the gravitational force acting on B.

DEFINITION Derivative of a Vector Function

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

This vector v′(t) is called the derivative of v(t). See Fig. 199.

EXAMPLE 4 Derivative of a Vector Function of Constant Length

Let v(t) be a vector function whose length is constant, say, |v(t)| = c. Then |v|² = v • v = c², and (v • v)′ = 2v • v′ = 0, by differentiation [see (11)]. This yields the following result. The derivative of a vector function v(t) of constant length is either the zero vector or is perpendicular to v(t).

EXAMPLE 5 Partial Derivatives

Let r(t₁, t₂) = a cos t₁ i + a sin t₁ j + t₂ k. Then and

PROBLEM SET 9.4

1–8 SCALAR FIELDS IN THE PLANE

Let the temperature T in a body be independent of z so that it is given by a scalar function T = T(x, t). Identify the isotherms T(x, y) = const. Sketch some of them.

1. T = x² − y²
2. T = xy
3. T = 3x − 4y
4. T = arctan (y/x)
5. T = y/(x² + y²)
6. T = x/(x² + y²)
7. T = 9x² + 4y²
8. CAS PROJECT. Scalar Fields in the Plane. Sketch or graph isotherms of the following fields and describe what they look like.
   1. x² − 4x − y²
   2. x²y − y³/3
   3. cos x sinh y
   4. sin x sinh y
   5. e^x sin y
   6. e^2x cos 2y
   7. x⁴ − 6x²y² + y⁴
   8. x² − 2x − y²

9–14 SCALAR FIELDS IN SPACE

What kind of surfaces are the level surfaces f(x, y, z) = const?

• 9. f = 4x − 3y + 2z
• 10. f = 9(x² + y²) + z²
• 11. f = 5x² + 2y²
• 12.
• 13. f = z − (x² + y²)
• 14. f = x − y²

15–20 VECTOR FIELDS

Sketch figures similar to Fig. 198. Try to interpet the field of v as a velocity field.

• 15. v = i + j
• 16. v = −yi + xj
• 17. v = xj
• 18. v = xi + yj
• 19. v = xi − yj
• 20. v = yi − xj
• 21. CAS PROJECT. Vector Fields. Plot by arrows:
  1. v = [x, x²]
  2. v = [1/y, 1/x]
  3. v = [cos x, sin x]
  4. 

22–25 DIFFERENTIATION

• 22. Find the first and second derivatives of r = [3 cos 2t, 3 sin 2t, 4t].
• 23. Prove (11)–(13). Give two typical examples for each formula.
• 24. Find the first partial derivatives of v₁ = [e^x cos y, e^x sin y] and v₂ = [cos x cosh y, −sin x sinh y].
• 25. WRITING PROJECT. Differentiation of Vector Functions. Summarize the essential ideas and facts and give examples of your own.

EXAMPLE 1 Circle. Parametric Representation. Positive Sense

The circle x² + y² = 4, z = 0 in the xy-plane with center 0 and radius 2 can be represented parametrically by

where 0 t 2π. Indeed, x² + y² = (2 cos t)² + (2 sin t)² = 4(cos² t + sin² t) = 4, For t = 0 we have r(0) = [2, 0], for we get and so on. The positive sense induced by this representation is the counterclockwise sense.

If we replace t with t* = −t, we have t = −t* and get

This has reversed the orientation, and the circle is now oriented clockwise.

EXAMPLE 2 Ellipse

The vector function

represents an ellipse in the xy-plane with center at the origin and principal axes in the direction of the x- and y- axes. In fact, since cos² t + sin² t = 1, we obtain from (3)

If b = a, then (3) represents a circle of radius a.

EXAMPLE 3 Straight Line

A straight line L through a point A with position vector a in the direction of a constant vector b (see Fig. 203) can be represented parametrically in the form

If b is a unit vector, its components are the direction cosines of L. In this case, |t| measures the distance of the points of L from A. For instance, the straight line in the xy-plane through A: (3, 2) having slope 1 is (sketch it)

EXAMPLE 4 Circular Helix

The twisted curve C represented by the vector function

is called a circular helix. It lies on the cylinder x² + y² = a². If c > 0, the helix is shaped like a right-handed screw (Fig. 204). If c < 0 it looks like a left-handed screw (Fig. 205). If c = 0, then (5) is a circle.

EXAMPLE 5 Tangent to an Ellipse

Find the tangent to the ellipse at P: .

Solution. Equation (3) with semi-axes a = 2 and b = 1 gives r(t) = [2 cos t, sin t]. The derivative is r′ (t) = [−2 sin t, cos t]. Now P corresponds to t = π/4 because

Hence From (9) we thus get the answer

To check the result, sketch or graph the ellipse and the tangent.

EXAMPLE 6 Circular Helix. Circle. Arc Length as Parameter

The helix r(t) = [a cos t, a sin t, ct] in (5) has the derivative r′ (t = [−a sin t, a cos t, c]. Hence r′ • r′ = a² + c², a constant, which we denote by K². Hence the integrand in (11) is constant, equal to K, and the integral is s = Kt. Thus t = s/K, so that a representation of the helix with the arc length s as parameter is

A circle is obtained if we set c = 0. Then K = a, t = s/a, and a representation with arc length s as parameter is

EXAMPLE 7 Centripetal Acceleration. Centrifugal Force

The vector function

(with fixed i and j) represents a circle C of radius R with center at the origin of the xy-plane and describes the motion of a small body B counterclockwise around the circle. Differentiation gives the velocity vector

v is tangent to C. Its magnitude, the speed, is

Hence it is constant. The speed divided by the distance R from the center is called the angular speed. It equals ω, so that it is constant, too. Differentiating the velocity vector, we obtain the acceleration vector

Fig. 210. Centripetal acceleration a

This shows that a = −ω²r (Fig. 210), so that there is an acceleration toward the center, called the centripetal acceleration of the motion. It occurs because the velocity vector is changing direction at a constant rate. Its magnitude is constant, |a| ω²|r| = ω²R. Multiplying a by the mass m of B, we get the centripetal force ma. The opposite vector −ma is called the centrifugal force. At each instant these two forces are in equilibrium.

We see that in this motion the acceleration vector is normal (perpendicular) to C; hence there is no tangential acceleration.

EXAMPLE 8 Superposition of Rotations. Coriolis Acceleration

A projectile is moving with constant speed along a meridian of the rotating earth in Fig. 211. Find its acceleration.

Fig. 211. Example 8. Superposition of two rotations

Solution. Let x, y, z be a fixed Cartesian coordinate system in space, with unit vectors i, j, k in the directions of the axes. Let the Earth, together with a unit vector b, be rotating about the z-axis with angular speed ω > 0 (see Example 7). Since b is rotating together with the Earth, it is of the form

Let the projectile be moving on the meridian whose plane is spanned by b and k (Fig. 211) with constant angular speed ω > 0. Then its position vector in terms of b and k is

We have finished setting up the model. Next, we apply vector calculus to obtain the desired acceleration of the projectile. Our result will be unexpected—and highly relevant for air and space travel. The first and second derivatives of b with respect to t are

The first and second derivatives of r(t) with respect to t are

By analogy with Example 7 and because of b″ = −ω²b in (20) we conclude that the first term in a (involving ω in b″!) is the centripetal acceleration due to the rotation of the Earth. Similarly, the third term in the last line (involving γ!) is the centripetal acceleration due to the motion of the projectile on the meridian M of the rotating Earth.

The second, unexpected term −2γR sin γt b′ in a is called the Coriolis acceleration³ (Fig. 211) and is due to the interaction of the two rotations. On the Northern Hemisphere, sin γt > 0 (for t > 0; also γ > 0 by assumption), so that a_cor has the direction of −b′, that is, opposite to the rotation of the Earth. |a_cor| is maximum at the North Pole and zero at the equator. The projectile B of mass m₀ experiences a force −m₀ a_cor opposite to m₀ a_cor, which tends to let B deviate from M to the right (and in the Southern Hemisphere, where sin γt < 0, to the left). This deviation has been observed for missiles, rockets, shells, and atmospheric airflow.

PROBLEM SET 9.5

1–10 PARAMETRIC REPRESENTATIONS

What curves are represented by the following? Sketch them.

1. [3 + 2 cos t, 2 sin t, 0]
2. [a + t, b + 3t, c − 5t]
3. [0, t, t³]
4. [−2, 2 + 5 cos t, −1 + 5 sin t]
5. [2 + 4 cos t, 1 + sin t, 0]
6. [a + 3 cos πt, b − 2 sin πt, 0]
7. [4 cos t, 4 sin t, 3t]
8. [cosh t, sinh t, 2]
9. [cos t, sin 2t, 0]
10. [t, 2, 1/t]

11–20 FIND A PARAMETRIC REPRESENTATION

• 11. Circle in the plane z = 1 with center (3, 2) and passing through the origin.
• 12. Circle in the yz-plane with center (4, 0) and passing through (0, 3). Sketch it.
• 13. Straight line through (2, 1, 3) in the direction of i + 2j.
• 14. Straight line through (1, 1, 1) and (4, 0, 2). Sketch it.
• 15. Straight line y = 4x − 1, z = 5x.
• 16. The intersection of the circular cylinder of radius 1 about the z-axis and the plane z = y.
• 17. Circle , z = y.
• 18. Helix x² + y² = 25, z = 2 arctan (y/x).
• 19. Hyperbola 4x² − 3y² = 4, z = −2.
• 20. Intersection of 2x − y + 3z = 2 and x + 2y − z = 3.
• 21. Orientation. Explain why setting t = −t* reverses the orientation of [a cos t, a sin t, 0].
• 22. CAS PROJECT. Curves. Graph the following more complicated curves:

  (a) r(t) = [2 cos t + cos 2t, 2 sin t − sin 2t] (Steiner's hypocycloid).

  (b) r(t) = [cos t + k cos 2t, sin t − k sin 2t] with k = 10, 2, 1, , 0, , −1.

  (c) r(t) = [cos t, sin 5t] (a Lissajous curve).

  (d) r(t) = [cos t, sin kt]. For what k's will it be closed?

  (e) r(t) = [R sin ωt + ωRt, R cos ωt + R] (cycloid).

• 23. CAS PROJECT. Famous Curves in Polar Form. Use your CAS to graph the following curves⁴ given in polar form ρ = ρ(θ), ρ² = x² + y², tan θ = y/x, and investigate their form depending on parameters a and b.

  

24–28 TANGENT

Given a curve C: r(t), find a tangent vector r′(t), a unit tangent vector u′(t), and the tangent of C at P. Sketch curve and tangent.

• 24. P: (2, 2, 1)
• 25. r(t) = [10 cos t, 1, 10 sin t], P: (6, 1, 8)
• 26. r(t) = [cos t, sin t, 9t], P: (1, 0, 18π)
• 27. r(t) = [t, 1/t, 0],
• 28. r(t) = [t, t², t³], P: (1, 1, 1)

29–32 LENGTH

Find the length and sketch the curve.

• 29. Catenary r(t) = [t, cosh t] from t = 0 to t = 1.
• 30. Circular helix r(t) = [4 cos t, 4 sin t, 5t] from (4, 0, 0) to (4, 0, 10π).
• 31. Circle r(t) = [a cos t, a sin t] from (a, 0) to (0, a).
• 32. Hypocycloid r(t) = [a cos³ t, a sin³ t], total length.
• 33. Plane curve. Show that Eq. (10) implies dx for the length of a plane curve C: y = f(x), z = 0, and a = x = b.
• 34. Polar coordinates = arctan (y/x) give

  

  where ρ′ = dp/dθ. Derive this. Use it to find the total length of the cardioid ρ = a(1 − cos θ). Sketch this curve. Hint. Use (10) in App. 3.1.

35–46 CURVES IN MECHANICS

Forces acting on moving objects (cars, airplanes, ships, etc.) require the engineer to know corresponding tangential and normal accelerations. In Probs. 35–38 find them, along with the velocity and speed. Sketch the path.

• 35. Parabola r(t) = [t, t², 0]. Find v and a.
• 36. Straight line r(t) = [8t, 6t, 0]. Find v and a.
• 37. Cycloid r(t) = (R sin ωt + Rt)i + (R cos ωt + R)j. This is the path of a point on the rim of a wheel of radius R that rolls without slipping along the x-axis. Find v and a at the maximum y-values of the curve.
• 38. Ellipse r = [cos t, 2 sin t, 0].

39–42 THE USE OF A CAS may greatly facilitate the investigation of more complicated paths, as they occur in gear transmissions and other constructions. To grasp the idea, using a CAS, graph the path and find velocity, speed, and tangential and normal acceleration.

• 39. r(t) = [cos t + cos 2t, sin t − sin 2t]
• 40. r(t) = [2 cos t + cos 2t, 2 sin t − sin 2t]
• 41. r(t) = [cos t, sin 2t, cos 2t]
• 42. r(t) = [ct cos t, ct sin t, ct] (c ≠ 0)
• 43. Sun and Earth. Find the acceleration of the Earth toward the sun from (19) and the fact that Earth revolves about the sun in a nearly circular orbit with an almost constant speed of 30 km/s.
• 44. Earth and moon. Find the centripetal acceleration of the moon toward Earth, assuming that the orbit of the moon is a circle of radius 239,000 miles = 3.85 · 10⁸ m, and the time for one complete revolution is 27.3 days 2.36 = 10⁶ s.
• 45. Satellite. Find the speed of an artificial Earth satellite traveling at an altitude of 80 miles above Earth's surface, where g = 31 ft/sec². (The radius of the Earth is 3960 miles.)
• 46. Satellite. A satellite moves in a circular orbit 450 miles above Earth's surface and completes 1 revolution in 100 min. Find the acceleration of gravity at the orbit from these data and from the radius of Earth (3960 miles).

47–55 CURVATURE AND TORSION

• 47. Circle. Show that a circle of radius a has curvature 1/a.
• 48. Curvature. Using (22), show that if C is represented by r(t) with arbitrary t, then

  

• 49. Plane curve. Using (22*), show that for a curve y = f(x),

  

• 50. Torsion. Using b = u × p and (23), show that (when κ > 0)

  

• 51. Torsion. Show that if C is represented by r(t) with arbitrary parameter t, then, assuming κ > 0 as before,

  

• 52. Helix. Show that the helix [a cos t, a sin t, ct] can be represented by [a cos (s/K) a sin (s/K), cs/K], where and s is the arc length. Show that it has constant curvature κ = a/K² and torsion τ = c/K².
• 53. Find the torsion of C: r(t) = [t, t², t³], which looks similar to the curve in Fig. 212.
• 54. Frenet⁵ formulas. Show that

  

• 55. Obtain κ and τ in Prob. 52 from (22*) and (23***) and the original representation in Prob. 54 with parameter t.

THEOREM 1 Chain Rule

Let w = f(x, y, z) be continuous and have continuous first partial derivatives in a domain D in xyz-space. Let x = x(u, v), y = y(u, v), z = z(u, v) be functions that are continuous and have first partial derivatives in a domain B in the uv-plane, where B is such that for every point (u, v) in B, the corresponding point [x(u, υ), y(u, υ), z(u, υ)] lies in D. See Fig. 213. Then the function

is defined in B, has first partial derivatives with respect to u and v in B, and

EXAMPLE 1 Chain Rule

If w = x² − y² and we define polar coordinates r, θ by x = r cos θ, y = r sin θ, then (2) gives

EXAMPLE 2 Partial Derivatives on Surface

Let w = f + x³ + y³ + z³ and let z = g = x² + y². Then (6) gives

We confirm this by substitution, using w(x, y) = x³ + y³ + (x² + y²)³, that is,

THEOREM 2 Mean Value Theorem

Let f(x, y, z) be continuous and have continuous first partial derivatives in a domain D in xyz-space. Let P₀: (x₀, y₀, z₀) and P: (x₀ + h, y₀ + k, z₀ + l) be points in D such that the straight line segment P₀P joining these points lies entirely in D. Then

the partial derivatives being evaluated at a suitable point of that segment.

DEFINITION 1 Gradient

The setting is that we are given a scalar function f(x, y, z) that is defined and differentiable in a domain in 3-space with Cartesian coordinates x, y, z. We denote the gradient of that function by grad f or ∇f (read nabla f). Then the qradient of f(x, y, z) is defined as the vector function

DEFINITION 2 Directional Derivative

The directional derivative D_bf or df/ds of a function f(x, y, z) at a point P in the direction of a vector b is defined by (see Fig. 215)

Here Q is a variable point on the straight line L in the direction of b, and |s| is the distance between P and Q. Also, s > 0 if Q lies in the direction of b (as in Fig. 215), s < 0 if Q lies in the direction of −b, and s = 0 if Q = P.

EXAMPLE 1 Gradient. Directional Derivative

Find the directional derivative of f(x, y, z) = 2x² + 3y² + z² at P: (2, 1, 3) in the direction of a = [1, 0, −2].

Solution. grad f = [4x, 6y, 2z] gives at P the vector grad f(P) = [8, 6, 6,]. From this and (5*) we obtain, since ,

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

THEOREM 1 Use of Gradient: Direction of Maximum Increase

Let f(P) = f(x, y, z) be a scalar function having continuous first partial derivatives in some domain B in space. Then grad f exists in B and is a vector, that is, its length and direction are independent of the particular choice of Cartesian coordinates. If grad f(P) ≠ 0 at some point P, it has the direction of maximum increase of at P.

PROOF

From (5) and the definition of inner product [(1) in Sec. 9.2] we have

where γ is the angle between b and grad f. Now f is a scalar function. Hence its value at a point P depends on P but not on the particular choice of coordinates. The same holds for the arc length s of the line L in Fig. 215, hence also for D_bf. Now (6) shows that D_bf is maximum when cos γ = 1, γ = 0, and then D_bf = |grad f|. It follows that the length and direction of grad f are independent of the choice of coordinates. Since γ = 0 if and only if b has the direction of grad f, the latter is the direction of maximum increase of f at P, provided grad f ≠ 0 at P. Make sure that you understood the proof to get a good feel for mathematics.

THEOREM 2 Gradient as Surface Normal Vector

Let f be a differentiable scalar function in space. Let f(x, y, z) = c = const represent a surface S. Then if the gradient of at a point P of S is not the zero vector, it is a normal vector of S at P.

EXAMPLE 2 Gradient as Surface Normal Vector. Cone

Find a unit normal vector n of the cone of revolution z² = 4(x² + y²) at the point P: (1, 0, 2).

Solution. The cone is the level surface f = 0 of f(x, y, z) = 4(x² + y²) − z². Thus (Fig. 217)

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

THEOREM 3 Gravitational Field. Laplace's Equation

The force of attraction

between two particles at points P₀: (x₀, y₀, z₀) and P: (x, y, z) (as given by Newton's law of gravitation) has the potential f(x, y, z) = c/r, where r (> 0) is the distance between P₀ and P.

Thus p = grad f = grad (c/r). This potential is a solution of Laplace's equation

[∇²f (read nabla squared f) is called the Laplacian of f.]

PROOF

That distance is r = ((x − x₀)² + (y − y₀)² + (z − z₂)²)^1/2. The key observation now is that for the components of p = [p₁, p₂, p₃] we obtain by partial differentiation

and similarly

From this we see that, indeed, p is the gradient of the scalar function f = c/r. The second statement of the theorem follows by partially differentiating (10), that is,

and then adding these three expressions. Their common denominator is r⁵. Hence the three terms −1/r³ contribute −3r² to the numerator, and the three other terms give the sum

so that the numerator is 0, and we obtain (9).

PROBLEM SET 9.7

1–6 CALCULATION OF GRADIENTS

Find grad f. Graph some level curves f = const. Indicate ∇f by arrows at some points of these curves.

1. f = (x + 1)(2y − 1)
2. f = 9x² + 4y²
3. f = y/x
4. (y + 6)² + (x − 4)²
5. f = x⁴ + y⁴
6. f = (x² − y²)/(x² + y²)

7–10 USEFUL FORMULAS FOR GRADIENT AND LAPLACIAN

Prove and illustrate by an example.

• 7. ∇(fⁿ) = nfⁿ⁻¹∇f
• 8. ∇(fg) = f∇g + g∇f
• 9. ∇(f/g) = (1/g²)(g∇f − f∇g)
• 10. ∇²(fg) = g∇²f + 2∇f · ∇g + f∇²g

11–15 USE OF GRADIENTS. ELECTRIC FORCE

The force in an electrostatic field given by f(x, y, z) has the direction of the gradient. Find ∇f and its value at P.

• 11. f = xy, P: (−4, 5)
• 12. f = x/(x² + y²), P: (1, 1)
• 13. f = ln(x² + y²), P: (8, 6)
• 14. f = (x² + y² + z²)^−1/2 P: (12, 0, 16)
• 15. f = 4x² + 9y² + z², P: (5, −1, −11)
• 16. For what points P: (x, y, z) does ∇f with f = 25x² + 9y² + 16z² have the direction from P to the origin?
• 17. Same question as in Prob. 16 when f = 25x² + 4y².

18–23 VELOCITY FIELDS

Given the velocity potential f of a flow, find the velocity v = ∇f of the field and its value v(P) at P. Sketch v(P) and the curve f = const passing through P.

• 18. f = x² − 6x − y², P: (−1, 5)
• 19. f = cos x cosh y,
• 20. f = x(1 + (x² + y²)⁻¹), P: (1, 1)
• 21. f = e^x cos y,
• 22. At what points is the flow in Prob. 21 directed vertically upward?
• 23. At what points is the flow in Prob. 21 horizontal?

24–27 HEAT FLOW

Experiments show that in a temperature field, heat flows in the direction of maximum decrease of temperature T. Find this direction in general and at the given point P. Sketch that direction at P as an arrow.

• 24. T = 3x² − 2y², P: (2.5, 1.8)
• 25. T = z/(x² + y²), P: (0, 1, 2)
• 26. T = x² + y² + 4z², P: (2, −1, 2)
• 27. CAS PROJECT. Isotherms. Graph some curves of constant temperature (“isotherms”) and indicate directions of heat flow by arrows when the temperature equals (a) x³ − 3xy², (b) sin x sinh y, and (c) e^x cos y.
• 28. Steepest ascent. If z(x, y) = 3000 − x² − 9y² [meters] gives the elevation of a mountain at sea level, what is the direction of steepest ascent at P: (4, 1)?
• 29. Gradient. What does it mean if |∇f(P)| > |∇f(Q)| at two points P and Q in a scalar field?

THEOREM 1 Invariance of the Divergence

The divergence div v is a scalar function, that is, its values depend only on the points in space (and, of course, on v) but not on the choice of the coordinates in (1) , so that with respect to other Cartesian coordinates x*, y*, z* and corresponding components υ₁*, υ₂*, υ₃* of v,

EXAMPLE 1 Gravitational Force. Laplace's Equation

The gravitational force p in Theorem 3 of the last section is the gradient of the scalar function f(x, y, z) = c/r, which satisfies Laplaces equation ∇²f = 0. According to (3) this implies that div p = 0 (r > 0).

EXAMPLE 2 Flow of a Compressible Fluid. Physical Meaning of the Divergence

We consider the motion of a fluid in a region R having no sources or sinks in R, that is, no points at which fluid is produced or disappears. The concept of fluid state is meant to cover also gases and vapors. Fluids in the restricted sense, or liquids, such as water or oil, have very small compressibility, which can be neglected in many problems. In contrast, gases and vapors have high compressibility. Their density ρ (= mass per unit volume) depends on the coordinates x, y, z in space and may also depend on time t. We assume that our fluid is compressible. We consider the flow through a rectangular box B of small edges Δx, Δy, Δz parallel to the coordinate axes as shown in Fig. 218. (Here Δ is a standard notation for small quantities and, of course, has nothing to do with the notation for the Laplacian in (11) of Sec. 9.7.) The box B has the volume ΔV = Δx Δy Δz. Let v = [υ₁, υ₂, υ₃] = υ₁i + υ₂j + υ₃k be the velocity vector of the motion. We set

and assume that u and v are continuously differentiable vector functions of x, y, z, and t, that is, they have first partial derivatives which are continuous. Let us calculate the change in the mass included in B by considering the flux across the boundary, that is, the total loss of mass leaving B per unit time. Consider the flow through the left of the three faces of B that are visible in Fig. 218, whose area is Δx Δz. Since the vectors υ₁i and υ₃ k are parallel to that face, the components υ₁ and υ₃ of v contribute nothing to this flow. Hence the mass of fluid entering through that face during a short time interval Δt is given approximately by

where the subscript y indicates that this expression refers to the left face. The mass of fluid leaving the box B through the opposite face during the same time interval is approximately (u₂)_y+Δy Δx Δz Δt, where the subscript y + Δy indicates that this expression refers to the right face (which is not visible in Fig. 218). The difference

is the approximate loss of mass. Two similar expressions are obtained by considering the other two pairs of parallel faces of B. If we add these three expressions, we find that the total loss of mass in B during the time interval Δt is approximately

where

This loss of mass in B is caused by the time rate of change of the density and is thus equal to

Fig. 218. Physical interpretation of the divergence

If we equate both expressions, divide the resulting equation by ΔVΔt, and let Δx, Δy, Δz, and Δt approach zero, then we obtain

or

This important relation 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 ∂ρ/∂t = 0 and the continuity equation is

If the density ρ is constant, so that the fluid is incompressible, then equation (6) becomes

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. Clearly, the assumption that the flow has no sources or sinks in R is essential to our argument. v is also referred to as solenoidal.

From this discussion you should conclude and remember that, roughly speaking, the divergence measures outflow minus inflow.

PROBLEM SET 9.8

1–6 CALCULATION OF THE DIVERGENCE

Find div v and its value at P.

1. v = [x², 4y², 9z²],
2. v = [0, cos xyz, sin xyz],
3. v = (x² + y²)⁻¹ [x, y]
4. v [υ₁(y, z), υ₂(z, x), υ₃(x, y)], P: (3, 1, −1)]
5. v = x²y²z²[x, y, z], P: (3, −1, 4)
6. v = (x² + y² + z²)^−3/2[x, y, z]
7. For what υ₃ is v [e^x cos y, e^x sin y, v₃] solenoidal?
8. Let v = [x, y, υ₃]. Find a υ₃ such that (a) div v > 0 everywhere, (b) div v > 0 if |z| < 1 and div v < 0 if |z > 1.
9. PROJECT. Useful Formulas for the Divergence. Prove
   1. div (kv) = k div v (k constant)
   2. div (fv) = f div v + v • ∇f
   3. div (f∇g) = f∇²g + ∇f • ∇g
   4. div (f∇g) − div (g∇f) = f∇²g − g∇²f

   Verify (b) for f = e^xyz and v = axi + byj + czk. Obtain the answer to Prob. 6 from (b). Verify (c) for f = x² − y² and g = e^{x + y}. Give examples of your own for which (a)–(d) are advantageous.

10. CAS EXPERIMENT. Visualizing the Divergence. Graph the given velocity field v of a fluid flow in a square centered at the origin with sides parallel to the coordinate axes. Recall that the divergence measures outflow minus inflow. By looking at the flow near the sides of the square, can you see whether div v must be positive or negative or may perhaps be zero? Then calculate div v. First do the given flows and then do some of your own. Enjoy it.
    1. v = i
    2. v = xi
    3. v = xi − yj
    4. v = xi + yj
    5. v = −xi − yj
    6. v = (x² + y²)⁻¹ (−yi + xj
11. Incompressible flow. Show that the flow with velocity vector v = yi is incompressible. Show that the particles that at time t = 0 are in the cube whose faces are portions of the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1 occupy at t = 1 the volume 1.
12. Compressible flow. Consider the flow with velocity vector v = xi. Show that the individual particles have the position vectors r(t) = c₁e^ti + c₂j + c₃k with constant c₁, c₂, c₃. Show that the particles that at t = 0 are in the cube of Prob. 11 at t = 1 occupy the volume e.
13. Rotational flow. The velocity vector v(x, y, z) of an incompressible fluid rotating in a cylindrical vessel is of the form v = w × r, where w is the (constant) rotation vector; see Example 5 in Sec. 9.3. Show that div v = 0. Is this plausible because of our present Example 2?
14. Does div u = div v imply u = v or u = v + k (k constant)? Give reason.

15–20 LAPLACIAN

Calculate ∇²f by Eq. (3). Check by direct differentiation. Indicate when (3) is simpler. Show the details of your work.

• 15. f = cos² x + sin² y
• 16. f = e^xyz
• 17. f = ln (x² + y²)
• 18.
• 19. f = 1/(x² + y² + z²)
• 20. f = e^2x cosh 2y

EXAMPLE 1 Curl of a Vector Function

Let v = [yz, 3zx, z] = yzi + 3zxj + zk with right-handed x, y, z. Then (1) gives

EXAMPLE 2 Rotation of a Rigid Body. Relation to the Curl

We have seen in Example 5, Sec. 9.3, that a rotation of a rigid body B about a fixed axis in space can be described by a vector w of magnitude ω in the direction of the axis of rotation, where ω (> 0) is the angular speed of the rotation, and w is directed so that the rotation appears clockwise if we look in the direction of w. According to (9), Sec. 9.3, the velocity field of the rotation can be represented in the form

where r is the position vector of a moving point with respect to a Cartesian coordinate system having the origin on the axis of rotation. Let us choose right-handed Cartesian coordinates such that the axis of rotation is the z-axis. Then (see Example 2 in Sec. 9.4)

Hence

This proves the following theorem.

THEOREM 1 Rotating Body and Curl

The curl of the velocity field of a rotating rigid body has the direction of the axis of the rotation, and its magnitude equals twice the angular speed of the rotation.

THEOREM 2 Grad, Div, Curl

Gradient fields areirrotational. That is, if a continuously differentiable vector function is the gradient of a scalar function, then its curl is the zero vector,

Furthermore, the divergence of the curl of a twice continuously differentiable vector function v is zero,

PROOF

Both (2) and (3) follow directly from the definitions by straightforward calculation. In the proof of (3) the six terms cancel in pairs.

EXAMPLE 3 Rotational and Irrotational Fields

The field in Example 2 is not irrotational. A similar velocity field is obtained by stirring tea or coffee in a cup. The gravitational field in Theorem 3 of Sec. 9.7 has curl p = 0. It is an irrotational gradient field.

THEOREM 3 Invariance of the Curl

curl v is a vector. It has a length and a direction that are independent of the particular choice of a Cartesian coordinate system in space.

PROOF

The proof is quite involved and shown in App. 4.

PROBLEM SET 9.9

1. WRITING REPORT. Grad, div, curl. List the definitions and most important facts and formulas for grad, div, curl, and ∇². Use your list to write a corresponding report of 3–4 pages, with examples of your own. No proofs.
2. (a) What direction does curl v have if v is parallel to the yz-plane? (b) If, moreover, v is independent of x?
3. Prove Theorem 2. Give two examples for (2) and (3) each.

4–8 CALCULUTION OF CURL

Find curl v for v given with respect to right-handed Cartesian coordinates. Show the details of your work.

• 4. v = [2y², 5x, 0]
• 5. v = xyz[x, y, z]
• 6. v = (x² + y² + z²)^−3/2 [x, y, z]
• 7. v = [0, 0, e^−x sin y]
• 8.

9–13 FLUID FLOW

Let v be the velocity vector of a steady fluid flow. Is the flow irrotational? Incompressible? Find the streamlines (the paths of the particles). Hint. See the answers to Probs. 9 and 11 for a determination of a path.

• 9. v = [0, 3z², 0]
• 10. v = [sec x, csc x, 0]
• 11. v = [y, −2x, 0]
• 12. v = [−y, x, π]
• 13. v [x, y, −z]
• 14. PROJECT. Useful Formulas for the Curl. Assuming sufficient differentiability, show that
  1. curl (u + v) = curl u + curl v
  2. div (curl v) = 0
  3. curl (fv) = (grad f) × v + f curl v
  4. curl (grad f) = 0
  5. div (u × v) = v • curl u − u • curl v

15–20 DIV AND CURL

With respect to right-handed coordinates, let u = [y, z, x], v = [yz, zx, xy], f = xyz, and g = x + y + z. Find the given expressions. Check your result by a formula in Proj. 14 if applicable.

• 15. curl (u + v), curl v
• 16. curl (gv)
• 17. v • curl u, u • curl v, u • curl u
• 18. div (u × v)
• 19. curl (gu + v), curl (gu)
• 20. div (grad (fg))

All vectors of the form a = [a₁, a₂, a₃] = a₁i + a₂j + a₃k constitute the real vector space R³ with componentwise vector addition

and componentwise scalar multiplication (c a scalar, a real number)

For instance, the resultant of forces a and b is the sum a + b.

The inner product or dot product of two vectors is defined by

where γ is the angle between a and b. This gives for the norm or length |a| of a

as well as a formula for γ. If a • b = 0, we call a and b orthogonal. The dot product is suggested by the work W = p • d done by a force p in a displacement d.

The vector product or cross product v = a × b is a vector of length

and perpendicular to both a and b such that a, b, v form a right-handed triple. In terms of components with respect to right-handed coordinates,

The vector product is suggested, for instance, by moments of forces or by rotations. CAUTION! This multiplication is anti commutative, a × b = −b × a, and is not associative.

An (oblique) box with edges a, b, c has volume equal to the absolute value of the scalar triple product

Sections 9.4–9.9 extend differential calculus to vector functions

and to vector functions of more than one variable (see below). The derivative of v(t) is

Differentiation rules are as in calculus. They imply (Sec. 9.4)

Curves C in space represented by the position vector r(t) have r′(t) as a tangent vector (the velocity in mechanics when t is time), r′(s) (s arc length, Sec. 9.5) as the unit tangent vector, and |r″ (s)| = κ as the curvature (the acceleration in mechanics).

Vector functions v(x, y, z) = [υ₁ (x, y, z), υ₂ (x, y, z), υ₃ (x, y, z)] represent vector fields in space. Partial derivatives with respect to the Cartesian coordinates x, y, z are obtained componentwise, for instance,

The gradient of a scalar function f is

The directional derivative of f in the direction of a vector a is

The divergence of a vector function v is

The curl of v is

or minus the determinant if the coordinates are left-handed.

Some basic formulas for grad, div, curl are (Secs. 9.7–9.9)

For grad, div, curl, and ∇² in curvilinear coordinates see App. A3.4.