VECTOR CALCULUS

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VECTOR CALCULUS

B.Sc. Mathematics

(V SEMESTER)

CORE COURSE

(2011 ADMISSION ONWARDS)

UNIVERSITY OF CALICUT

SCHOOL OF DISTANCE EDUCATION

Calicut University, P.O. Malappuram, Kerala, India-673 635

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UNIVERSITY OF CALICUT

SCHOOL OF DISTANCE EDUCATION

B.S

C

.

MATHEMATICS

(2011

A

DMISSION

ONWARDS

)

V

SEMESTER CORE COURSE:

VECTOR CALCULUS

Prepared by: Sri. Nandakumar. M. Assistant Professor, NAM College, Kallikkandi, Kannur. Scrutinized by: Dr. Anil Kumar. V Head of the Dept. Dept. of Maths, University of Calicut.

Layout & Settings

:

Computer Section, SDE

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CONTENTS PAGES

MODULE - I

_{5 ‐ 70}

MODULE - II

_{71 ‐ 137}

MODULE - III

_{138 ‐ 166}

MODULE - IV

_{167 ‐ 222}

SYLLABUS

223

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MODULE - 1

ANALYTIC GEOMETRY IN SPACE

VECTORS

1. A vector is a quantity that is determined by both its magnitude and its direction; thus

it is an arrow or a directed line segment. For example force is a vector. A velocity is a

vector giving the speed and direction of motion. We denote vectors by lowercase

boldface letter a,b,v, etc.

2. A scalar is a quantity that is determined by its magnitude, its number of units

measured on a scale. For Problem, length temperature, and voltage are scalars.

3. A vector has a tail, called its initial point, and a tip, called its terminal point.

4. The length of a vector a is the distance between its initial point and terminal point.

5. The length (or magnitude) of a vector a is also called the norm (or Euclidean norm)

of a and is denoted by .

6. A vector of length 1 is called a unit vector.

7. Two vector and are equal, written, a=b, if they have the same length and the

same direction. Hence a vector can be arbitrarily translated, that is, its initial point

can be chosen arbitrarily.

COMPONENTS OF A VECTOR

We consider a

Cartesian coordinate system in space, that is, a usual rectangular

coordinate system with the same scale of measurement on the three mutually perpendicular

coordinate axes. Then if a given vector has initial point

and terminal point

, then the three numbers

………..(1)

Are called the components of the vector a with respect to that coordinated system, and we

write simply

In terms of components, length of a is given by

…….(2)

Problem

Find the components and length of the vector a with initial point

and terminal point

(6)

Solution.

The components of a are

Hence

. Using (2), the length of the vector is

POSITION VECTOR

A Cartesian coordinate system being given, the position vector r of a point

is the vector with the origin

as the initial point and as the terminal point. From (1),

the components are given by

So that

Theorem I (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

there corresponds precisely one vector

, with

corresponding to the zero vector 0, which has length 0 and no direction.

VECTOR ADDITION

Definition (Addition of Vectors)

The sum

of two vectors

and

is obtained by adding

the corresponding components.

....(3)

Basic Properties of Vector Addition

a) (commutativity) b) (associativity) c)

d)

where denotes the vector having the length and the direction opposite to that of

SCALAR MULTIPLICATION

Definition (Scalar Multiplication by a Number)

The product ca of any vector

and any scalar c (real number c) is the

vector obtained by multiplying each component of a by c. That is

,

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…...(4)

Geometrically, if

, then with

has the direction of and with

has

the direction opposite to . In any case, the length of is

and

if or (or both) Basic Properties of Scalar Multiplication

(a) (b)

(c) (written cka) …(5) (d)

Remarks (4) and (5) imply for any vector a

(a) (b)

Instead of we simply write

Problem Given the vectors and Find and

Solution:

; ; ;

Unit Vectors i,j,k

A

vector

can also be represented as

a=a

1

i+a

2

j+a

3

k …… (6)

In this representation i,j,k are the unit vectors in the positive directions of the axes of

a Cartesian coordinate system . Hence

…..(7)

Problem The vectors and can also be written as and

Inner Product

Definition (Inner Product (Dot Product) of vectors) The inner product or dot product (read “a dot b”) of two vectors a and b is the product of their lengths times the cosine of their angle.

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if …(1) if

The angle

between a and b in measured when the vectors have their initial

points coinciding. In components,

…...(2)

Definition A vector a is called orthogonal to a vector b if

Then b is also orthogonal to a and we call these vectors orthogonal vectors.

1. The zero vector is orthogonal to every vector.

2. For nonzero vectors

if and only if

thus

Theorem 1 (Orthogonality) The inner product of two nonzero vector is zero if and only if

these vectors are perpendicular.

Length and Angle in terms of inner Product

From

(1),

with we get

. Hence

…(3)

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

……(4)

Problem Find the inner product and the lengths of

and

as well as

the angle between these vectors

Solution

and (4) gives the angle radians.

Properties of Inner Products

For any vectors and scalars

(a) (Linearity)

(b) (Symmetry)

(c) if and only if (Positive definiteness)

Hence dot multiplication is commutative and is distributive with respect to vector

addition, in fact from the above with

and

we have

(Distributivity) Furthermore, from (1) and we see that

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Result: Prove the following triangle inequality:

Proof

using (3)

since

using (3) and Schwarz inequality

Taking square roots on both sides, we obtain

Result: Prove the Parallelogram equality (parallelogram identity)

Proof

, using (3)

Derivation of (2) from (1)

We can write the given vectors a and b in components as

and

Since i,j and k are unit vectors, we have from (3)

Since they are orthogonal (because the coordinate axes perpendicular) Orthogonality

Theorem gives

Hence if we substitute those representations of a and b into

and use Distributivity and

Symmetry, we first have a sum of nine inner products.

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APPLICATIONS OF INNER PRODUCTS

Work done by a force as inner product

Consider a body which a constant force p acts. Let the body be given a displacement

d. Then the work done by p in the displacement d is defined as

that is, magnitude of the force times length of the displacement times the cosine of the

angle between p and d. If

, then

. If p and d are orthogonal, then the work is

zero. If

, then

which means that in the displacement one has to do work

against the force.

PROJECTION OF A VECTOR IN THE DIRECTION OF ANOTHER NON ZERO VECTOR

Components or projection of a vector a in the direction of a vector

is defined by

……..(5)

where is the angle between a and b.

Thus is the length of the orthogonal projection o a on a straight line l parallel to b, taken

with the plus sign if b has the direction of b and with the minus sign if has the direction

opposite to b

Multiplying (5) by , we have

ie

(b

……(6)

if b is a unit vector,as it is often used for fixing a direction then (6) simply gives

p=a.b (|b|=1)

Definition An orthonormal basis I,j k associated with a Cartesian coordinate system. Then

{i, j, k} form an orthonormal basis, called standard basis

An orthonormal basis has the advantage that the determination of the coefficients in

representation

v=l

1

a+l

2

b+l

3

c (v a given vector)

is very simple. This is illustrated in the following Problem.

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Solution

, since

and

Similarly, it can be shown that

and

Normal Vector to a given line

• Two non-zero vectors

and

in the plane are perpendicular (or

orthogonal) if

i,e, if

• Consider a line

The line

though the origin and parallel to is

when can also be written

where

and

.

Now

implies that a is perpendicular to position vector of each points on the

line

. Hence a is perpendicular to the line

and also to because and

are parallel

is called a normal vector to (and to

).

is another normal vector to (and to

).

• Two straight lines

and

are perpendicular if their

normal vectors are perpendicular. Since

and

are normal

vectors of and respectively

and are perpendicular if

Problem Find a normal vector to the line

Solution

Let given line be

. Then the

through the origin and parallel to is

which can also be written as

where

and

Hence by

the discussion above a normal vector to the line

is

Problem

Find the straight line through the point

in the xy-plane and

perpendicular to the straight line

Also find the point of intersection of the lines and

Solution

Suppose the required straight line be

. Then

is a normal

vector to and is perpendicular to the normal vector

of the line

. That is

i.e,

….(8)

Now, if we take

and

we have

is a normal vector to and

hence

Since it passes through

by substituting

in the

equation of we have

or

Hence the equation of the required line is

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or

Now the point of intersection of of

and

can be

obtained by solving the following systems of equations.

Solving, we obtain

and

Hence the point of intersection of the lines

and is

NORMAL VECTOR TO A PLANE

Let

be a plane in space. It can also be writer as

….(9)

where

and

The unit vector in the direction of a is

Dividing (9) by , we get

…..(10)

where

Representation (10) is called Hesse’s normal form of a plane.

In (10), p is the projection of r in the direction of n. Note that the projection has the same constant value for the position vector r of any point in the plane.

(10) holds if and only if n is perpendicular to the plane. n is called unit normal vector to the plane (the other being –n)

Remark

From the above discussion it follows that is the distance of the plne from the

origin.

Problem 6 Find a unit vector perpendicular to the plane

. Also find the

distance of the plane from the origin.

Solution

A normal vector to the given line is

(13)

and since

we have

and the plane has distance

from the origin

Assignments

For the vectors

,

and

find

1. ,

2 ,

,

3.

4.

5 Find the work done by the force p acting on a body if the body displaced from a point A to a

point B along the straight segment AB. Sketch p and AB.

6

7

8 Can work be zero or negative? In what cases?

Let

. Find the angle between

9

10

11 Find the angle between the straight lines

and

12 Find the angle between the planes

and

13 Find the angles of the triangle with vertices

14. Find the angles of the parallelogram with vertices

Find the component of a in the direction of b:

15

16.

17 Vector Product

Definition The

vector product (cross product)

of two vector

and

is a vector

as follows:

• If a and b have the same direction,

• If a and b have the opposite direction,

• In any, other case,

has the length

…..(1)

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where is the angle between a and b. This is the area of the parallelogram with a and b as

adjacent sides The direction of

is perpendicular to both a and b and such that

a,b,v in this order, form a right handed triple.

CROSED PRODUCT IN COMPONENTS

In components, the cross product is given by ….(2) Notice that in (2)

Hence is the expansion of the determinant

by the first row.

Definition A Cartesian coordinate system is called right-handed if the corresponding

unit vectors I,j,k in the positive directions of the axes form a right handed triple. The system

is called left-handed if the sense of k is reversed.

Problem Find the vector product of and in right-handed coordinates.

Solution

or

Problem With respect to a right-handed Cartesian coordinate system, let and Then

Vectors Product and Standard basis vectors

Since i,j,k are orthogonal (mutually perpendicular) unit vectors, the definition of

vector product gives some useful formulas for simplifying vector products; in right-handed

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coordinates these are

…… (3)

For left-handed coordinates, replace k by –k , Thus

GENERAL PROPERTIES OF VECTOR PRODUCT

Cross multiplication has the following properties: • For every scalar l

…(4)

• It is distributive with respect to vector addition, that is, (a)

(b) ….(5)

• Vector product is not commutative but anticommutative, that is,

…(6)

• It is not associative, that is in general

so that the parentheses cannot be omitted.

Proof.

(4) follows directly from the definition

.

…(8)

The sum of the two determinants in (8) is the first component of the right side of (5a). For the other components in (5a) and in (5b), equality follows by the same idea.

Now to get (6), note that

, using (2**)

, as the interchange of rows 2

And 3 multiples the determinants by -1

again using (2**)

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We can confirm this geometrically if we set

and

then

by

(11) and for b, a,w to from a right handed triple, we must have

For the proof of (7), note that

, where as

SCALAR TRIPLE PRODUCT

Definition (Scalar Triple Product) The scalar product of three vectors a,b and c, denoted by

, is defined as :

The scalar triple product

is also denoted by

and is also called the box product

of the three vectors

Remarks

• The scalar triple product is a scalar quantity.

• Since the scalar triple product involves both the signs of ‘cross’ and ‘dot’ it is some

times called the mixed product.

Geometrical meaning of Scalar triple product

The scalar triple product has a geometrical interpretation. Consider the

parallelepiped with a, b and c as co terminus edges. Its height is the length of the

component of a on

. To be precise, we should say that this height is the magnitude of

, where is the angle between a and

.

Now,

Where the sign or depends on

which is positive or negative according is acute or

obtuse that is according as a, b, c is right handed or left handed.

Hence the volume of the parallelepiped with co terminal edges a, b and c is

,

up to sign, the scalar triple product.

Expression for the scalar triple product as a determinant

Let and

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Problem Compute if and

Solution

Problem Find the volume of the parallelepiped whose co-terminal edges are arrows representing the vectors and

Solution

Problem Find the volume of the tetrahedron with co terminal edges representing the vectors and

Solution

The volume of the tetrahedron

Hence, the volume of the tetrahedron is

Theorem (Linear independence of three vectors) Three vectors form a linearly independent

set if and only if their scalar triple product is not zero.

The following is restatement of the above Theorem.

Theorem

Scalar triple product of three coplanar vectors is zero.

Proof.

Let

be three coplanar vectors. Now

represent a vector which is

perpendicular to the plane containing b and c in which also lies the vector a and hence

is perpendicular to a. Therefore

Thus

when three vectors are

coplanar.

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Conversely, suppose that

. That is

, which shows that

is

perpendicular to a. But

is vector perpendicular to the plane containing b and c hence a

should also lie in the plane of b and c. That is a, b, c are coplanar vectors.

Problem 11 Show that the vectors

are linearly

independent.

Solution

By the Theorem, it is enough to show that the scalar triple product of the given

vectors is not zero. It can be seen that the scalar triple product is not zero. Hence the given

vectors are linearly independent.

Problem Find the constant so that the vectors

are coplanar.

Solution Three vectors a, b, c are coplanar if

i.e, if or if or if

Problem Prove that the points and are

coplanar.

Solution

Let the given points be

respectively and if these four points are coplanar

then the vectors

are coplanar, so that their scalar triple product is zero. i.e

Now sition vector of position vector of

Similarly Now

Hence the given points are coplanar.

Equation of a Plane with three points

Let

and c=x

3

i+y

3

j+z

3

k be the position vectors of three points

and

Let us assume that the three points

and do not lie in the same straight line.

Hence they determine a plane. Let

be the position vector of any point

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in the plane. Consider the vectors

which all lie in the plane. That is

and

are coplanar vectors. Now we apply the

condition for coplanar vectors.

or

Problem Find the equation for the plane determine by the points

and

Solution

The equation of a plane with three points

and

is

given by

Hence here, the equation of the plane with the points

and

is given by

or

i.e, the equation of the plane is

or

Assignments

In Assignments 1-9 with respect to a right-handed Cartesian coordinate system, let Find the following expressions.

1. 2.

3 4 .

5 6

7 8

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10

What properties of cross multiplication do Assignments 1,4 and 5 illustrate ?

11 A wheel is rotating about the x-axis with angular speed 3

The rotation appears

clockwise if one looks from the origin in the positive x-direction. Find the velocity

and the speed at the point

.

12 What are the velocity and speed in Exercise 11 at the point

if the wheel rotates

about the y-axis and

?

A force p acts on a line through a point A. Find the moment vector m of p about a point are 13

14

15 Find the area of the parallelogram if the vertices are

16 Find the area of the triangle in space if the vertices are

17 Find the plane through

18

Find the volume of the parallelepiped if the edge vectors are

19 Find the volume of the tetrahedron with the vertices

20 Are the vectors

linearly independent ?

LINES AND PLANES IN SPACE

In this chapter we show how to use scalar and vector products to write equations for

lines, line segments, and planes in space.

Lines and Line Segments in Space

Suppose is line in space passing through a point

and is parallel to a vector

Then is the set of all points

for which

is parallel to v. That

is, lies on if and only if

is a scalar multiple of .

Vector equation for the line through and parallel to v is given by …(1)

Expanding Eq. (1), we obtain

Equating the corresponding components of the two sides gives three scalar equations

involving the parameter t:

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When rearranged, these equations give us the standard parameterization of the line

for the interval

as follows :

……(2)

Standard parameterization of the line through

and parallel to

is given by

above.

Problem Find parametric equations for the line through and parallel to .

Solution

With

equal to

and

equal to

Eq (2)

become

Problem Find parametric equations for the line through

and

.

Solution

The

vector

is parallel to the line , and Eg.(2) with “base point”

give

…..(3)

If we choose

as the “base point” we obtain

…..(4)

The equations in (4) serve as well as the equations in (3); they simply place you at a different

point for a given value of .

Line Segment Joining Two Points

To parameterize a line segment joining two points, we first parameterize the line

through the points. We then find the values for the end points and restrict to lie in the

closed interval bounded by these values. The line equations together with this added

restriction parameterize the segment.

Problem

Parameterize the line segment joining the points

and

.

Solution

We begin with equations for the line through and , which obtained in Problem 2:

…..(5)

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We observe that the point

Passes through

at

and

at

We add the restriction

to Eq (3) to parameterize the line segment:

…..(6)

The Distance from a Point to a Line in Space

To find the distance from a point to to a line that passes through a point parallel

to a vector v, we find the length of the component of normal to the line . In the notation

of the figure, the length is

which is

Distance from a Point to a Line Through parallel to is given by

...(7)

Problem

Find the distance from the point

to the line

...(8)

Solution

Putting we see from the equations for that passes through and is parallel to (v is obtained by comparing (8) with (2) to get With

and

Eq. (7) gives

Equations for planes in Space

Suppose plane passes through a point

and is normal (perpendicular)

to the nonzero vector

Then is the set of all points

for which

is orthogonal to

That is, lies on if and only if

This equation is equivalent to

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i.e. Plane Through

normal to

is given by the

following equivalent equatations:

Vector equation :

….(9)

Component equatation :

….(10)

Problem Find an equatation for the plane through perpendicular to .

Solution

Using Eq. (10),

Problem Find the plane through

Solution

We find a vector normal to the plane and use it with one of the points (it does not

matter which) to write an equatation for the plane.

The cross Product is normal to the plane. Note that Similiarly,

Hence

We substitute the components of this normal vector and the co ordinates of the point (0,0,1) into Eq. (10) to get

i.e

Problem Find the point where the line

Intersects the plane

Solution

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…(12)

Lies in the plane if its coordinates satisfy the equatation of the plane; that is, if

Putting the point of intersection is

The Distance from a Point to a Plane

ProblemFind the distance from to the plane

Solution

We follow the above algorithm.

Using (11) vector normal; to the given plane is given by

The poins on the plane easiest to find from the plane’s equation are the intercepts .If we take to be the y-intercept, then putting and in the equation of the plane or

Hence is , and then

The distance from to the plane is

Angles Between Planes; Lines of Intersection

The angle between two intersecting planes is defined to be the ?(acute) angle determined by their normalvectors

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Problem Find the angle between the planes and

Solution

Using (11), it can be seen that the vectors

are normals to the given planes and respectively. The angle between them (using the definition of dot product) is

Problem Find the vector parallel to the line of intersection of the planes and .

Solution

The line of intersection of two planes is perpendicular to the plane’s normal vector’s

and , and therefore parallel to

.In particular

is a vector parallel to the

plane’s line of intersection. In our case,

,

We note that any nonzero scalar multiple of

is also a vector parallel to the

line of intersection of the planes

and

.

Problem Find parametric equations for the line in which the planes

and

intersect.

Solution

v =14i+2j+15k as a vector parallel to the line. To find a point on the line, we can take any

point common to the two planes. Substituting z=0 in the plane equations we obtain

and solving for the x and y simultaneously gives x=3, y= -1. Hence one of the point common

to the plane is (3,-1,10)

The line is [Using eq.(2)]

Assignments

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1. The line through the point

parallel to the vector i+j+k

2. The line through

and

3. The line through the origin parallel to the vector 2j+k

4. The line through (1,1,1) parallel to the z-axis

5. The line through

perpendicular to the plane

6. The x-axis

Find the parametrizations for the line segments joining the points in Assignments 7-10.

Draw coordinate axes anD sketch each segments indicating the direction of increasing t

for your parametrization.

7. ,

8. ,

9 ,

.

10 ,

Find equations for the planes in Assignments 11-13.

11. The plane through

normal to

12 The plane through

and

13 The plane through

perpendicular to the line

14 Find the point of intersection of the lines

and

15 Find the plane determined by the intersection of the lines:

16 Find a plane through

and perpendicular to the line of intersection of the

planes

In Assignments 17-19, find the distance from the point to the line.

17

18

19 In Exercise 20-22 , find the distance from the point to the plane.

20

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22 23 Find the distance from the plane

to the plane,

24 Find the angles between the following planes:

Use a calculator to find the acute angles between planes in Assignments 25-26 to the nearest

hundredth of a radial.

25.

26

In Exercise 27-28 , find the point in whixh the line meets the given planes 27

28.

Find parametrizations for the lines in which the planes in Assignments 29 intersect.

29. 30.

Given two lines in space, either they are parallel, or they intersect or they are skew (imagine,

for Problem the flight paths of two planes in the sky). Exercise 31 give three lines. Determine

whether the lines, taken two at a time, are parallel, intersect or are skew. If they interesect,

find the point of intersection.

31

CYLINDERS,SPHERE,CONE AND QUADRIC SURFACES

Definition A cylinder is a surface generated by a line which is always parallel to a fixed

line passes through (intersects) a given curve.

The fixed line is called the axis of the cylinder and the given curve is called a guiding

curve or generating curve

Remark

• If the guiding curve is a circle, the cylinder is called a right circular cylinder.

• Since the generator is a straight line, it extends on either side infinitely. As

such , a cylinder is an infinite surface.

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• The degree of the equation of a cylinder depends on the degree of the equation

of the guiding curve.

• A cylinder, whose equation is of second degree, is called a quadric cylinder.

When graphing a cylinder or other surface by hand or analyzing one generated by a

computer, it helps to look at the curves formed by intersecting the surface with planes

parallel to the coordinate planes. These curves are called cross sections or traces.

We now consider a cylinder generated by a parabola.

Problem Find an equation for the cylinder made by the lines parallel to the z-axis that pass

through the parabola

Solution

Suppose that the point

lies on the parabola

in the

plane. Then,

for any value of z, the point

will lie on the cylinder because it lies on the line

through parallel to the z-axis. Conversely any point

whose

y-coordinate is the squre of it x-y-coordinate lie on the cylinder because it lies on the line

through parallel to the z-axis

Remark

Regardless of the value of z, therefore, the points on the surface are the

points whose coordinates satisfy the equation

. This makes

an equation for the

cylinder. Because of this, we call the cylinder “the cylinder

’’.

As Problem 1 or the Remark follows it suggests, any curve

in the - plane

defines a cylinder parallel to the z-axis whose equation is also

Problem The

equation

defines the circular cylinder made by the lines

parallel to the z-axis that pass through the circle

in the xy-plane.

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Problem

The equation

defines the

elliptical cylinder made by the lines parallel to the z-axis that passes through the ellipse

in the

In the similar way, we have the following:

• Any curve

in the

defines a cylinder parallel to the y-axis whose

space equation is also

.

• Any curve

defines a cylinder parallel to the x-axis whose space equation is

also

.

We summarize the above as follows :

Problem

The equation

defines surface made by the lines parallel to the

x-axis that passes through the hyperbola

in the plane

Quadric Surfaces

A quadric surface is the graph in space of a second-degree equation in the

and z.

The most general form is

Where A,B,C and so on are constants, but the equation can be simplified y translation and

rotation, as in the two-dimensional case. We will study only the simpler equations.

Although the definition did not require it, the cylinders considered so far in this chapter

were also Problem of quadric surfaces. We now examine ellipsoids (these include spheres as

a special case), paraboloids, cone, and hyperboloids.

Problem

The ellipsoid

……(1)

cuts the coordinate axes at

and

. It lies within the rectangular

box defined by the inequalities

The surface is symmetric with

respect to each of the coordinate planes because the variables in the defining equation are

squared.

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The section cut from the surface by the plane

is the ellipse

…..(2)

Special Cases: If any two of the semi axes a, b and c are equal, the surface is an ellipsoid of

revolution. If all three are equal, the surface is sphere.

Problem

The elliptic paraboloid

…..(3)

is symmetric with respect to the planes

and

as the variables x and y in the

defining equation are squared. The only intercept on the axes is the origin (0,0,0). Except for

this point, the surface lies above or entirely below the xy-plane, depending on the sign of c.

The sections cut by the coordinate planes are

…..(4)

Each plane

above the xy-plane cuts the surface in the ellipse

Problem The

circular paraboloid or paraboloid of revolution

…..(5)

is obtained by taking by b=a in Eg. (3) for the elliptic paraboloid. The cross sections of the

surface by planes perpendicular to the z-axis are circles centered on the z-axis. The cross

sections by planes containing the z-axis are congruent parabolas with a common focus at the

point (0,0,

.

Application : Shapes cut from circular paraboloids are used for antennas in radio telescopes,

satellite trackers, and microwave radio links.

Definition A cone is a surface generated by lines all of which pass through a fixed point

(called vertex) and

(i) all the lines intersect a given curve (called guiding curve)

or (ii)

all the lines touch a given surface

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or (iii)

all the lines are equally inclined to a fixed line through the fixed point.

The moving lines which generate a cone are known as its generators. When the moving

lines satisfy condition (ii) in the definition of a cone, we term the cone as enveloping cone.

Problem The elliptic cone

…..(6)

is symmetric with respect to the three coordinate planes Fig (7). The sections cut by

coordinate planes are

…(7) ……(8)

The secions cut by planes

above and below the xy-plane are ellipse whose enters lie

on the z-axis and whose vertices lie on the lines in Eq.(7) and (8).

If a=b, the cone is a right circular cone.

Problem 9

The hyperboloid of one sheet

….(9)

is symmetric with respect to each of the three coordinate planes . The sections cut out by the

coordinate planes are

…..(10)

The plane

cuts the surface in an ellipse with center on the z-axis and vertices on one

of the hyperbolas in (10)

If a=b, the hyperboloid is a surface of revolution

Remark : The surface in Problem 9 is connected, meaning that it is possible to travel from

one point on it to any other without leaving the surface. For this reason it is said to have one

sheet, in contrast to the hyperboloid in the next Problem, which as two sheets.

Problem

The hyperboloid of two sheets

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is symmetric with respect to the three coordinate planes. The plane z=0 does not intersect

the surface; in fact, for a horizontal plane to intersect the surface, we must have

The

hyperbolic sections

have their vertices and foci on the

. The surface is separated into two portions, one

above the plane

and the other below the plane

. This accounts for the name,

hyperboloid of two sheets.

Eq.(9) and (11) have different numbers of negative terms. The number in each case is

the same as the number of sheets of the hyperboloid. If we replace the I on the right side of

either Eq.(9) or Eq.(11) by 0, we obtain the equation

for an elliptic cone (Eq.6). The hyperboloids are asymptotic to this cone in the same way that the hyperbolas

are asymptotic to the lines In the xy-plane.

Problem The hyperbolic paraboloid

…..(12)

Has symmetry with respect to the planes

and

. The sections in these planes are

………. (13)

….(14)

In the plane x=0, the parabola opens upward from the origin. The parabola in the

plane y=0 opens downward.

If we cut the surfaces by a plane

, the section is a hyperbola.

……(15)

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If is negative, the focal axis is parallel to the x-axis and the vertices lie on the

parabola in (14).

Near the origin, the surface is shaped like a saddle. To a person travelling along the

surface in the yz-plane, the origin looks like a minimum. To a person travelling along the

surface in the xz-plane, the origin looks like a maximum. Such a point is called a minimax

or saddle point of surface.

Assignments

Sketch the surfaces in Assignments 1-32

1. 2. 3 . 4 5. 6 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17 18. 19 20. 21 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32

CYLINDRICAL AND SPHERICAL COORDINATES

Cylindrical and Spherical Coordinates

This section introduces two new coordinate systems for space: the cylindrical

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simplify the equations of cylinders. Spherical

coordinates simplify the equations of

spheres and cones.

Cylindrical Coordinates

We obtain cylindrical coordinates for space by combining polar coordinates in the

xy-plane with the usual z-axis. This assigns to every point in space

one or more coordinates

triples of the form (r, θ, z), as shown in Fig 1.

Definition

Cylindrical coordinates represent a point P in space by ordered triples (r, θ , z) in which

1. r and θ are polar coordinates for the vertical projection of P on the

xy - plane,

2. z is the rectangular vertical coordinate.

The values of x, y, r , and θ in rectangular and cylindrical coordinates are related by the

usual equations.

Equations Relating Rectangular (x,y,z) and Cylindrical (r, θ, z) Co ordinates

x = r cos θ, y = r sin θ, z = z ; r

2

_{= x}

2

_{+ y}

2

_{tan θ =}

In cylindrical coordinates , the equation r = a describes not just a circle in

the

xy-plane but an entire cylinder about the z-axis . The z-axis is given by r = 0. The equation

θ= θ

0

describes the plane that contains the z-

axis and makes an angle θ

0

with the positive

x-axis. And, just as in

rectangular coordinates, the equation z= z

0

describes a plane

perpendicular

to the z-axis.

Problem

What points satisfy the equations

r = 2, θ =

Solution

These points make up the line in which the cylinder r = 2 cuts the portion of

the

plane

θ =

where r is positive. This is the line through

the point (2,

,

0)parallel to the z -axis.

Along this line, z varies while r and θ have the constant values r = 2 and θ =

Problem Sketch the surface r = 1 + cos θ

Solution

The equation involves only r and θ; the coordinate variable z is missing .

Therefore, the surface is a cylinder of lines that pass through the cardioid r = 1

+

cos

θ in the r θ-plane and lie parallel to the z-axis. The rules for sketching

the cylinder are the same as always: sketch the x-, y-, and z-axes, draw a few

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perpendicular cross sections, connect the cross sections with parallel lines ,

and darken the exposed parts.

Problem Find a Cartesian equation for the surface z = r

2

_{and identity the surface.}

Solution From Eqs. (1) we have z=r2_{= x}2_{+ y}2_{. The surface is the circular paraboloid x}2_{+ y}2_{= z}

Problem Find an equation for the circular cylinder 4x2_{+ 4y}2_{= 9 in cylindrical coordinates.}

Solution

The cylinder consists of the points whose distance from the z-axis is

=

The corresponding equation in cylindrical coordinates is

r =

Problems Find an equation for the cylinder x

2

_{+ (y−3)}

2

_{= 9}

_{in cylindrical coordinates}

Solution The equation for the cylinder in cylindrical coordinates is the same as the

polar equation for the cylinder's base in the xy- plane:

Spherical Coordinates

Definition Spherical coordinates represent a point P in space by ordered triples

( ,

in which

1. is the distance from P to the origin.

2. is the angle

makes with the positive z-axis (0

),

3. is the angle from cylindrical coordinates.

The equation = a describes the sphere of radius a

centered at the origin. The equation

=

0

describes a single cone whose vertex lies at the origin and whose axis lies along

the z-axis. (We

broaden our interpretation to include the xy-plane as the cone

)

If

0

is greater than

, the cone =

0

opens downward.

Equations Relating Spherical Coordinates to Cartesian and Cylindrical Coordinates

r = , x = r cos = ,

r = , y = r sin = , (2)

Problem Find a spherical coordinate equation for the sphere

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1

Problem Find a spherical coordinate equation for the cone z =

Solution

Use geometry. The cone is symmetric with respect to the z-axis and cuts the

first quadrant of the yz-plane along the line z = y. The angle between the cone and the

positive z-axis is therefore

radians. The cone consists of the points whose spherical

coordinates have equal to

so its equations

is

Assignments

In Assignments 1-26, translate the equations and inequalities from the given

coordinates system (rectangular, cylindrical, spherical ) into equations and inequalities in

the other two systems. Also , identify the figure being defined.

1. r = 0

2. 3. z = 0

4. z = −2

5. z =

6. z =

7.

8. = 1

9.

10. 11. = 5 cos

12. = −6 cos

13. r = csc θ

14. r = −3 sec θ

15.

16.

17.

18.

19.

20.

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21. z = 4−4 , 0

22. z = 4−r, 0

23. , 0

24. , 0

25. z +

= 0

26. 27. Find the rectangular coordinates of the center of the sphere

___________________________________

VECTOR- VALUED FUNCTIONS

AND SPACE CURVES

Space Curves

In this chapter, we shall consider the equations of the form

…..(1)

where,

and

are real valued functions of the scalar variable t. As t increases from its

initial value to the value the point

trace out some geometric object in space;

it may be straight line or curve. This geometric object is called space curve or arc. Simply, the

equations

represent a curve in space. A space curve is the locus of the point

whose co‐ordinates are

functions of a single variable

Definitions

When a particle moves through space during a time interval I, then the particle’s coordinates

can be considered as functions defined on I.

The points

make up the curve in space that we call the particle’s

path. The equations and interval in (1) parameterize the curve.

The vector

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from the origin to the particle’s position

at time is the particle’s position

vector. The functions

and h are the component functions (components) of the position vector.

We think of the particle’s path as the curve traced by during the time interval

Equation (1) defines as a vector function of the real variable on the interval More

generally, a vector function or vector‐valued function on a domain set D is a rule that assigns a

vector in space to each element in D. For now the domains will be intervals of real numbers. There

are situations when the domains will be regions in the plane or in space. Vector functions will then

be called “vector fields”. A detailed study on this will be done in later chapter.

We refer to real‐valued functions as scalar functions to distinguish them from vector

functions. The components of r are scalar functions of t. When we define a vector‐valued function

by giving its component functions, we assume the vector function’s domain to be the common

domain of the components.

Problem Consider the circle

It is most convenient to use the

trigonometric functions with t interpreted as the angle that varies from

. Then we

have

In this chapter we show that curves in space constitute a major field of applications of vector

calculus. To track a particle moving in space, we run a vector r from the origin to the particle and

study the changes in r.

Problem A straight line L through a point A with position vector

in the

direction of constant vector

can be represented in the form

……(2)

If b is a unit vector, its components are the direction cosines of L. In this case

measures the

distance of the points of L from A.

Problem The vector function

is defined for all real values of t. The curve traced by r is a helix (from an old Greek word for “spiral”)

that winds around the circular cylinder

. The curve lies on the cylinder because

the i‐ and j‐ components of r, being the x‐ and y‐ coordinates of the tip of r, satisfy the cylinder’s

equation:

The curve rises as the k‐component

increase. Each time increases by

the curve

completes one turn around the cylinder. The equations

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Limits

The way we define limits of vector‐valued functions is similar to the way we define limits of real‐

valued functions.

Definition

Let

be a vector function and a vector. We say that has limit as

approaches and write

If, for every number

, there exists a corresponding number

such that for all

If

, then

precisely when

The equation provides a practical way to calculate limits of vector functions.

Problem If then

and

We define continuity for vector functions the same way we define continuity for scalar functions.

Continuity

Definition

A vector function

is continuous at a point

in its domain if

. The function is continuous if it is continuous at every point in its domain.

Component Test for Continuity at a Point

Since limits can be expressed in terms of components, we can test vector functions for continuity by

examining their components. The vector function

is continuous at

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Problem The function

is contitinuous because the components

,cost,sint and t are continuous

Consider the function

is discontinuous at every integer. We note that the components

and

are continuous

everywhere. But the function

is discontinuous at every integer

Hence

is discontinuous at every integer.

Derivatives and Motion

Suppose that

is the position vetor of a particles moving along a curve

in space and that

are differentiable functions of Then the difference between the

particle’s positions at time and time

is

In terms of components,

As approaches zero, three things seem to happen simultaneously.

• First, approaches along the curve.

• Second, the secant line

seems to approach a limiting position tangent to the curve at

• Third, the quotient

approaches the following limit

We are therefore led by past experience to the following definitions.

Definitions

The

vector

function

is

differentiable

at

.Also, is said to be differentiable if it is differentiable at

every point of its domain. At any point at which is differentiable, its derivative is the vector

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Problem

If

find

Solution

Given

Hence

Definition

The curve traced by is smooth if

is continuous and never 0, i.e., if

have continuous first derivatives that are not simultaneously 0.

Remark

The vector

when different from0, is also a vector tangent to the curve.

Definition

The tangent line to the curve at a point

is defined to be the line

through the point parallel to

at

.

Remark

We require

for a smooth curve to make sure the curve has a continuously

turning tangent at each point. On a smooth curve there are no sharp corners of cusps.

Definition

A curve that is made up of a finite number of smooth curves pieced together in a

continuous fashion is called piecewise smooth

Definitions

If r is the position vector of a particle moving along smooth curve in space, then

is the particle’s velocity vector, tangent to the curve. At any time the direction of is the

direction of motion, the magnitude of v is the particle’s speed, and the derivative

when

it exists, is the particle’s acceleration vector. In short,

1. Velocity is the derivative of position :

2 Speed is the magnitude of velocity :

Speed=

3 Acceleration is the derivative of velocity

:

4 The vector

is the direction of motion at time .

We can express the velocity of a moving particle as the product of its speed and direction.

Problem

The vector

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gives the position of a moving body at time . Find the body’s speed and acceleration when

At

times, if any, are the body’s velocity and acceleration orthogonal ?

Solution

At

, the body’s speed and direction are

Speed :

Direction

To find the times when and are orthogonal, we look for values of for which

The only value is

Problem 8

A particle moves along the curve

Find the velocity and acceleration at

Solution

Here the position vector of the particle at time is given by

Then the velocity is given by

and the acceleration a is given by

When

,

and

Problem

Show that if b, c, d are constant vectors, then

is a path of a point moving with constant acceleration.

Solution

The velocity v is given by

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since the derivative of the constant vector d is 0

The acceleration a is given by

The above is a constant vector, being a scalar multiple of the constant vector b. Hence the result.

Problem

A particle moves so that its position vector is given by

where is a constant. Show that

(i) the velocity of the particle is perpendicular to r

(ii) the acceleration is directed towards the origin and has magnitude proportional to the

distance from the origin.

(iii)

is a constant vector.

Solution

(i) The velocity v is given by

Now

Hence velocity of the particles is perpendicular to r.

(ii) The acceleration a is given by

Thus direction of acceleration is opposite to that vector r and as such

it is directed towards the origin and the magnitude is proportional

to

. i.e., the acceleration is directed towards the origin and has magnitude

proportional to the distance from the origin

(iii)

a constant vector.

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Problem

(Centripetal acceleration) suppose a particle moves along a circle C having radius R in

the counter clock wise sense. Then its motion is given by the vector function

…. (1)

The velocity vector

Is a tangent at each point to the circle C and it magnitude

is constant, Hence

so that is the called the angular speed.

The acceleration vector is

with magnitude

. Since

are constants this implies that there is an

acceleration of constant magnitude

towards the origin.( due to negative sign). This acceleration

is called centripetal acceleration. It results from the fact that the velocity vector is changing

direction at a constant rate. The centripetal force is

Where is the mass of . The opposite

vector

is called centrifugal force, and the two forces are in equilibrium at each instant of the

motion.

Differentiation Rules

Because the derivatives of vector functions may be computed component by component,

the rules for differentiating vector functions have the same form as the rules for differentiating

scalar functions They are :

Constant Function Rule :

(any constant vector C)

If and are differentiable vector functions of t, then

Scalar Multiple Rules :

(any number C)

(any differentiable scalar function f(t))

Sum Rule :

Difference Rule :

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Cross product rule

Chain Rule (Short Form) : If r is a differentiable function of t and t is a differentiable function of s, then We will prove the dot and cross product rules and Chain Rule and leaving the others as Assignments . Proof of the Dot Product Rule Suppose that and Then u'.v u.v’ Proof of the Cross Product Rule According to the definition of derivative, To change this fraction into an equivalent one that contains the difference quotients for the derivates of u and v, we subtract and add in the numerator. Then

The last of these equalities holds because the limit of the cross product of two vector functions is

the cross product of their limits if the latter exist. As approaches zero,

approaches

because v, being differentiable at is continuous at The two fractions approach the values of

and

at In short

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Proof the Chain Rule Suppose that is a differentiable vector function of and that t is a differentiable scalar functions of some other variable . Then are differentiable functions of and the Chain Rule for differentiable real‐valued functions gives Problem Show that Solution Let then and Also or

Constant Vectors

A vector changes if either its magnitude changes or its direction changes or both direction and

magnitude change.

Theorem A

The necessary and sufficient condition for the vector function

to be constant is

that

the zero vector,

Solution

Necessary Part

If

is constant, then

.

Hence

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Sufficiency Part

Conversely, suppose that

Let

Then

Hence using the assumption

we have

Equating the coefficients, we get

and this implies that

and

are constants, (in other words this means that

and

are independent of

.

Therefore

is a constant vector.

Theorem B

The necessary and sufficient condition for the vector function

to have constant

magnitude is that

Solution

Let F be a vector function of the scalar variable

Suppose

to have constant magnitude, say F, so that

Then

Therefore,

or

Conversely, suppose that

. Then

or

and this implies that

is a constant or is a constant. i.e. the vector function

have constant

magnitude.

Theorem C

The necessary and sufficient condition for the vector function

to have constant

direction is that

Solution

Let be a vector function of the scalar variable and n be a unit vector in the direction of If be the magnitude of then