Vector Differential Calculus



Vector Differential Calculus

2.1 Vector Valued Functions

In this section we consider functions whose domain consists of real numbers and whose range consists of vectors.

| |

|Definition 2.1 A vector valued function consists of two parts; a domain, which |

|is a collection of numbers and a rule which assigns to each number in the |

|domain one and only one vector. |

The numbers in the domain of the vector valued function are usually denoted by t. The set of all vectors assigned by a vector valued function to members of its domain is called the range of the function.

Example 1 Let [pic]. Determine the domain of F.

Solution [pic] and [pic] ( ( 1 ≤ t ≤ 1.

Therefore, domain of F = [pic].

Note that: We refer any function whose domain and range are sets of real numbers as a real valued

function.

A vector valued function F can be written as

[pic]

[pic], [pic] and[pic] are real valued functions called the component functions of F.

For the function given in example 1,

[pic]=[pic], [pic]=[pic]and [pic]= 1

are the component functions of F.

Graphs of Vector Valued Functions

We usually show a vector valued function F pictorially by drawing only its range. If we think F (t) as a point in space, then as t increases, F (t) traces out a curve in space.

[pic]

If[pic], then the parametric equations of the curve C are:

[pic], [pic]and [pic].

Example 2 Let[pic]. Sketch the curve traced out by F.

Solution If we let (x, y, z) be a point on the curve traced out by F (t), then

[pic], [pic] and z = 3t.

But these are the parametric equations of the line containing (( 1, 1, 0) and parallel to the vector [pic].

[pic]

Example 3 Let[pic]. Then

[pic] = 1 for all t ( (.

Thus, F (t) traces out all points in the xy plane that are at a distance of one unit from the origin.

Moreover; as t increases, the vector F (t) moves around the circle in a counterclockwise direction.

Example 4 Let[pic]. Then sketch the curve traced out by F.

Solution Let (x, y, z) be a point on the curve traced out by F (t), then

[pic], [pic]and[pic]

Thus, [pic] = [pic]= 4 and x = z.

Therefore, the curve traces out by F (t) is the intersection of the sphere

[pic] = 4 and the plane x = z.

[pic]

Combinations of Vector Valued Functions

| |

|Definition 2.2 Le F and G be vector valued functions and let f and g be real valued functions. |

|Then the functions[pic], [pic], [pic], [pic], [pic] and [pic] |

|are defined as follows: |

|i) [pic]=[pic] ii) [pic]= [pic] |

|iii) [pic]=[pic] iv) [pic]= [pic] |

|v) [pic]=[pic] |

Example 5 Let [pic] and [pic]. If g (t) = cos t,

then find

i) [pic] ii) [pic] iii) [pic]

Solution From the above definition we have

i) [pic]= [pic]= [pic]

= [pic]= [pic].

Therefore, [pic]= [pic].

ii) [pic]= [pic]= [pic]

= [pic]

= [pic]

Therefore, [pic]= [pic]

iii) [pic]=[pic]= [pic]= [pic].

Therefore, [pic]= [pic].

Example 6 Let[pic]. Then sketch the curve traced out by H.

|Solution Let[pic]and[pic]. |[pic] |

|Then H (t) = F (t) + G (t). | |

|The curve traced out by F (t) is a unit circle with center at the | |

|origin. Thus the point corresponding to H (t) lies[pic] units | |

|above or below the points corresponding to F (t). | |

|Therefore, the curved traced out by H (t) is a circular Helix. | |

Note that: If [pic], then we get a circular helix oriented clockwise as t increases

Example 7 Cycloid

A cycloid is a curve traced out by a point on the circumference of a circle as the circle rolls along a straight line. Suppose a circle of radius r rolls along the x-axis in the positive direction. Let P be the point that is at the origin. Find the vector equation of the cycloid traced out by P.

|Solution Suppose the circle rolled through an angle t |[pic] |

|radians. Then | |

|[pic] | |

|Hence, [pic] | |

|and [pic] | |

|Thus, [pic] | |

|Therefore, the vector equation of the cycloid is: | |

|[pic] | |

|and the parametric equations are: | |

|x (t) = [pic]and y (t) = [pic] | |

| |[pic] |

2.2 Calculus of Vector Valued Functions

| |

|Definition 2.3 Let F be a vector valued function defined at each point in some open |

|interval containing [pic], except possibly at [pic]itself. A vector L is the limit of |

|F (t) as t approaches [pic](or L is the limit of F at [pic]) if for every ( ( 0 there is |

|a ( ( 0 such that |

|if [pic], then [pic] ( ( |

|In this case we write |

|[pic] and say that [pic]exists. |

| |

|Theorem 2.1 Let [pic]. Then F has a limit at [pic]if and |

|only if [pic], [pic] and[pic] have limits at [pic]. In this case |

|[pic] |

Example 8 Evaluate [pic].

Solution Let F (t) = [pic]. Then

[pic]= [pic]

Therefore, [pic] = [pic].

| |

|Theorem 2.2 Let F and G be vector valued functions, and let f and g be real valued functions. |

|Assume that [pic],[pic]and [pic] exist and that |

|[pic]= [pic]. Then |

|i) [pic][pic]= [pic][pic] |

|ii) [pic][pic]= [pic][pic] |

|iii) [pic][pic]= [pic][pic] |

|iv) [pic][pic]= [pic][pic] |

|v) [pic][pic]= [pic] [pic] |

Example 9 Let F (t) = [pic] and G (t) = [pic].

Then find i) [pic][pic] ii) [pic][pic]

Solutions i) [pic][pic]= [pic][pic]= [pic]

= [pic]

Therefore, [pic][pic]= [pic].

ii) [pic][pic]= [pic][pic]= [pic]

= [pic]

Therefore, [pic][pic]= [pic].

| |

|Definition 2.4 A vector valued function F is continuous at a point [pic]in its domain if |

|[pic]F (t) = F ([pic]) |

|Theorem 2.3 A vector valued function F is continuous at a point [pic]if and only if each of |

|its component functions is continuous at [pic]. |

| |

|Definition 2.5 Let [pic]be a number in the domain of a vector valued function F. |

|If [pic] exists, we call the derivative of F at [pic]and write |

|[pic]. |

|In this case we say F has derivative at [pic], or F is differentiable at [pic]or that |

|[pic]exists. |

Geometric Interpretation of [pic]

Let C be the curve traced out by F, and let [pic]and P be points on the curve corresponding to [pic]and F(t) respectively.

|[pic] |[pic] |[pic] |

Now consider the vector [pic].

[pic] = F (t) ( [pic]

The vector [pic] has the same direction as [pic] if t > [pic], and has opposite direction to [pic] if

t < [pic]. Hence [pic]exists, then it points in the same direction in which C is traced out by F. [pic]is tangent to the curve C at [pic].

| |

|Theorem 2.4 Let [pic]. Then F is differentiable at [pic] |

|if and only if [pic], [pic] and[pic] are differentiable at [pic]. |

|In this case, [pic] |

Example 10 Let [pic]. Then [pic].

Example 11 Let [pic]. Then

[pic].

Note that: For any a, b, c (( and for all t ( (, [pic]is a constant vector valued function.

Example 12 Show that the derivative of a linear vector valued function is a constant vector valued function.

Solution Let [pic] be any linear vector valued function.

Then [pic].

Therefore, [pic]is a constant vector valued function.

Remark: Let F be a vector valued function defined on an interval I.

i) If the components of F are differentiable on I, then we say that F is differentiable on I.

ii) If I is a closed interval [a, b], then F is differentiable on [a, b] if and only if the component

functions are differentiable on (a, b) and have appropriate one-sided limits at a and b.

Example 13 Let[pic]. Then F is differentiable on [0, 1] but not on [( 1, 1], since [pic]

is not differentiable at t = 0.

Solution Left to the reader.

| |

|Theorem 2.5 Let F, G and f be differentiable at [pic] and let g be differentiable |

|at [pic]with g ([pic]) = [pic]. Then the following holds |

|i) [pic]=[pic] |

|ii) [pic]= [pic] |

|iii) [pic]=[pic] |

|iv) [pic]= [pic] |

|v) [pic]=[pic]= [pic] |

Example 14 Let [pic] and[pic]. Then find

[pic]and [pic]

Solution Now [pic] and[pic].

Therefore, [pic]= [pic].

Since, [pic]= [pic]

and [pic]= [pic].

[pic]

Example 15 Let [pic] and[pic]. Then find

[pic]and [pic]

Solution Now [pic] and[pic].

Then, [pic]=1 and[pic]= 0, [pic]= [pic]

and [pic]= [pic].

Therefore, [pic]= 1and [pic]= [pic].

| |

|Corollary 2.6 Let F be differentiable on an interval I and assume that there is a |

|non-zero number c such that |

|[pic] = c for all t in I. |

|Then [pic] = 0 for all t in I. |

Proof: Since, [pic] = c by hypothesis, it follows that

[pic]for all t in I.

Thus,[pic]is a constant real valued function on I.

Hence, [pic]= 0.

Therefore, [pic] = 0 for all t in I.

Note that: If [pic] is a constant real valued function, then for each t in the domain of F, one and only one

of the following is true.

i) F (t) = 0 ii) [pic]= 0 iii) [pic] and [pic]are orthogonal.

Let [pic]. The second derivative of F is defined to be the derivative of [pic], denoted by [pic]is given by:

[pic]

Example 16 Let [pic]. Then find [pic].

Solution [pic]= [pic]and [pic]= [pic].

Therefore, [pic]= [pic].

Velocity and Acceleration

As an object moves through space, the coordinates x, y and z of its location are functions of time.

Let as assume that these functions are twice differentiable. Then we define

Position: r(t) = [pic]

Velocity: v (t) = [pic]

Speed:[pic]

Acceleration: [pic]

Note that: i) The position vector r (t) is called the radial vector or radius vector.

ii) If [pic]is fixed and t ( [pic], then the vector r (t) ( r ([pic]) is called the

displacement vector.

iii) The average velocity is defined as

[pic]

and hence the velocity is the limit of the average velocity.

Example 17 Suppose the position of an object is given by

r(t) = [pic].

Determine the velocity, speed and acceleration of the object.

Solution v (t) = [pic], [pic]

and [pic].

Example 18 Let r(t) = [pic]. Find the set of all possible values of t

for which[pic]= 0.

Solution v (t) = [pic].

Hence, [pic]= 0 ( [pic]= 0 and [pic]= 0.

( [pic]and [pic]= 0 ( t = 2n( for any integer n.

Therefore, (t ( (: t = 2n( for any natural number n( is the required solution.

Example 19 An object moves counterclockwise along a circle of radius [pic]> 0 with a constant

speed [pic]> 0. Find formulas for the position, velocity and acceleration of the object.

Solution Let us set up a coordinate system so that the circle lies in the xy plane with center at

the origin so that the object is on the positive x-axis at t = 0.

|[pic] |Then r(t) = [pic]and hence, |

| |v (t) = [pic]. |

| |Since the object moves counterclockwise, ( is increasing and hence [pic]>|

| |0. |

| |Thus, [pic] = [pic]and hence [pic] |

Now since ( (0) = 0 we get [pic].

Therefore, r(t) = [pic], v(t) = [pic]

and a(t) = [pic].

Integration of Vector valued Functions

| |

|Definition 2.6 Let [pic], where [pic],[pic] and[pic]are |

|continuous real valued functions on [a, b]. Then the definite integral |

|[pic] and the indefinite integral [pic]are defined by |

|[pic] |

|and[pic] |

Example 20 Let [pic]. Find[pic] and [pic].

Solution [pic]

= [pic]

=[pic]

Therefore, [pic] = [pic]

[pic] =[pic]

= [pic], where C is a constant vector.

Therefore, [pic] =[pic], where C is a constant vector.

If [pic], where [pic],[pic] and[pic]are continuously differentiable, then

[pic]

= [pic]

= [pic]

= [pic] + [pic].

Therefore,[pic] = [pic], where C is a constant vector.

Space Curve and Their Lengths

| |

|Definition 2.6 A space curve (or simply curve) is the range of a continuous |

|vector valued function on an interval of real numbers. |

Notation: We will generally use C to denote a curve and [pic]to denote a vector valued function

whose range is a curve C. In this case we say that C is parameterized by [pic]or that [pic]

is a parameterization of C.

Suppose [pic]. Then x =[pic], y = [pic]and z = [pic]are parametric equations of the vector valued function [pic].

Let f be a continuous real valued function on an interval I. Now we need to show that the graph of f is a curve and find a parameterization of the curve.

Now let [pic]for all t in I. Then [pic]is continuous and traces out the graph of f. Thus, the graph of f is the range of [pic], so the graph of f is a curve.

Hence, [pic]is its parameterization.

Note that: The vector valued function

[pic], c ≠ 0

represents a curve called circular helix. It lies on the cylinder [pic] .

Properties of Space Curves

| |

|Definition 2.8 A curve C is closed if it has a parameterization [pic]whose domain |

|is a closed and bounded interval [a, b] such that [pic]. |

|[pic] |[pic] |

Example 21 Let [pic] for t ( [0, 2(].

Now [pic]is continuous on [0, 2(] and [pic]= [pic]= [pic].

Therefore, the curve traced out by [pic]is closed curve.

Example 22 Let [pic] for t ( [0, 2(].

Now [pic]is continuous on [0, 2(] and [pic]= [pic]while [pic].

Therefore, the curve traced out by [pic]is not closed.

| |

|Definition 2.9 a) A vector valued function [pic] defined on an interval I is smooth if [pic] |

|has a continuous derivative on I and [pic]0 for each interior point t of I. |

|A curve C is smooth if it has a smooth parameterization. |

|b) A continuous vector valued function [pic] defined on an interval I is |

|piecewise smooth if I is composed of a finite number of subintervals |

|on each of which [pic] is smooth and if [pic]has one sided derivatives at |

|each interior point of I. |

|A curve C is piecewise smooth if it has a piecewise smooth parameterization. |

Example 23 Show that [pic] is smooth.

Solution [pic], [pic]is continuous and [pic]0 for every t ( (.

Therefore, [pic] is smooth.

Example 24 Show that [pic] is piecewise smooth.

Solution [pic] for all t ( ( and hence [pic] is continuous for all t ( (. But [pic]= 0 only for t = 0.

Therefore, [pic] is piecewise smooth on (– (, ().

Example 25 Find a smooth parameterization of the line segment from [pic] to [pic].

Solution Suppose [pic] and [pic] are distinct points in space.

x = [pic]+ (x – [pic]) t, y = [pic]+ (y – [pic]) t and z = [pic]+(z –[pic]) t

are parametric equations for the line through [pic] and [pic].

Now (x, y, z) = [pic] for t = 0 and (x, y, z) = [pic] for t = 1.

Therefore, [pic]= [pic] for 0 ≤ t ≤ 1 is a

smooth parameterization of the segment from [pic] to [pic].

Length of a Curve

Let C be the line segment joining the points [pic] to [pic] in space. Then the length L of C is given by:

[pic].

Now, let C be a smooth curve and [pic]= [pic] a ≤ t ≤ b, be a smooth parameterization of C. Let T = {[pic]} be any partition of [a, b], and let [pic]be the length of the [pic]portion of C.

If [pic]is small, then

[pic].

By the mean-value theorem there are numbers [pic], [pic] and [pic] in [[pic],[pic]] such that

[pic], [pic]

and [pic], where [pic].

Hence [pic].

Therefore, the total length L of C is:

L = [pic]

and for small [pic], [pic]. But

[pic].

Therefore, [pic].

| |

|Definition 2.10 Let C be a curve with piecewise smooth parameterization |

|[pic]defined on [a, b]. Then the length L of the curve C is defined by: |

|[pic] |

Example 26 Find the length L of the segment of the circular helix

[pic]

for 0 ≤ t ≤ 2(.

Solution By definition 2.10,

[pic] = [pic] = [pic].

Therefore, L = [pic] units.

Example 27 Find the length L of the curve

[pic]

for 1 ≤ t ≤ 2.

Solution Now [pic] and [pic].

Hence, [pic] = [pic]= [pic]= 3 + ℓn 2.

Therefore, L = 3 + ℓn 2 units.

Note that: i) Every curve has many parameterizations.

ii) The length of a curve is independent of the parameterization of the curve.

The Arc Length Function

Let C be a smooth curve parameterized on an interval I = [a, b] by

[pic] for t in I.

Let a be a fixed number in I. We define the arc length function s by:

[pic] = [pic] for t in I. (1)

If r (t) denotes the position of an object at time t ≥ a, then s (t) is the distance traveled by the object between time a and time t.

If we differentiate (1) with respect to t, we obtain

[pic] = [pic]

[pic] > 0, since [pic]is smooth for all t in I.

Thus, s (t) is increasing and hence s has an inverse.

Example 28 Find [pic] if [pic].

Solution [pic] and hence [pic] = [pic].

Therefore, [pic] = [pic]for t ≥ 0.

Note that: Any quantity depending on t also depends on s.

Arc Length s as parameter

If a smooth curve C is parameterized by r (t) for t in [a, b] and if C has length L, then C can be parameterized by r (t (s)) for s in [0, L].

Example 29 Consider the Circular helix

[pic].

Then [pic] and [pic].

Thus, the arc length function s (t) is given by:

[pic]( [pic].

Therefore, a formula for the helix using the arc length function s as a parameter is

[pic].

If we set c = 0, then we get [pic] which is a parameterization for a circle.

[pic]

Arc length in Polar form

Let C be a smooth curve parameterized on an interval I by

[pic] for t in I.

[pic] and y = [pic]for t in I.

Hence, [pic] and [pic]

Thus, [pic]

and [pic]

Consequently, Or equivalently, [pic]

Tangents and Normals to Curves

Tangents to Curves

| |

|Definition 2.11 Let C be a smooth curve and [pic]a (smooth) parameterization of |

|C defined on an interval I. Then for any interior point t of I, the tangent |

|vector [pic]at the point r (t) is defined by |

|[pic] |

Note that: T (t) is a unit vector along [pic].

Example 30 Find a formula for the tangent T (t) to the circular helix

[pic]

Solution [pic] and [pic] = [pic].

Therefore, [pic].

Note that: T (t) and r (t) are not orthogonal, since [pic] ( constant for all t.

Example 31 Find the tangent vector T (t) to the curve parameterized by r (t), where

[pic] for [pic] ≤ t ≤ [pic].

Solution [pic]and [pic] = [pic].

Therefore, [pic].

Normals to Curves

Let C be a smooth curve and r (t) be a parameterization of C. Let [pic]be also smooth. Then the tangent vector T (t) is differentiable. Moreover; if[pic] = 1 for all t in the domain of T. But then

[pic] = 0. Hence if [pic] ( 0, then [pic]and T (t) are orthogonal.

| |

|Definition 2.12 Let C be a smooth curve and [pic]a smooth parameterization of |

|C defined on an interval I such that [pic] is smooth. Then for any interior |

|point t of I for which [pic]( 0, the normal vector N (t) at the point r (t) is |

|defined by |

|[pic] |

Example 32 Find a formula for the normal N (t) of the curve parameterized by

[pic]

Solution [pic] and [pic] = [pic].

Thus, [pic]

Hence [pic] and [pic].

Therefore, N (t) = [pic].

Example 33 Find a formula for the normal N (t) of the curve parameterized by

[pic] for [pic] ≤ t ≤ [pic].

Solution [pic]and [pic] = [pic]

Thus, [pic] and [pic]. But [pic].

Therefore, N (t) = [pic].

Example 34 Find a formula for the normal N (t) of the curve parameterized by

[pic] for 0 ≤ t ≤ [pic].

Solution [pic] and [pic] = [pic].

Thus, [pic] 0 < t < [pic] and hence [pic]

Consequently, [pic]

Therefore, N (t) = [pic] for 0 < t < [pic].

Tangential and Normal Components of Acceleration

Note that: Since the tangent vector T and the normal vector N at any point on a smooth curve C are orthogonal, any vector b in the plane determined by T and N can be expressed in the form

| [pic] |[pic] |

|Where [pic] is the tangential component of [pic]. | |

|[pic] is the normal component of [pic]. | |

Suppose an object moves along a curve C. The velocity and acceleration vectors lie in the plane determined by

T and N. Let [pic]be the position vector of the object that moves along the curve C and suppose T and N exist. Then

[pic]

Thus the tangential component of ( is [pic], the speed of the object, and the normal component of the velocity is 0.

Furthermore; the acceleration vector [pic]

[pic]=[pic]. Since [pic] = [pic],

[pic]= [pic], where [pic]and [pic].

Therefore, [pic]= [pic].

The real valued functions[pic] and [pic] are the tangential and normal components of acceleration.

Hence [pic]= [pic] = [pic] = [pic]

Therefore, [pic].

Example 35 Let [pic]. Then find the normal and the tangential components of

acceleration.

Solution [pic] and [pic]

= [pic]=[pic].

Then[pic]= [pic]and [pic].

Thus, [pic]= [pic].

Consequently,[pic]= [pic]= [pic]= 1

Therefore, [pic]= [pic]and [pic]= 1.

Orientation of Curves

Let [pic]be a piecewise smooth parameterization of the curve C. Since each tangent vector points in the direction in which the curve is traced out by [pic], we say that [pic] determines the orientation (or direction) of the curve C.

Note that: Once a piecewise smooth curve C has a given orientation, the tangent vectors to the curve C are

uniquely defined, independent to any parameterization [pic] of C.

Suppose [pic]is a piecewise smooth parameterization of the curve C on [a, b] and let

[pic] for a ≤ t ≤ b.

Then [pic] is a piecewise smooth parameterization of C and determines an orientation opposite to the orientation determined by [pic]. Furthermore; an oriented curve is a piecewise smooth curve with a

particular orientation associated with it.

Example 36 Find a piecewise smooth parameterization for the circle in the plane x = ( 2 centered at the point

(( 2, ( 2, ( 1) with radius 3 whose orientation is clockwise as viewed from the yz-plane.

Solution The parameterization of the circle in the yz-plane centered at the origin with radius 3 units oriented in

a clockwise direction as viewed from the positive x axis direction is:

[pic] for 0 ≤ t ≤ 2(.

Thus, [pic]= [pic]+ [pic]( [pic]= [pic]

Now [pic] is continuous on [0, 2(] and [pic]= [pic].

Since [pic]= 3 ( 0 for 0 ( t ( 2(, [pic] is smooth.

Therefore, [pic] is a piecewise smooth parameterization of the required curve.

Curvature

Let C be a smooth curve. The direction of the tangent vector can vary from point to point according to the nature of the curve.

Example 37

|[pic] | |

| |i) If the curve is a straight line, then T (t) is a constant |

| |vector valued function, and hence |

| |[pic]= 0 |

| | |

| | |

| | |

| | |

| |ii) If the curve undulates gently, then the tangent vector |

| |T (t) changes direction slowly along the curve, and |

| |hence [pic] changes but gently. |

| | |

| | |

| |iii) If the curve twisted, then the tangent vector T (t) |

| |changes rapidly and hence [pic] changes rapidly. |

|[pic] | |

|[pic] | |

Thus, the rate of change of the tangent vector, [pic] is closely related to the rate at which the curve twists and turns.

Since [pic] = [pic] it follows that [pic] = [pic]= [pic].

|Definition 2.13 Let a curve C have a smooth parameterization [pic]such that [pic] |

|is differentiable. Then the curvature k of C is defined by the formula |

|k (t) = [pic]= [pic] |

Example 38 Find the curvature k of the curve traced out by

[pic]

Solution [pic]and [pic]= 1.

Thus, T (t) = [pic]

[pic]and [pic] = 1.

Therefore, k (t) = 1.

Example 39 Find the curvature k of the graph of y = sin x.

Solution The graph of y = sin x is the range of the continuous vector valued function

[pic]

Thus, [pic]and [pic]= [pic].

Hence, T (t) =[pic], [pic]=[pic] and[pic].

Therefore, k (t) =[pic].

Example 40 Find the curvature k of the graph of y = [pic] for x > 0.

Solution The graph of y = [pic] for x > 0 is the range of the continuous vector valued function

[pic] for t > 0

Thus, [pic]and [pic]= [pic]

Now let u = [pic]. Then [pic].

Hence, [pic]and [pic]

Therefore, k (t) =[pic].

| |

|Definition 2.14 The radius of curvature ( (t) of a curve at a point P corresponding |

|to t is given by |

|[pic] |

Example 41 Find the radius of curvature of the curve traced out by

[pic]

Solution [pic] and [pic]= [pic]= [pic]

Then, [pic].

Thus, [pic] and [pic]

Therefore, [pic].

Alternative Formulas for Curvature

Let r be a smooth parameterization of a curve C with tangent T and normal N. Then the velocity and acceleration of an object moving along the curve C with position r are given by:

[pic]and a = [pic]

Hence, [pic]=[pic]=[pic]

Thus, [pic], since [pic]= 1, [pic]= [pic], since[pic]= [pic]

Hence, [pic].

Therefore, k = [pic].

Example 42 Show that the helix [pic]has constant curvature.

Solution [pic]and hence [pic]and a (t) =[pic].

Then [pic]and hence [pic].

Therefore, k (t) = [pic]that is constant.

If [pic] represents an object moving along a curve C in the xy plane, we have

[pic], [pic] and [pic].

Then k = [pic] = [pic].

Example 43 Find the curvature k of the plane curve C parameterized by

[pic]

Solution The parametric equations are: x = 2 cos t, y = 3 sin t

Then [pic] = – 2 sin t, [pic]= 3 cos t, [pic]= – 2 cos t and [pic] = – 3 sin t .

Thus, k = [pic].

Therefore, k = [pic].

Example 44 Find the curvature k of the graph of y = sin x.

Solution [pic] = cos x and [pic] = – sin x.

Therefore, k = [pic].

2.4 Calculus of Vector Fields

In this section we study calculus of a type of functions called a Vector Fields, which assigns vectors to points in space.

| |

|Definition 2.15 A vector field F consists of two parts: a collection D of points in |

|space, called the domain, and a rule, which assigns to each point (x, y, z) |

|in D one and only one vector F (x, y, z). In other words, a vector field is |

|a vector valued function of three variables. |

A vector field F is graphically represented by drawing the vector F (x, y, z) as an arrow emanating from (x, y, z).

|[pic] |[pic] |

Example 45 The gravitational force F (x, y, z) exerted by a point mass m at the origin on a unit mass located

at point (x, y, z) ( (0, 0, 0) is given by:

[pic]

where G is a gravitational constant, [pic]is the unit vector emanating from (x, y, z) and directed

towards the origin. Hence the vector field is called the gravitational field of the point mass.

Note that: [pic] has the same direction as [pic].

Then [pic] = [pic].

Hence, F (x, y, z) =[pic].

If a point (x, y, z) in space is represented by the vector [pic], then the gravitational field can be written as:

F (x, y, z) =[pic], where [pic]= [pic].

A vector field F can be expressed in terms of its components, say M, N and P as follows:

F (x, y, z) = [pic]

In short we can write

F (x, y, z) = [pic].

Note that: M, N and P are scalar fields.

Let F (x, y, z) = [pic] be a vector field, we say F is continuous at (x, y, z) if and only if M, N and P are continuous at (x, y, z).

The Gradient as a Vector Field

Suppose f is a differentiable function of three variables. Then the gradient of f, is a vector field, denoted by grad f or [pic]f and is given by:

grad f (x, y, z) = [pic]= [pic] + [pic] + [pic].

If a vector field F is equal to the gradient of some differentiable function f of several variables, then F is called a conservative vector field, and f is a potential function for F.

Example 46 Show that the gravitational field F of a point mass is a conservative vector field.

Solution F (x, y, z) =[pic], where [pic]= [pic].

Then F (x, y, z) = [pic]for some scalar fields M, N and P.

We need to show that there is a differentiable function f of several variables such that

F = [pic]f.

Now M = [pic], N = [pic] and P = [pic]

Then [pic]= [pic] + k (y, z) = [pic]+ k (y, z)

[pic]= [pic] + ℓ(x, z) = [pic]+ ℓ(x, z)

and [pic]= [pic] + q (x, y) = [pic]+ q (x, y)

Now let f (x, y, z) =[pic], then F = grad f.

Therefore, F is a conservative vector field.

Recovering a Function from its Gradient

A function of several variables can sometimes be recovered from its gradient by successive integration.

Example 47 Find a function f of three variables such that

[pic]

Solution [pic], [pic] and [pic] (*)

Integrating both sides of the last equation in (*) with respect to z we get:

[pic] (**)

where g (x, y) is constant with respect to z.

Now taking partial derivatives of both sides of (**) with respect to x and y respectively, we find that

[pic] + [pic] and [pic] + [pic]

comparing these with the first and the second equations in (*) respectively we get:

[pic] = [pic] = 0

Thus, g (x, y) = c with respect to x, y and z.

Therefore, [pic] , where c is a real number.

Example 48 Find a function f of three variables such that

[pic]

Solution [pic] = [pic], [pic][pic] and [pic] (*)

Integrating both sides of the second equation in (*) with respect to y we get:

[pic] (**)

where g (x, z) is constant with respect to y.

Now taking partial derivatives of both sides of (**) with respect to x and z respectively, we find that

[pic] + [pic] and [pic] [pic]

comparing these with the first and the last equations in (*) respectively we get:

[pic] and [pic] (***)

Integrating the second equation in (***) with respect to z, we get

g (x, z) = [pic] (****)

where k (x) is constant with respect to z.

Differentiating both sides of (****) with respect to x and comparing with the first equation in (***)

we get:

[pic], hence k (x) = c , constant.

Therefore,[pic] , where c is a real number.

Derivatives of a vector Field

There are two types of derivatives of a vector field, one that is a real valued function and the other one is

a vector valued function.

The Divergence of a Vector Field

|Definition 2.16 Let F =[pic]be a vector field such that [pic], |

|[pic] and [pic]exists. Then the divergence of F, denoted div F or |

|[pic]is the function defined by |

|div F (x, y, z) = [pic] |

|= [pic] |

Example 49 Find the divergence of the vector field F, where

[pic]

Solution div F (x, y, z) = [pic] = 0.

Therefore, div F = 0.

Note that: If div F = 0, then F is said to be divergence free or solenoidal.

Example 50 Find the div F, if [pic].

Solution div F (x, y, z) = [pic].

= [pic]

Therefore, div F = [pic].

The Curl of a Vector Field

|Definition 2.17 Let F =[pic]be a vector field such that the first |

|partial derivatives of M, N and P all exist. Then the curl of F, which |

|is denoted curl F or [pic]is the function defined by |

|curl F (x, y, z) = [pic] |

|= [pic] |

The Curl of F is symbolically expressed as:

Curl F = [pic]

Example 51 Find curl F if [pic]

Solution M = y + z, N = x + z and P = x + y.

Then [pic] = [pic]= 1, [pic] = [pic] = 1 and [pic] = [pic]= 1.

Therefore, curl F = 0.

Note that: If curl F = 0, then F is said to be irrotational.

Example 52 Find curl F if [pic].

Solution M = cos x, N = siny and P = [pic].

Then [pic] = [pic], [pic] = [pic]and [pic]= [pic] = [pic] = [pic]= 0.

Therefore, curl F = [pic].

Let f be a scalar field, then

[pic] = div (grad f) = [pic]

The right side of this formula is the Laplacian of f usually denoted by [pic]. A function that satisfies the equation

[pic]= 0

which is known as the Laplace’s equation is said to be harmonic.

Let f, M and N be functions of two variables, and let F (x, y) = [pic], then

grad f (x, y) = [pic], curl F (x, y) = [pic]

div F (x, y) = [pic] and [pic](x, y) = [pic]

Suppose F = [pic] is a vector field such that M, N and P have continuous partial derivatives and if there is a function f such that F = grad f, then curl F = curl (grad f) = 0, But curl F = 0 is equivalent to:

[pic], [pic] and [pic] (*)

Note that: (*) holds for a vector field F = [pic]need not imply that F is conservative.

| |

|Theorem 2. 6 Let F = [pic] be a vector field. If there is a function f |

|having a continuous mixed partial derivatives whose gradient is F, then |

|[pic], [pic] and [pic] |

|If the domain of F is [pic]and if (*) holds, then there is a function f such |

|that F = grad f. |

In case a vector field F is given by

F (x, y) = [pic]

the conditions in (*) reduce to

[pic]

and the corresponding statements in the theorem holds for such vector fields.

Example 53 Let [pic]

and [pic].

Show that F is the gradient of some function but G is not the gradient of any function.

Solution For F we have

[pic], [pic] and [pic].

Since the domain of F is [pic], F is the gradient of some function f.

For G we have

[pic] and [pic], so that the first equation in (*) is not satisfied.

Therefore, G is not the gradient of any function.

Example 54 Let [pic]and [pic].

Show that F is the gradient of some function but G is not the gradient of any function.

Solution For F we have

[pic]

Since the domain of F is [pic], F is the gradient of some function f.

For G we have

[pic] and [pic].

Therefore, G is not the gradient of any function.

Note that: (*) is a necessary condition for a vector field to be a conservative field.

If (*) does not hold, then there is no scalar field f such that F = grad f.

Vector Identities

The notation ( that we saw in the div F and curl F is said to be the del operator and

[pic]

Note that: The derivative operations appearing in the del operator act only on functions appearing

to the right of the del operator.

Let F and G be vector fields having continuous partial derivatives and let f and g be scalar fields. Then

i) div (curl F ) = 0 and curl (grad f) = 0

ii) div (f F) = f div F + [pic]and curl (f F) = f (curl F) + [pic]

iii) div ([pic]) = [pic] – [pic]and div ([pic]) = 0

Exercise

Show that the arc length of a polar graph is given by

[pic]

-----------------------

y

z

x

F(t)

curve C

[pic]

x

y

z

(( 1, 1, 0)

y

x

z

plane x = z

Sphere [pic]= 4

[pic]

y

x

z

H(t)

Circular Helix

[pic]

y

y

x

C

B

P

O

y

t

P

P0

F(t)

F(t0)

F(t)

F(t0)

F'(t0)

F(t0)

y

x

z

y

x

z

x

y

z

t < t0

t > t0

r(t)x

yx

x

((t)

r (b)

z

x

y

r (a) = r (b)

z

x

y

r (a)

Closed curve

curve not closed

N

[pic]

[pic]

T

T

y

x

z

z

x

y

z

x

y

z

y

x

F (x, y, z) = [pic]

x

y

F (x, y, z) = [pic]

P

P0

P

P0

................
................

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