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Unit II 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 . Determine the domain of F.Solution and 1 ≤ t ≤ 1.Therefore, domain of F = .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

, and are real valued functions called the component functions of F.For the function given in example 1, = , = and = 1are the component functions of F.

Graphs of Vector Valued FunctionsWe 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.

Prepared by Tekleyohannes Negussie 1


If , then the parametric equations of the curve C are: , and .Example 2 Let . 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 , and z = 3t.But these are the parametric equations of the line containing ( 1, 1, 0) and parallel to the vector .

Example 3 Let . Then = 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 . 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 , andThus, = = 4 and x = z.Therefore, the curve traces out by F (t) is the intersection of the sphere

Prepared by Tekleyohannes Negussie

y

z

x

F(t)

curve C

kji 32 x

y

z

( 1, 1, 0)

2


= 4 and the plane x = z.

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 , , , , and are defined as follows: i) = ii) = iii) = iv) =

v) =

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

then find i) ii) iii) Solution From the above definition we have i) = = = =

. Therefore, = .


y

x

z

plane x = z

Sphere = 4

ktjtittF cos2sin2cos2)(

3


ii) = = =

=

Therefore, = iii) = = = . Therefore, = .Example 6 Let . Then sketch the curve traced out by H.Solution Let and. 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 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 , then we get a circular helix oriented clockwise as t increasesExample 7 CycloidA 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.


y

x

z

H(t)

Circular Helix

4


Solution Suppose the circle rolled through an angle t radians. Then Hence, and Thus,

Therefore, the vector equation of the cycloid is:

and the parametric equations are: x (t) = and y (t) =

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 , except possibly at itself. A vector L is the limit of F (t) as t approaches (or L is the limit of F at ) if for every 0 there is a 0 such that if , then In this case we write and say that exists.

Theorem 2.1 Let . Then F has a

limit at if and

only if , and have limits at . In this case


y

y

x

C

B

P

O

y

t

5


Example 8 Evaluate .

Solution Let F (t) = . Then

=

Therefore, = .

Theorem 2.2 Let F and G be vector valued functions, and let f and g be real valued functions. Assume that , and exist and that = . Then

i) =

ii) =

iii) =

iv) =

v) =

Example 9 Let F (t) = and G (t) =

.

Then find i) ii)

Solutions i) = =

=



Therefore, = .

ii) = =

=

Therefore, = .

Definition 2.4 A vector valued function F is continuous at a point in its domain if F (t) = F ( )

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

its component functions is continuous at .

Definition 2.5 Let be a number in the domain of a vector valued function F.

If exists, we call the derivative

of F at and write

.

In this case we say F has derivative at , or F is differentiable at or that exists.

Geometric Interpretation of

Let C be the curve traced out by F, and let and P be points on the curve

corresponding to and F(t) respectively.


z z


Now consider the vector . = F (t)

The vector has the same direction as if t > , and has

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

Example 10 Let . Then .Example 11 Let . Then

.

Note that: For any a, b, c and for all t , 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 be any linear vector valued function. Then . Therefore, is a constant vector valued function. Remark: Let F be a vector valued function defined on an interval I.


F(t0)F(t)

xy

t > t0

P0

P

F(t0)F(t)

yx t < t0

P

P0 P0

PF'(t0)

F(t0)y

x

z

Theorem 2.4 Let . Then F is

differentiable at

if and only if , and are differentiable at .

In this case,

8


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 . Then F is differentiable on [0, 1] but not on [ 1, 1], since is not differentiable at t = 0.Solution Left to the reader.

Theorem 2.5 Let F, G and f be differentiable at and let g be differentiable at with g ( ) = . Then the following holds

i) =

ii) =

iii) =

iv) =

v) = =

Example 14 Let and . Then find

and Solution Now and

.

Therefore, = .



Since, =

and = .

Example 15 Let and . Then find and

Solution Now and . Then, =1 and = 0, =

and = .

Therefore, = 1and =

.

Corollary 2.6 Let F be differentiable on an interval I and assume that there is a non-zero number c such that = c for all t in I. Then = 0 for all t in I.

Proof: Since, = c by hypothesis, it follows that for all t in I. Thus, is a constant real valued function on I. Hence, = 0. Therefore, = 0 for all t in I. Note that: If 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) = 0 iii) and are orthogonal. Let . The second derivative of F is defined to be the derivative of , denoted by is given by:

Example 16 Let . Then find .

Solution = and = .

Therefore, = . 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) =

Velocity: v (t) =

Speed:

Acceleration:

Note that: i) The position vector r (t) is called the radial vector or radius vector. ii) If is fixed and t , then the vector r (t) r ( ) is called the displacement vector. iii) The average velocity is defined as

and hence the velocity is the limit of the average velocity.Example 17 Suppose the position of an object is given by r(t) = . Determine the velocity, speed and acceleration of the object.Solution v (t) = ,

and .



Example 18 Let r(t) = . Find the set of all possible values of t for which = 0.Solution v (t) = . Hence, = 0 = 0 and = 0. and = 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 > 0 with a constant speed > 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.

Then r(t) = and hence, v (t) = . Since the object moves counterclockwise, is increasing and hence > 0.Thus, = and hence

Now since (0) = 0 we get .

Therefore, r(t) = , v(t) =

and a(t) = .


r(t)x

yx

x

(t)

12


Integration of Vector valued Functions

Definition 2.6 Let , where ,

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

and the indefinite integral are

defined by

and

Example 20 Let . Find and .

Solution

=

=

Therefore, =

= = , where C is a constant vector. Therefore, = , where C is a constant vector.



If , where , and are continuously differentiable, then

=

=

= + .

Therefore, = , 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 to denote a vector valued function whose range is a curve C. In this case we say that C is parameterized by or that is a parameterization of C.

Suppose . Then x = , y = and z = are parametric equations of the vector valued function .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 for all t in I. Then is continuous and traces out the graph of f. Thus, the graph of f is the range of , so the graph of f is a curve.Hence, is its parameterization.Note that: The vector valued function , c ≠ 0



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

Properties of Space Curves

Definition 2.8 A curve C is closed if it has a parameterization whose domain is a closed and bounded interval [a, b] such that .

Example 21 Let for t [0, 2]. Now is continuous on [0, 2] and = = .Therefore, the curve traced out by is closed curve.Example 22 Let for t [0, 2]. Now is continuous on [0, 2] and = while .Therefore, the curve traced out by is not closed.

Definition 2.9 a) A vector valued function defined on an interval I is smooth if has a continuous derivative on I and 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 defined on an interval I is piecewise smooth if I is composed of a finite number of subintervals on each of which is smooth and if has one


yx

z

r (a) = r (b)

Closed curve

yx

z

r (b)r (a)

curve not closed

15


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 is smooth.Solution , is continuous and 0 for every t .Therefore, is smooth.Example 24 Show that is piecewise smooth.

Solution for all t and hence is

continuous for all t . But = 0 only for t = 0.Therefore, is piecewise smooth on (– , ).Example 25 Find a smooth parameterization of the line segment from

to .

Solution Suppose and are distinct points in space.

x = + (x – ) t, y = + (y – ) t and z = +(z – ) t

are parametric equations for the line through and .

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

Therefore, = for 0 ≤ t ≤ 1 is a

smooth parameterization of the segment from to .Length of a Curve

Let C be the line segment joining the points to in space. Then the length L of C is given by: .Now, let C be a smooth curve and = a ≤ t ≤ b, be a smooth parameterization of C. Let T = { } be any partition of [a,

b], and let be the length of the portion of C.

If is small, then

.



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

and , where .

Hence .

Therefore, the total length L of C is:

L =

and for small , . But

.

Therefore, .

Definition 2.10 Let C be a curve with piecewise smooth parameterization defined on [a, b]. Then the length L of the curve C is defined by:

Example 26 Find the length L of the segment of the circular helix for 0 ≤ t ≤ 2. Solution By definition 2.10,

= = .

Therefore, L = units.Example 27 Find the length L of the curve for 1 ≤ t ≤ 2.



Solution Now and .

Hence, = = = 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 FunctionLet C be a smooth curve parameterized on an interval I = [a, b] by for t in I.Let a be a fixed number in I. We define the arc length function s by:

= 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

=

> 0, since is smooth for all t in I.Thus, s (t) is increasing and hence s has an inverse.

Example 28 Find if .

Solution and hence = .

Therefore, = for t ≥ 0.Note that: Any quantity depending on t also depends on s. Arc Length s as parameterIf 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

.



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

.

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

.

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

Arc length in Polar formLet C be a smooth curve parameterized on an interval I by for t in I. and y = for t in I.

Hence, and

Thus,

and

Consequently, Or equivalently,

Tangents and Normals to CurvesTangents to Curves

Definition 2.11 Let C be a smooth curve and a (smooth) parameterization of C defined on an interval I. Then for any interior point t of I, the tangent vector at the point r (t) is defined by

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



Example 30 Find a formula for the tangent T (t) to the circular helix Solution and = .Therefore, .Note that: T (t) and r (t) are not orthogonal, since constant for all t.Example 31 Find the tangent vector T (t) to the curve parameterized by r (t), where for ≤ t ≤ . Solution and = .Therefore, .Normals to CurvesLet C be a smooth curve and r (t) be a parameterization of C. Let be also smooth. Then the tangent vector T (t) is differentiable. Moreover; if = 1 for all t in the domain of T. But then = 0. Hence if 0, then and T (t) are orthogonal.

Definition 2.12 Let C be a smooth curve and a smooth parameterization of C defined on an interval I such that is smooth. Then for any interior point t of I for which 0, the normal vector N (t) at the point r (t) is defined by

Example 32 Find a formula for the normal N (t) of the curve parameterized by Solution and = . Thus,

Hence and . Therefore, N (t) = .Example 33 Find a formula for the normal N (t) of the curve parameterized by

for ≤ t ≤ . Solution and =



Thus, and . But .

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

for 0 ≤ t ≤ . Solution and = .

Thus, 0 < t < and hence

Consequently,

Therefore, N (t) = for 0 < t < .

Tangential and Normal Components of AccelerationNote 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

Where is the tangential component of . is the normal component of

.

Suppose an object moves along a curve C. The velocity and acceleration vectors lie in the plane determined by T and N. Let be the position vector of the object that moves along the curve C and suppose T and N exist. Then

Thus the tangential component of is , the speed of the object, and the normal component of the velocity is 0.Furthermore; the acceleration vector

= . Since = ,


N NbTbT

T

21


= , where and .Therefore, = . The real valued functions and are the tangential and normal components

of acceleration. Hence = = =

Therefore, .Example 35 Let . Then find the normal and the tangential components of acceleration.Solution and = = . Then = and . Thus, = . Consequently, = = = 1Therefore, = and = 1.Orientation of Curves

Let 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 , we say that 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 of C. Suppose is a piecewise smooth parameterization of the curve C on [a, b] and let for a ≤ t ≤ b.Then is a piecewise smooth parameterization of C and determines an orientation opposite to the orientation determined by . 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: for 0 ≤ t ≤ 2. Thus, = + =

Now is continuous on [0, 2] and = . Since = 3 0 for 0 t 2, is smooth. Therefore, 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

i) If the curve is a straight line, then T (t) is a constant vector valued function, and hence

= 0

ii) If the curve undulates gently, then the tangent vector T (t) changes direction slowly along the curve, and

hence changes but gently.

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

changes rapidly and hence changes

rapidly.


yx

z

yx

z

yx

z

23


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

Since = it follows that = = .

Definition 2.13 Let a curve C have a smooth parameterization such that

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

k (t) = =

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

Solution and = 1.

Thus, T (t) =

and = 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 Thus, and = .

Hence, T (t) = , = and

.

Therefore, k (t) = .

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



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

Thus, and =

Now let u = . Then .

Hence, and

Therefore, k (t) = .

Definition 2.14 The radius of curvature (t) of a curve at a point P corresponding to t is given by

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

Solution and = =

Then, .

Thus, and

Therefore, .

Alternative Formulas for CurvatureLet 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: and a =



Hence, = =

Thus, , since = 1, = , since=

Hence, .

Therefore, k = .

Example 42 Show that the helix has constant curvature.Solution and hence and a (t) =

. Then and hence .Therefore, k (t) = that is constant. If represents an object moving along a curve C in the xy plane, we have

, and .

Then k = = .

Example 43 Find the curvature k of the plane curve C parameterized by Solution The parametric equations are: x = 2 cos t, y = 3 sin t

Then = – 2 sin t, = 3 cos t, = – 2 cos t and = – 3 sin t .

Thus, k = .

Therefore, k = .

Example 44 Find the curvature k of the graph of y = sin x.Solution = cos x and = – sin x.

Therefore, k = .

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).

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:

where G is a gravitational constant, 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: has the same direction as .

Then = .

Hence, F (x, y, z) = .


z

y

xF (x, y, z) =

y

x

27

F (x, y, z) =


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

F (x, y, z) = , where = .A vector field F can be expressed in terms of its components, say M, N and P as follows: F (x, y, z) = In short we can write F (x, y, z) = .Note that: M, N and P are scalar fields.Let F (x, y, z) = 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 f and is given by:

grad f (x, y, z) = = + +

.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) = , where = .

Then F (x, y, z) = 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 = f. Now M = , N = and P =

Then = + k (y, z) = + k (y,

z)



= + ℓ(x, z) = + ℓ(x,

z)

and = + q (x, y) = + q

(x, y) Now let f (x, y, z) = , then F = grad f. Therefore, F is a conservative vector field.

Recovering a Function from its GradientA 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 Solution , and

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

(**) 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

+ and + comparing these with the first and the second equations in (*)

respectively we get:

= = 0 Thus, g (x, y) = c with respect to x, y and z. Therefore, , where c is a real number.Example 48 Find a function f of three variables such that

Solution = , and (*)



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

(**) 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

+ and comparing these with the first and the last equations in (*) respectively

we get:

and (***)

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

(****) 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:

, hence k (x) = c , constant. Therefore, , where c is a real number.

Derivatives of a vector FieldThere 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 = be a vector field such that ,

and exists. Then the divergence of F, denoted div F or is the function defined by div F (x, y, z) = =



Example 49 Find the divergence of the vector field F, where Solution div F (x, y, z) = = 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 .

Solution div F (x, y, z) = .

= Therefore, div F = . The Curl of a Vector Field

Definition 2.17 Let F = 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 is the function defined by curl F (x, y, z) = =

The Curl of F is symbolically expressed as:

Curl F =

Example 51 Find curl F if Solution M = y + z, N = x + z and P = x + y. Then = = 1, = = 1 and = = 1. Therefore, curl F = 0.

Note that: If curl F = 0, then F is said to be irrotational.Example 52 Find curl F if .



Solution M = cos x, N = siny and P = .

Then = , = and = = = = 0. Therefore, curl F = . Let f be a scalar field, then

= div (grad f) =

The right side of this formula is the Laplacian of f usually denoted by . A function that satisfies the equation = 0which 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) =

, then

grad f (x, y) = , curl F (x, y) =

div F (x, y) = and (x, y) =

Suppose F = 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:

, and (*)Note that: (*) holds for a vector field F = need not imply that F is conservative.

Theorem 2. 6 Let F = be a vector field. If there is a function f having a continuous mixed partial derivatives whose gradient is F, then



, and If the domain of F is 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) = the conditions in (*) reduce to and the corresponding statements in the theorem holds for such vector fields.

Example 53 Let and . Show that F is the gradient of some function but G is not the gradient of any function.Solution For F we have

, and . Since the domain of F is , F is the gradient of some function f.

For G we have and , so that the first equation in (*) is not satisfied. Therefore, G is not the gradient of any function.

Example 54 Let and . Show that F is the gradient of some function but G is not the gradient of any function.Solution For F we have Since the domain of F is , F is the gradient of some function f.

For G we have and . 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 IdentitiesThe notation that we saw in the div F and curl F is said to be the del operator and 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 + and curl (f F) = f (curl F) +

iii) div ( ) = – and div ( ) = 0

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


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