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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