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 F (t )  1  t i  1  t j  k . Determine the domain of F.
Solution 1  t  0 and 1  t  0   1 ≤ t ≤ 1.
Therefore, domain of F = 1, 1 .
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
F (t )  f1 (t ) i  f 2 (t ) j  f3 (t ) k
f1 , f 2 and f 3 are real valued functions called the component functions of F.
For the function given in example 1,
f1 (t ) = 1  t , f 2 (t ) = 1  t and f 3 (t ) = 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.
z
curve C
F(t)
y
x
If F (t )  f1 (t ) i  f 2 (t ) j  f 3 (t ) k , then the parametric equations of the curve C are:
x  f1 (t ) , y  f 2 (t ) and z  f 3 (t ) .
Example 2 Let F (t )  (2t 1) i  (1 t ) j  3t k . Sketch the curve traced out by F.
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Solution If we let (x, y, z) be a point on the curve traced out by F (t), then
x  2t 1 , y 1  t and z = 3t.
But these are the parametric equations of the line containing ( 1, 1, 0) and parallel to the vector
2i  j  3k .
z
( 1, 1, 0)
2i  j  3k
y
x
Example 3 Let F (t )  cos t i  sin t j . Then
F (t )  cos 2 t  sin 2 t = 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 F (t )  2 cos t i  2 sin t j  2 cos t k . 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
x  2 cos t , y   2 sin t and z  2 cos t
Thus, x 2  y 2  z 2 = 2 cos 2 t  4 sin 2 t  2 cos 2 t = 4 and x = z.
Therefore, the curve traces out by F (t) is the intersection of the sphere
x 2  y 2  z 2 = 4 and the plane x = z.
z
Sphere
x2  y2  z 2 = 4
plane x = z
x
y
F (t )  2 cos t i  2 sin t j  2 cos t k
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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 F  G , F  G , f F , F  G , F  G and F  g
are defined as follows:
i) ( F  G ) (t ) = F (t )  G (t )
iii) ( F  G ) (t ) = F (t )  G (t )
ii) ( f F ) (t ) = f (t ) F (t )
iv) ( F  G ) (t ) = F (t )  G (t )
v) ( F  g ) (t ) = F ( g (t ))
Example 5 Let F (t )  2t i  (1  t ) j  3t k and G(t )  (1  t ) i  (3t  2) j  t k . If g (t) = cos t,
then find
i) ( F  G ) (t )
ii) ( F  G ) (t )
iii) ( F  g ) (t )
Solution From the above definition we have
i) ( F  G ) (t ) = F (t )  G (t ) = (2t ,1  t ,  3t )  (1  t , 3t  2, t )
= 2t (1  t )  (1  t ) (3t  2)  3t (t ) = 2t 2  3t  2 .
Therefore, ( F  G ) (t ) = 2t 2  3t  2 .
ii) ( F  G ) (t ) = F (t )  G (t ) = (2t ,1  t ,  3t )  (1  t , 3t  2, t )
= 5t (2t 1) i  t (5t  3) j  (5t 2  6t 1) k
= (10t 2  5t ) i  (5t 2  3t ) j  (5t 2  6t 1) k
Therefore, ( F  G ) (t ) = (10t 2  5t ) i  (5t 2  3t ) j  (5t 2  6t 1) k
iii) ( F  g ) (t ) = F ( g (t )) = F (cos t ) = 2 cos t i  (1  cos t ) j  3 cos t k .
Therefore, ( F  g ) (t ) = 2 cos t i  (1  cos t ) j  3 cos t k .
Example 6 Let H (t )  cos t i  sin t j  t k . Then sketch the curve traced out by H.
Solution Let F (t )  cos t i  sin t j and G(t )  t k .
z
Then H (t) = F (t) + G (t).
H(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 t units
y
above or below the points corresponding to F (t).
x
Therefore, the curved traced out by H (t) is a circular Helix.
Circular Helix
H (t )  cos t i  sin t j  t k
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Note that: If H (t )  cos t i  sin t j  t k , 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
radians. Then
y
OP  OC  CP
Hence, OC  r t i  r j
y
and CP   r sin t i  r cos t j
P
Thus, OP  r (t  sin t ) i  r (1  cos t ) j
Therefore, the vector equation of the cycloid is:
y
t
OP  r (t  sin t ) i  r (1  cos t ) j
C
and the parametric equations are:
x (t) = r (t  sin t ) and y (t) = r (1  cos t )
O
B
x
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 t 0 , except possibly at t 0 itself. A vector L is the limit of
F (t) as t approaches t 0 (or L is the limit of F at t 0 ) if for every   0 there is
a   0 such that
if 0  t  t 0   , then F (t )  L  
In this case we write
im F (t )  L and say that im F (t ) exists.
t  t0
t  t0
Theorem 2.1 Let F (t )  f1 (t ) i  f 2 (t ) j  f 3 (t ) k . Then F has a limit at t 0 if and
only if f1 (t ) , f 2 (t ) and f 3 (t ) have limits at t 0 . In this case
im F (t )   im f (t ) i   im f (t )  j   im f (t ) k






1
2
3
t  t0
t  t0

t  t0

 t  t0

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 sin t

i  et j  (t  2 ) k  .
Example 8 Evaluate im 
t 0 t

sin t
i  et j  (t  2 ) k . Then
Solution Let F (t) =
t
im F (t )   im sin t  i   im et  j   im (t  2 ) k = i  j  2 k






t  t0
t  0 t 
t  0 
 t  t0

 sin t

i  et j  (t  2 ) k  = i  j  2 k .
Therefore, im 
t 0 t

Theorem 2.2 Let F and G be vector valued functions, and let f and g be real valued functions.
Assume that im F (t ) , im G (t ) and im f (t ) exist and that
t  t0
t  t0
t  t0
im g ( s ) = t . Then
0
s  s0
im
i) im ( F  G ) (t ) = im F (t ) 
G (t )
t  t0
t  t0
t  t0
im ( f F ) (t ) = im f (t ) im F (t )
t  t0
t  t0
t  t0
im
iii) im ( F  G ) (t ) = im F (t ) 
G (t )
t  t0
t  t0
t  t0
ii)
iv) im ( F  G ) (t ) =
t  t0
v)
Example 9 Let F (t) =
im F (t )  im G (t )
t  t0
t  t0
im
im
( F  g ) (t ) =
F ( g (t ))
t  t0
t  t0
sin (t 1)
t 3
t 3
i
j  cos  t k and G (t) = (1  t 2 ) i 
j  t 2 1 k .
t 1
t 2
t 3
Then find i) im ( F  G ) (t )
ii) im ( F  G ) (t )
t 1
t 1
1
Solutions i) im ( F  G ) (t ) = im F (t )  im G (t ) = ( i ,  4 j ,  k )  ( 2i , j ,  2 k )
t 1
t 1
t 1
=1
4
2
Therefore, im ( F  G ) (t ) = 1 
t 1
2.
im
1
G (t ) = ( i ,  4 j ,  k )  ( 2i , j ,  2 k )
ii) im ( F  G ) (t ) = im F (t ) 
t 1
t 1
= (4 2 
t 1
4
1
33
) i  ( 2  2) j 
k
4
4
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Vector Differential Calculus
1
33
k .
Therefore, im ( F  G ) (t ) = (4 2  ) i  ( 2  2) j 
t 1
4
4
Definition 2.4 A vector valued function F is continuous at a point t 0 in its domain if
im
F (t) = F ( t 0 )
t  t0
Theorem 2.3 A vector valued function F is continuous at a point t 0 if and only if each of
its component functions is continuous at t 0 .
Definition 2.5 Let t 0 be a number in the domain of a vector valued function F.
If
im F (t )  F (t 0 )
exists, we call the derivative of F at t 0 and write
t  t0
t  t0
F ' (t 0 ) 
im F (t )  F (t 0 )
.
t  t0
t  t0
In this case we say F has derivative at t 0 , or F is differentiable at t 0 or that
F ' (t 0 ) exists.
Geometric Interpretation of F ' (t 0 )
Let C be the curve traced out by F, and let P0 and P be points on the curve corresponding to F (t 0 ) and F(t)
respectively.
z
z
P
P0
P
F(t)
F(t0)
x
P
P0
y
t > t0
F(t)
F(t0)
x
y
F'(t0)
P0
F(t0)
t < t0
z
y
x
Now consider the vector P0 P .
P0 P = F (t)  F (t 0 )
The vector
F (t )  F (t 0 )
has the same direction as P0 P if t > t 0 , and has opposite direction to P0 P if
t  t0
t < t 0 . Hence F ' (t 0 ) exists, then it points in the same direction in which C is traced out by F. F ' (t 0 ) is
tangent to the curve C at P0 .
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Theorem 2.4 Let F (t )  f1 (t ) i  f 2 (t ) j  f 3 (t ) k . Then F is differentiable at t 0
if and only if f1 (t ) , f 2 (t ) and f 3 (t ) are differentiable at t 0 .
In this case, F ' (t 0 )  f1 ' (t 0 ) i  f 2 ' (t 0 ) j  f 3 ' (t 0 ) k
Example 10 Let F (t )  sin t 2 i  cos t j . Then F ' (t )  2 t cos t 2 i  sin t j .
sin t
j  t 2 k . Then
t
t cos t  sin t
G ' (t )   2 sin t i 
j  2t k .
t2
Note that: For any a, b, c  and for all t  , F (t )  a i  b j  c k is a constant vector valued function.
Example 11 Let G (t )  2 cos t i 
Example 12 Show that the derivative of a linear vector valued function is a constant vector valued function.
Solution Let F (t )  ( x0  a t ) i  ( y0  b t ) j  ( z 0  c t ) k be any linear vector valued function.
Then F ' (t )  a i  b j  c k .
Therefore, F ' (t ) 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 F (t )  t i  1  t j . Then F is differentiable on [0, 1] but not on [ 1, 1], since t
is not differentiable at t = 0.
Solution Left to the reader.
Theorem 2.5 Let F, G and f be differentiable at t 0 and let g be differentiable
at s 0 with g ( s 0 ) = t 0 . Then the following holds
i) ( F  G ) ' (t 0 ) = F ' (t 0 )  G ' (t 0 )
ii) ( f F ) ' (t 0 ) = f ' (t 0 ) F (t 0 )  f (t 0 ) F ' (t 0 )
iii) ( F  G ) ' (t 0 ) = F ' (t 0 )  G (t 0 )  F (t 0 )  G ' (t 0 )
iv) ( F  G ) ' (t 0 ) = F ' (t 0 )  G (t 0 )  F (t 0 )  G ' (t 0 )
v) ( F  g ) ' ( s 0 ) = F ' ( g ( s 0 )) g ' ( s 0 ) = F ' (t 0 ) g ' ( s 0 )
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Example 14 Let F (t )  i  t j  t 5 k and G (t )  e t cos t i  e t sin t k . Then find
( F  G) ' (t ) and ( F  G) ' (t )
Solution Now F ' (t )  j  5 t 4 k and G ' (t )  et ( cos t  sin t ) i  et (sin t  cost ) k .
Therefore, ( F  G) ' (t ) = (1  t 5 ) e t cos t  (t 5  5 t 4  1) e t sin t .
Since, F ' (t )  G (t ) =  e t (sin t i  5 t 4 cos t j  cos t k )


and F (t )  G ' (t ) =  e t t (sin t  cos t ) i  (( t 5 1) sin t  ( t 5  1) cos t ) j  (cos t  sin t ) k .



Therefore, ( F  G ) ' (t )   e t (t  1) sin t  t cos t ) i  ( t 5  1) sin t  ( t 5  5t 4  1) cos t j

 e t 2 cos t  sin t  k

Example 15 Let F (t )  t i  5 cos t k and G (t )  i  n t j . Then find
( F  G) ' (t ) and ( F  G) ' (t )
1
j .
t
Then, F ' (t )  G (t ) =1 and F (t )  G ' (t ) = 0, ( F ' G) (t ) = 5 n t sin t i  5 sin t j  n t k
Solution Now F ' (t )  i  5 sin t k and G ' (t ) 
and ( F  G' ) (t ) = 
5
cos t i  k .
t


Therefore, ( F  G) ' (t ) = 1and ( F  G) ' (t ) =  5 n t sin t 
5

cos t  i  5 sin t j  (1  n t ) k .
t

Corollary 2.6 Let F be differentiable on an interval I and assume that there is a
non-zero number c such that
F (t ) = c for all t in I.
Then F (t )  F ' (t ) = 0 for all t in I.
Proof: Since, F (t ) = c by hypothesis, it follows that
( F  F ) (t )  F (t )  F (t )  F (t )
2
 c 2 for all t in I.
Thus, F  F is a constant real valued function on I.
Hence, ( F  F ) ' (t ) = 0.
Therefore, F (t )  F ' (t ) = 0 for all t in I.
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Note that: If F (t ) 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) F ' (t ) = 0
iii) F (t ) and F ' (t ) are orthogonal.
Let F (t )  f1 (t ) i  f 2 (t ) j  f 3 (t ) k . The second derivative of F is defined to be the derivative of
F ' (t ) , denoted by F ' ' (t ) is given by:
F ' ' (t )  f1 ' ' (t ) i  f 2 ' ' (t ) j  f 3 ' ' (t ) k
Example 16 Let F (t )  i  n t j  2 t k . Then find F ' ' (1) .
1
1
j  2 k and F ' ' (t ) =  2 j .
t
t
Therefore, F ' ' (t ) =  j .
Solution F ' (t ) =
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) = x (t ) i  y (t ) j  z (t ) k
Velocity: v (t) =
Speed: v (t ) 
dx
dy
dz
i
j
k
dt
dt
dt
 dx  2  dy  2  dz  2
     
 dt 
 dt 
 dt 
Acceleration: a (t ) 
dv d 2 r
d 2x
d2y
d 2z
 2 
i

j

k
dt
dt
dt 2
dt 2
dt 2
Note that: i) The position vector r (t) is called the radial vector or radius vector.
ii) If t 0 is fixed and t  t 0 , then the vector r (t)  r ( t 0 ) is called the
displacement vector.
iii) The average velocity is defined as
r (t )  r (t 0 )
t  t0
and hence the velocity is the limit of the average velocity.
Example 17 Suppose the position of an object is given by
r(t) = cosh t i  sinh t j  t k .
Determine the velocity, speed and acceleration of the object.
Solution v (t) = sinh t i  cosh t j  k , v (t ) 
cosh 2 t  sinh 2 t 1  2 cosh t
and a (t )  cosh t i  sinh t j .
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Example 18 Let r(t) = r (t  sin t ) i  r (1  cos t ) j . Find the set of all possible values of t
for which v (t ) = 0.
Solution v (t) = r (1  cos t ) i  r sin t j .
Hence, v (t ) = 0  r (1  cos t ) = 0 and r sin t = 0.
 cos t  1 and sin t = 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 r0 > 0 with a constant
speed v 0 > 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.


v (t) = r  ' (t ) sin  (t ) i  cos  (t ) j .
Then r(t) = r0 cos  (t ) i  sin  (t ) j and hence,
y
x
(t)
r(t)
x
0
Since the object moves counterclockwise,  is increasing
x
and hence  ' (t ) > 0.
v
Thus, v (t ) = r0  ' (t ) and hence  ' (t )  0
r0
v
Now since  (0) = 0 we get  (t )  0 t .
r0




v
v
v
v
Therefore, r(t) = r0 cos 0 t i  sin 0 t j  , v(t) = v0  sin 0 t i  cos 0 t j 
r0
r0
r0
r0





v 
v
v
and a(t) = 0  cos 0 t i  sin 0 t j  .
2
r0 
r0
r0

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Integration of Vector valued Functions
Definition 2.6 Let F (t )  f1 (t ) i  f 2 (t ) j  f 3 (t ) k , where f1 , f 2 and f 3 are
continuous real valued functions on [a, b]. Then the definite integral
b
a F (t ) dt and the indefinite integral  F (t ) dt are defined by
b

b

b

 f (t ) dt  i   f (t ) dt  j   f (t ) dt  k
F
(
t
)
dt

a
 1

 2

 3

a

a

a

b

and F (t ) dt 
 f1 (t) dt  i   f 2 (t) dt  j   f3 (t) dt k
1
Example 20 Let F (t )  cosh t i  sinh t j  k . Find  F (t ) dt and  F (t ) dt .
0
1


 1 


F
(
t
)
dt

cosh
t
dt
i

sinh
t
dt
j





 dt  k
0
 0

 0

 0 
1
1
1
= sinh t  i  cosh t  j   t  k
0
0
0
1
Solution
1
 e 2  1   (e  1) 2 
i 

  2e  j  t k
2
e

 

1
 e 2  1   (e  1) 2 
i 

Therefore,  F (t ) dt = 
  2e  j  t k
2
e

 

0
= 
 F (t ) dt =  cosh t dt i   sinh t dt  j   dtk
= sinh t  i  cosh t  j   t  k  C , where C is a constant vector.
Therefore,
 F (t ) dt = sinh t  i  cosh t  j  t  k
 C , where C is a constant vector.
If F (t )  f1 (t ) i  f 2 (t ) j  f 3 (t ) k , where f1 , f 2 and f 3 are continuously differentiable, then
 F ' (t ) dt    f1 ' (t ) i  f 2 ' (t )
=
=
j  f 3 ' (t ) k

 f1 '(t) dt  i   f 2 '(t) dt  j   f3 ' (t) dt k
 f1 (t )  c1  i   f 2 (t )  c2  j   f 3 (t )  c3 k


= f1 (t ) i  f 2 (t ) j  f 3 (t ) k + c1 i  c2 j  c3 k .
Therefore,  F ' (t ) dt = f1 (t ) i  f 2 (t ) j  f 3 (t ) k  C , where C is a constant vector.
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Unit II
Vector Differential Calculus
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 r to denote a vector valued function
whose range is a curve C. In this case we say that C is parameterized by r or that r
is a parameterization of C.
Suppose r (t )  x (t ) i  y (t ) j  z (t ) k . Then x = x (t ) , y = y (t ) and z = z (t ) are parametric
equations of the vector valued function r (t ) .
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 r (t )  t i  f (t ) j for all t in I. Then r (t ) is continuous and traces out the graph of f. Thus, the
graph of f is the range of r (t ) , so the graph of f is a curve.
Hence, r (t )  t i  f (t ) j is its parameterization.
Note that: The vector valued function
r (t )  a cos t i  a sin t j  c t k , c ≠ 0
represents a curve called circular helix. It lies on the cylinder x 2  y 2  a 2 .
Properties of Space Curves
Definition 2.8 A curve C is closed if it has a parameterization r (t ) whose domain
is a closed and bounded interval [a, b] such that r (a)  r (b) .
z
z
r (a)
r (b)
r (a) = r (b)
y
x
Closed curve
y
x
curve not closed
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Unit II
Vector Differential Calculus
Example 21 Let r (t )  cos t i  sin t j for t  [0, 2].
Now r (t ) is continuous on [0, 2] and r (0) = i = r (2  ) .
Therefore, the curve traced out by r (t ) is closed curve.
Example 22 Let r (t )  cos t i  sin t j  t k for t  [0, 2].
Now r (t ) is continuous on [0, 2] and r (0) = i while r (2  )  i  2 k .
Therefore, the curve traced out by r (t ) is not closed.
Definition 2.9 a) A vector valued function r (t ) defined on an interval I is smooth if r (t )
has a continuous derivative on I and r ' (t )  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 r (t ) defined on an interval I is
piecewise smooth if I is composed of a finite number of subintervals
on each of which r (t ) is smooth and if r (t ) 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 r (t )  sin t i  cos t j  t 2 k is smooth.
Solution r ' (t )  cos t i  sin t j  2 t k , r ' (t ) is continuous and r ' (t )  0 for every t  .
Therefore, r (t ) is smooth.
Example 24 Show that r (t )  (et  t ) i  t 2 j  t 3 k is piecewise smooth.
Solution r ' (t )  (et 1) i  2 t j  3 t 2 k for all t   and hence r ' (t ) is continuous for all t  . But
r ' (t ) = 0 only for t = 0.
Therefore, r (t ) is piecewise smooth on (– , ).
Example 25 Find a smooth parameterization of the line segment from x0 , y 0 , z 0  to x1, y1, z1  .
Solution Suppose x0 , y 0 , z 0  and x1, y1, z1  are distinct points in space.
x = x0 + (x – x0 ) t, y = y0 + (y – y0 ) t and z = z 0 +(z – z 0 ) t
are parametric equations for the line through x0 , y 0 , z 0  and x1, y1, z1  .
Now (x, y, z) = x0 , y 0 , z 0  for t = 0 and (x, y, z) = x1, y1, z1  for t = 1.
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Unit II
Vector Differential Calculus
Therefore, r (t ) = x0  ( x  x0 )i  y0  ( y  y0 ) j  z0  ( z  z0 )k for 0 ≤ t ≤ 1 is a
smooth parameterization of the segment from x0 , y 0 , z 0  to x1, y1, z1  .
Length of a Curve
Let C be the line segment joining the points x0 , y 0 , z 0  to x1, y1, z1  in space. Then the length L of C is
given by:
L  ( x1  x0 ) 2

( y1  y0 ) 2

( z1  z 0 ) 2 .
Now, let C be a smooth curve and r (t ) = x (t ) i  y (t ) j  z (t ) k a ≤ t ≤ b, be a smooth
parameterization of C. Let T = { t0 , t1, t2 , ..., tn } be any partition of [a, b], and let Lk be the length of the
k th portion of C.
If t k is small, then
L k 
 x (t
k
 y (t
)  x(tk  1 )
2
k
 z (t
)  y (tk  1 )
2
k

)  z (tk  1 ) 2 .
By the mean-value theorem there are numbers u k , v k and k in [ t k  1 , t k ] such that
x (t k )  x(tk  1 )  x ' (u k )  t k , y (t k )  y (tk  1 )  y ' (v k )  t k
and z (t k )  z (t k  1 )  z ' ( k )  t k , where  tk  tk  tk  1 .
 x ' (u k ) 2  y ' (v k ) 2  z ' ( k ) 2 t k .
Hence L k 
Therefore, the total length L of C is:
n
L=
 L k
k 1
and for small t k , L 
b
a  x ' (t ) 2  y ' (t ) 2  z ' (t ) 2 dt . But
 x ' (t ) 2  y ' (t ) 2  z ' (t ) 2 
Therefore, L 
r ' (t ) .
b
a
r ' (t ) dt .
Definition 2.10 Let C be a curve with piecewise smooth parameterization
r (t ) defined on [a, b]. Then the length L of the curve C is defined by:
L
b
a
r ' (t ) dt
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Unit II
Vector Differential Calculus
Example 26 Find the length L of the segment of the circular helix
r (t )  cos t i  sin t j  t k
for 0 ≤ t ≤ 2.
Solution By definition 2.10,
2
 sin t  2  cos t  2
0
L
2
 1 dt =
0
2 dt = 2 2  .
Therefore, L = 2 2  units.
Example 27 Find the length L of the curve
r (t )  2t i  t 2 j  n t k
for 1 ≤ t ≤ 2.
Solution Now r ' (t )  2 i  2 t j 
Hence, L 
2
1
4  4t 2 
1
k and r ' (t ) 
t
1
t2
dt = L 
2
1
4  4t 2 
1
.
t2

 2t  1  dt = L  t 2 

t 

n t

2
= 3 + ℓn 2.
1
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
r (t )  x (t ) i  y (t ) j  z (t ) k for t in I.
Let a be a fixed number in I. We define the arc length function s by:
s (t ) 
t
a
t
r ' (u ) du =
2
2
2
a x ' (u )  y ' (u )  z ' (u )
du 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
s ' (t )  r ' (t ) =
x ' (t )2  y ' (t )2  z ' (t )2
s (t )
r (t )

> 0, since r (t ) is smooth for all t in I.
dt
dt
Thus, s (t) is increasing and hence s has an inverse.
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Unit II
Vector Differential Calculus
3
2
s (t )
if r (t )  sin 2t i  cos 2t j  t 2 k .
3
dt
1
r (t )
Solution r ' (t )  2 cos 2t i  2 sin 2t j  t 2 k and hence
=
dt
s (t )
Therefore,
= 4  t for t ≥ 0.
dt
Example 28 Find
4t.
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
r (t )  a cos t i  a sin t j  c t k .
Then r ' (t )   a sin t i  a cos t j  c k and
r ' (t ) 
a2  c2 .
Thus, the arc length function s (t) is given by:
s (t ) 
t
0
a 2  c 2 du  t a 2  c 2  t 
s
a c
2
2
.
Therefore, a formula for the helix using the arc length function s as a parameter is





cs
s
s
 i  a sin 
 j 
r ( s )  a cos 
 2




2
2
2
2
2
 a c 
 a c 
 a c
s
If we set c = 0, then we get t 
which is a parameterization for a circle.
a

k .


 s 


 i  a sin  s  j
r (t )  a cos 

 a 
 a 


Arc length in Polar form
Let C be a smooth curve parameterized on an interval I by
r (t )  x (t ) i  y (t ) j for t in I.
x  r cos  and y = r sin  for t in I.
Hence,
dx dr
dy dr
d
d

cos   r sin 

sin   r cos 
and
dt dt
dt
dt dt
dt
dr
d
 dx  2  dr  2
 d  2
2
cos  sin 
 r 2 sin 2  
 
 cos   2 r

dt
dt
 dt 
 dt 
 dt 
Thus, 
dr
d
 dy  2  dr  2
 d  2
2
cos  sin 
 r 2 cos 2  
 
 sin   2 r

dt
dt
 dt 
 dt 
 dt 
and 
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Unit II
Vector Differential Calculus
Consequently, Or equivalently, L 
b
a
2  d  2
 dr  2
  r 
 dt
 dt 
 dt 
Tangents and Normals to Curves
Tangents to Curves
Definition 2.11 Let C be a smooth curve and r (t ) a (smooth) parameterization of
C defined on an interval I. Then for any interior point t of I, the tangent
vector T (t ) at the point r (t) is defined by
T (t ) 
r ' (t )
r ' (t )
Note that: T (t) is a unit vector along r ' (t ) .
Example 30 Find a formula for the tangent T (t) to the circular helix
r (t )  2 cos t i  2 sin t j  t k
Solution r ' (t )   2 sin t i  2 cos t j  k and
r ' (t ) =
5.
2
2
1
sin t i 
cos t j 
k.
5
5
5
Note that: T (t) and r (t) are not orthogonal, since r (t )  constant for all t.
Therefore, T (t )  
Example 31 Find the tangent vector T (t) to the curve parameterized by r (t), where


≤t≤
.
6
3
Solution r ' (t )   3 cos 2 t sin t i  3 sin 2 t cos t j and r ' (t ) = 3 sin t cos t .
r (t )  cos3 t i  sin 3 t j for
Therefore, T (t )   cos t i  sin t j .
Normals to Curves
Let C be a smooth curve and r (t) be a parameterization of C. Let r ' (t ) be also smooth. Then the tangent vector
T (t) is differentiable. Moreover; if T (t ) = 1 for all t in the domain of T. But then
T ' (t )  T (t ) = 0. Hence if T ' ( t )  0, then T ' (t ) and T (t) are orthogonal.
Definition 2.12 Let C be a smooth curve and r (t ) a smooth parameterization of
C defined on an interval I such that r ' (t ) is smooth. Then for any interior
point t of I for which T ' (t )  0, the normal vector N (t) at the point r (t) is
defined by
N (t ) 
T ' (t )
T ' (t )
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Unit II
Vector Differential Calculus
Example 32 Find a formula for the normal N (t) of the curve parameterized by
r (t )  2 cos t i  2 sin t j  t k
Solution r ' (t )   2 sin t i  2 cos t j  k and
r ' (t ) =
5.
2
2
1
sin t i 
cos t j 
k
5
5
5
2
2
2
Hence T ' (t )  
.
cos t i 
sin t j and T ' (t ) 
5
5
5
Thus, T (t )  
Therefore, N (t) =  cos t i  sin t j .
Example 33 Find a formula for the normal N (t) of the curve parameterized by
r (t )  cos3 t i  sin 3 t j for


≤t≤
.
6
3
Solution r ' (t )   3 cos 2 t sin t i  3 sin 2 t cos t j and
r ' (t ) = 3 sin t cos t
Thus, T (t )   cos t i  sin t j and T (t )  sin t i  cos t j . But T ' (t )  1 .
Therefore, N (t) = sin t i  cos t j .
Example 34 Find a formula for the normal N (t) of the curve parameterized by
r (t )  t i  cos t j for 0 ≤ t ≤
Solution r ' (t )  i  sin t j and
Thus, T (t ) 
i  sin t j
Consequently, T ' (t ) 
Therefore, N (t) = 
r ' (t ) = 1  sin 2 t .
0<t<
1  sin 2 t

.
2

 sin t cos t i  cos t j
and hence T ' (t ) 
3
2
1  sin 2 t 2


cos t
.
1  sin 2 t
sin t i  j
1  sin t
2
for 0 < t <

.
2
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
b  bT T  bN N
T
Where bT is the tangential component of b .
bN is the normal component of b .
bT
T
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N
bN
18
Unit II
Vector Differential Calculus
Suppose an object moves along a curve C. The velocity and acceleration vectors lie in the plane determined by
T and N. Let r (t ) be the position vector of the object that moves along the curve C and suppose T and N exist.
Then
v (t ) 
dr
dr

T  v T
dt
dt
Thus the tangential component of  is
is 0.
v , the speed of the object, and the normal component of the velocity
Furthermore; the acceleration vector a (t )
dT
dT
dv d v
dT
N,
. Since
=

T  v
dt
dt
dt
dt
dt
d v
dT
and a N  v
.
a (t ) = aT T  aN N , where aT 
dt
dt
a (t ) =
Therefore, a (t ) = aT T  aN N .
The real valued functions aT and a N are the tangential and normal components of acceleration.
Hence
a
2
= aT T  aN N   aT T  aN N  = aT
Therefore, a N 
a
2
2
2
T
 aN
2
N
2
= aT
2
 aN
2
2
 aT .
Example 35 Let r (t )  cosh t i  sinh t j  t k . Then find the normal and the tangential components of
acceleration.
Solution v (t )  sinh t i  cosh t j  k and v (t ) 
=
Then aT =
2 cosh 2 t =
sinh 2 t  cosh 2 t  1
2 cosh t .
2 sinh t and a (t )  cosh t i  sinh t j .
Thus, a (t ) 2  cosh 2 t  sinh 2 t = 2 cosh 2 t  1 .
Consequently, a N = 2 cosh 2 t  1  2 sinh 2 t =
Therefore, aT =
2 sinh t and a N = 1.
2 (cosh 2 t  sinh 2 t )  1 = 1
Orientation of Curves
Let r (t )  x (t ) i  y (t ) j  z (t ) k 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 r (t ) , we say that r (t ) 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 r (t ) of C.
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Unit II
Vector Differential Calculus
Suppose r (t ) is a piecewise smooth parameterization of the curve C on [a, b] and let
r 1 (t )  r (a  b  t ) for a ≤ t ≤ b.
Then r 1 (t ) is a piecewise smooth parameterization of C and determines an orientation opposite to the
orientation determined by r (t ) . 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:
r 1 (t )  3 cos t j  3 sin t k for 0 ≤ t ≤ 2.
Thus, r (t ) =  2 i  2 j  k + r 1 (t )  r (t ) =  2 i  ( 2  3 cos t ) j  (1  3 sin t ) k
Now r (t ) is continuous on [0, 2] and r ' (t ) =  3 sin t j  3 cos t k .
Since r ' (t ) = 3  0 for 0  t  2, r (t ) is smooth.
Therefore, r (t ) 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
z
vector valued function, and hence
dT
=0
ds
y
x
ii) If the curve undulates gently, then the tangent vector
z
x
T (t) changes direction slowly along the curve, and
y
hence
dT
changes but gently.
ds
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Vector Differential Calculus
z
iii) If the curve twisted, then the tangent vector T (t)
changes rapidly and hence
y
x
dT
changes rapidly.
ds
dT
is closely related to the rate at which the curve twists and turns.
ds
dT
dT
dT
dT ds
dT
dt
dt .

Since
=
it follows that
=
=
ds
dr
dt
ds dt
ds
dt
dt
Thus, the rate of change of the tangent vector,
Definition 2.13 Let a curve C have a smooth parameterization r (t ) such that r ' (t )
is differentiable. Then the curvature k of C is defined by the formula
k (t) =
dT
T ' (t )
=
r ' (t )
dr
dt
dt
Example 38 Find the curvature k of the curve traced out by
4
3
r (t )  cos t i  (1  sin t ) j  cos t k
5
5
4
3
Solution r (t )   sin t i  cos t j  sin t k and r ' (t ) = 1.
5
5
4
3
Thus, T (t) =  sin t i  cos t j  sin t k
5
5
4
3
T ' (t )   cos t i  sin t j  cos t k and T ' (t ) = 1.
5
5
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
r (t )  t i  sin t j
Thus, r ' (t )  i  cos t j and r ' (t ) =
Hence, T (t) =
Therefore, k (t) =
i  cos t j
, T ' (t ) =
1  cos 2 t
sin t cos t i  sin t j
1  cos t 
2
sin t
1  cos 2 t 
1  cos 2 t .
3
2
3
2
and T ' (t ) 
sin t
1  cos 2 t
.
.
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Unit II
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Example 40 Find the curvature k of the graph of y = 3 x for x > 0.
Solution The graph of y = 3 x for x > 0 is the range of the continuous vector valued function
r (t )  t i  3 t j for t > 0
1
Thus, r ' (t )  i  t
3
Now let u =
9t
Hence, T ' (t )  u
Therefore, k (t) = 6u
4
3
2
3
1
j and r ' (t ) = t
3
 1 . Then T (t )  u
3
3

( 2t

1
3i
1

2
3
9t
4
3
1
1
( 3t
3i
 j).
1
 6t
3
j ) and T ' (t )  2u
2
t

1
3
1
t 3.
Definition 2.14 The radius of curvature  (t) of a curve at a point P corresponding
to t is given by
 (t ) 
1
k (t )
Example 41 Find the radius of curvature of the curve traced out by
r (t )  2 t i  t 2 j 
1 3
t k
3
4  4t 2  t 4 = 2  t 2
Solution r ' (t )  2 i  2t j  t 2 k and r ' (t ) =


 1 
 2 i  2t j  t 2 k .
Then, T (t )  
2
2  t 


 1 
2
  4 i  2 (2  t 2 ) j  4 t k and T ' (t ) 
Thus, T ' (t )  
2
2  t2
2  t 
Therefore,  (t ) 


2
1
2  t2 .
2
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:
v  v T and a = aT T  a N N
Hence, v  a  ( v T )  (aT T  a N N ) = ( v T  aT T )  ( v T  a N N ) = ( v a N ) ( T  N )
Thus, v  a  v aN , since T  N = 1, v  a  v aN = v 2 T ' , since a N =
Prepared by Tekleyohannes Negussie
v
T'
22
Unit II
Vector Differential Calculus
v a
Hence,
v

3
T'
v
v a
Therefore, k =
v
3
T'

r'
.
.
Example 42 Show that the helix r (t )  cos t i  sin t j  t k has constant curvature.
Solution v (t )   sin t i  cos t j  k and hence v 
2 and a (t) =  cos t i  sin t j .
Then v  a  sin t i  cos t j and hence v  a  2 .
Therefore, k (t) =
1
that is constant.
2
If r (t ) represents an object moving along a curve C in the xy plane, we have
dx
dy
d 2x
d2y
i 
j and a (t )  2 i  2 j .
dt
dt
dt
dt
2
2
dx d y dy d x

dt dt 2
dt dt 2
r (t )  x (t ) i  y (t ) j , v (t ) 
Then k =
v a
v
3
=
 dx  2  dy  2 
 dt    dt  
  
 
3
2
.
Example 43 Find the curvature k of the plane curve C parameterized by
r (t )  2 cos t i  3 sin t j
Solution The parametric equations are: x = 2 cos t, y = 3 sin t
dx
dy
d 2x
d2y
Then
= – 2 sin t,
= 3 cos t,
= – 2 cos t and
= – 3 sin t .
dt
dt
dt 2
dt 2
6 sin 2 t  6 cos 2 t
Thus, k =
4 sin 2 t  9 cos2 t 
Therefore, k =
6
4  5 cos t 
2
3
2
3
2
.
.
Example 44 Find the curvature k of the graph of y = sin x.
Solution y ' = cos x and y ' ' = – sin x.
Therefore, k =
sin x
1 cos t 
2
3
2
.
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.
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Unit II
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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).
z
y
y
x
x
F (x, y, z) =
z j
F (x, y, z) = x i  y j
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:
F ( x, y , z ) 
Gm
u ( x, y , z )
x  y2  z2
2
where G is a gravitational constant, u ( x, y, z) 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: u ( x, y, z) has the same direction as  x i  y j  z k .
Then u ( x, y, z) =
Hence, F (x, y, z) = 
 xi  y j  z k
.
x2  y2  z 2
Gm
x
2
 y 
2

3
2 2
z
x i  y j  z k .
If a point (x, y, z) in space is represented by the vector r , then the gravitational field can be written as:
F (x, y, z) = 
Gm
r
3
r , where r = x i  y j  z k .
A vector field F can be expressed in terms of its components, say M, N and P as follows:
F (x, y, z) = M ( x, y, z) i  N ( x, y, z) j  P ( x, y, z) k
In short we can write
F (x, y, z) = M i  N j  P k .
Note that: M, N and P are scalar fields.
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Unit II
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Let F (x, y, z) = M i  N j  P k 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:
f
f
f
( x, y , z ) i +
( x, y , z ) k .
( x, y , z ) j +
x
z
y
grad f (x, y, z) =  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.
Gm
Solution F (x, y, z) = 
3
r
r , where r = x i  y j  z k .
Then F (x, y, z) = M i  N j  P k 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 = 
Gm
r
Then
3
x,N= 
Gm
r
 M dx =  G m 
3
y and P = 
r
x

x2  y 2  z 2
 N dy =  G m 

y
x2  y 2  z 2
and
 P dz =  G m 

z
x2  y 2  z 2
Now let f (x, y, z) =
Gm
x  y2  z2
2
Gm

3
2

3
2

3
2
3
z
dx + k (y, z) =
dy + ℓ(x, z) =
dz + q (x, y) =
Gm
x2  y2  z 2
Gm
x2  y2  z 2
Gm
x  y2  z2
2
+ k (y, z)
+ ℓ(x, z)
+ q (x, y)
, 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
grad f ( x, y, z)  yz e xy i  xz e xy j  (e xy  cos z) k
f
f
f
 yz e xy ,
 e xy  cos z
 xz e xy and
Solution
x
z
y
Prepared by Tekleyohannes Negussie
(*)
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Unit II
Vector Differential Calculus
Integrating both sides of the last equation in (*) with respect to z we get:
f ( x, y, z )  ze xy  sin z  g ( x, y )
(**)
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
f
g
f
g
 yz e xy +
and
 xz e xy +
x
x
y
y
comparing these with the first and the second equations in (*) respectively we get:
g
g
=
=0
x
y
Thus, g (x, y) = c with respect to x, y and z.
Therefore, f ( x, y, z )  ze
xy
 sin z  c , where c is a real number.
Example 48 Find a function f of three variables such that
grad f ( x, y, z)  (2 xy  z 2 ) i  x 2 j  (2 xz   cos z) k
Solution
f
f
f
 2 xz   cos  z
= 2 xy  z 2 ,
 x 2 and
x
z
y
(*)
Integrating both sides of the second equation in (*) with respect to y we get:
f ( x, y , z )  x 2 y  g ( x, z )
(**)
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
f
g
f
g
 2 xy +

and
x
x
z
z
comparing these with the first and the last equations in (*) respectively we get:
g
g
 z 2 and
 2 xz   cos  z
x
z
(***)
Integrating the second equation in (***) with respect to z, we get
g (x, z) = x z 2  sin  z  k ( x)
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:
dk
 0 , hence k (x) = c , constant.
dx
Therefore, f ( x, y, z )  x 2 y  xz 2  sin  z  c , 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.
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Unit II
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The Divergence of a Vector Field
Definition 2.16 Let F = M i  N j  P k be a vector field such that
M
,
x
P
N
and
exists. Then the divergence of F, denoted div F or
z
y
  F is the function defined by
div F (x, y, z) =   F ( x, y, z )
=
M
N
P
( x, y, z ) 
( x, y, z ) 
( x, y, z )
x
y
z
Example 49 Find the divergence of the vector field F, where
F ( x, y, z)  ( y  z) i  ( x  z) j  ( x  y) k
 ( y  z)  ( x  z )  ( x  y)
Solution div F (x, y, z) =
= 0.


x
y
z
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 F ( x, y, z )  xe i  ye x j  sin yz k .
y
 ( x e y )  ( y e x )  (sin yz )
Solution div F (x, y, z) =
.


x
y
z
= e  e x  y cos yz
y
y
Therefore, div F = e x  e  y cos yz .
The Curl of a Vector Field
Definition 2.17 Let F = M i  N j  P k 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   F is the function defined by
curl F (x, y, z) =   F ( x, y, z )
 P N   M P 
 N M 
 i  
k

  j  

x 
y 
 y z   z
 x
= 
The Curl of F is symbolically expressed as:
i

Curl F =
x
M
j

y
N
k

z
P
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Unit II
Vector Differential Calculus
Example 51 Find curl F if F ( x, y, z)  ( y  z) i  ( x  z) j  ( x  y) k
Solution M = y + z, N = x + z and P = x + y.
Then
N
M
P
N
P
M
=
= 1,
=
= 1 and
=
= 1.
z
x
z
x
y
y
Therefore, curl F = 0.
Note that: If curl F = 0, then F is said to be irrotational.
Example 52 Find curl F if F ( x, y, z )  cos x i  sin y j  e
xy
k .
xy
Solution M = cos x, N = siny and P = e .
Then
N  M
N
P
M
xy P
xy
= xe ,
= y e and
=
=
=
= 0.
x
z
z
x
y
y
Therefore, curl F = x e
xy
i  y e xy j .
Let f be a scalar field, then
  ( f ) = div (grad f) =
2 f
x 2

2 f
y 2

2 f
z 2
The right side of this formula is the Laplacian of f usually denoted by  2 f . A function that satisfies the
equation
2 f = 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) = M ( x, y) i  N ( x, y) j , then
 N M 

k

y 
 x
2 f
2 f
M
N

div F (x, y) =
and  2 f (x, y) =

x
y
x 2
y 2
grad f (x, y) =
f
f
i 
j , curl F (x, y) =
x
y
Suppose F = M i  N j  P k 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:
P
P
N M
N
M

,
and


x
y
z z
x
y
(*)
Note that: (*) holds for a vector field F = M i  N j  P k need not imply that F is conservative.
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Unit II
Vector Differential Calculus
Theorem 2. 6 Let F = M i  N j  P k be a vector field. If there is a function f
having a continuous mixed partial derivatives whose gradient is F, then
P
P
N M
N
M

,
and


x
y
z z
x
y
If the domain of F is 3 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) = M ( x, y) i  N ( x, y) j
the conditions in (*) reduce to
N
M

x
y
and the corresponding statements in the theorem holds for such vector fields.
Example 53 Let F ( x, y, z)  yz i  xz j  xy k
and G ( x, y, z)  ( x 2  y 2 ) i  ( y 2  z 2 ) j  ( x 2  z 2 ) k .
Show that F is the gradient of some function but G is not the gradient of any function.
Solution For F we have
P
P
N M
N
M
 y
,
and
.
x
 z
x
y
z z
x
y
Since the domain of F is 3 , F is the gradient of some function f.
For G we have
N
P
 2 z , so that the first equation in (*) is not satisfied.
 0 and
z
y
Therefore, G is not the gradient of any function.
x
y
xy
xy
Example 54 Let F ( x, y)  y 2 e i  (1  xy) e j and G ( x, y, z )  i 
j.
y
x
Show that F is the gradient of some function but G is not the gradient of any function.
Solution For F we have
M
N
 (2  xy) y e xy 
y
x
Since the domain of F is 2 , F is the gradient of some function f.
For G we have
M
x
N
y
  2 and
 2.
y
x
y
x
Therefore, G is not the gradient of any function.
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Unit II
Vector Differential Calculus
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




i 
j 
k
x
y
z
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 + ( grad f )  F and curl (f F) = f (curl F) + ( grad f )  F
iii) div ( F  G ) = (curl F )  G – F  (curl G ) and div ( grad f  grad g ) = 0
Exercise
Show that the arc length of a polar graph is given by

L


r
2
 r '2 d
Prepared by Tekleyohannes Negussie
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