Naval Architecture & Ocean Enginee ring
Engineering Mathematics 2
Prof. Kyu-Yeul Lee
Department of Naval Architecture and Ocean Engineering, Seoul National University of College of Engineering
[2008][06-1]
October, 2008
Naval Architecture & Ocean Enginee ring
Vector Calculus (2)
: Vector Functions, Vector Fields
Vector Functions Partial Derivatives
Directional Derivatives
Tangent Planes and Normal Lines
Vector Fields and Fluid Mechanics
Vector Functions
Parametric Curve
x
y
(x(t0),y(t0))c
( ) ( ), ( )0 0
0 x t y t
t r
x
y
)) ( ), ( ), (
(x t0 y t0 z t0
c
) ( ), ( ), ( )
(t0 x t0 y t0 z t0
z
r( ) ( ), ( ) ( ) ( ) t x t y t
f t g t
r
i j
Vector Function
( ) ( ), ( ), ( )
( ) ( ) ( )
t x t y t z t
f t g t h t
r
i j k
2-D
3-D
For a given parameter , is a position vector t r ( t ) of a point on a curve P C ( ) ( )
x t f t ( ) ( ) y t g t
( ) ( ) x t f t
( ) ( ) y t g t
( ) ( ) z t h t
P
0P
0o
o
Vector Functions
Limit of a Vector Function
If , , and exist, then Definition 9.1
) ( lim f t
a
t
lim g ( t )
a
t
lim h ( t )
a t
) ( lim ), ( lim ),
( lim )
(
lim t f t g t h t
a t a
t a
t a
t
r
Properties of Limits
If and , then Theorem 9.1
1 1
( )
lim r L
t
a
t
lim r
2( ) L
2
t
a t
scalar a
is ,
) ( lim
)
( i c
1t c
1c
a
t
r L
1( )
2( )
1 1lim )
( r r L L
t t
ii
t a) ( )
( lim )
( iii r t r t L L
Vector Functions
Continuity
A vector function is said to be continuous at if is defined
exists, and Definition 9.2
) ( )
( i r a
r
) ( )
( lim )
( iii t a
a
t
r r
Derivatives of Vector Function The derivative of a vector is
for all for which the limit exists Definition 9.3
r t
( ) ( )
lim 1 )
(
0t t t
t t
t
r r
r
a t )
( ) ( ii r t
dt
t d r
r ( )
Vector Functions
Differentiation of Components
If where and are differentiable, then Theorem 9.2
) ( ), ( ), ( )
( t f x g x h x
r f , g h
) ( ), ( ), ( )
( t f x g x h x r
Proof
t
t h t
t h f
t
t g t
t g t
t f t
t f
t h t g t f t
t h t t
g t t
t f
t t
t t t
t t t
) ( )
, ( ) ( )
, ( ) ( )
lim (
) ( ), ( ), ( )
( ), (
), 1 (
lim
) ( )
1 ( lim )
(
0 0
0
r r
r
Vector Functions
Smooth curve
When the component function of a vector function have continuous first derivatives and for all in the open interval , then
is said to be a smooth function and the curve traced by is called a smooth curve
r 0
) (
t
r t ( a , b )
r C r
Vector Functions
Not Smooth curve ex.)
3
3
21
( ) 2 (4 ) 4 2 cos 2 (sin )
2 2
F t t
t
t
i
tj
tk
Vector Functions
Differentiation of Components
If where and are differentiable, then Theorem 9.2
) ( ), ( ), ( )
( t f x g x h x
r f , g h
) ( ), ( ), ( )
( t f x g x h x r
High-Order Derivatives
) ( ), ( ),
( )
( t f x g x h x r
Also obtained by differentiating its components.
Ex) Second-order
Vector Functions
Example 4
Tangent Vectors
Graph the curve C that is traced by a point P whose position is given by r(t)=cos2ti+sintj, 0≤t≤2π.
Graph r′(0) and r′(π/6).
t
t t
t t
t x
2
2 2
2 2
sin 2 1
sin )
sin 1
(
sin cos
2 cos
x
y r ( 0 )
2 ,1 2 1
) 0 , 1 (
6 r
t y sin
) 2 0
( t
1 1
, 2
1
2
y x
x
j i
r ( t ) 2 sin 2 t cos t j
r ( 0 )
j i
r 2
3 3 6
Solution)
Vector Functions
Example 5 Tangent Line
Find parametric equations of the tangent line to the graph of the
curve C whose parametric equations are x=t
2, y=t
2-t, z=-7t at t=3.
k j
i
r ( t ) t
2 ( t
2 t ) 7 t
k j
i
r ( t ) 2 t ( 2 t 1 ) 7 k j
i
r ( 3 ) 6 5 7 k j
i
r ( 3 ) 9 6 21
i j i k j i j k k
r r
t t
t
t t
7 21 5
6 6
9
7 5 6 21
6 9 )
3 ( )
3 (
t z
t y
t
x 9 6 , 6 5 , 21 7
Solution)
Vector Functions
Example 6
Derivative of a Vector Function
If r(t)=(t 3 -2t 2 )i+4tj+e -t k,
then r′ (t)=(3t 2 -4t)i+4j-e -t k
and r″(t)=(6t-4)i+e -t k.
Vector Functions
Chain Rule
If is a differentiable vector function and is a differentiable scalar function, then, the derivatives of with respect to is
Theorem 9.3
r s u (t )
) (s r
) ( ) ( s u t dt
ds ds d dt
d r r r
t
Vector Functions
Example 7 Chain Rule
If r(s)=cos2si+sin2sj+e
-3sk, where s=t
4, then
. 12
) 2 cos(
8 ) 2 sin(
8
4 3
2 cos 2 2
sin 2
3 4
3 4
3 4
3
3 3
k i
k j
r i
t s
e t j
t t
t t
t e
s dt s
d
Vector Functions
Rules of Differentiation
Let and be differentiable vector functions a differentiable scalar function.
Theorem 9.4
r
1u (t )
) ( )
( )]
( )
( [ )
(
1t
2t
1t
2t
dt
i d r r r r r
2) ( ) ( )
( ) ( )]
( ) ( [ )
( u t
1t u t
1t u t
1t dt
ii d r r r
) ( )
( )
( )
( )]
( )
( [ )
(
1t
2t
1t
2t
1t
2t
dt
iii d r r r r r r
) ( )
( )
( )
( )]
( )
( [ )
(
1t
2t
1t
2t
1t
2t
dt
iv d r r r r r r
Differentiation of Cross Product
Vector Functions
Integral of Vector Functions
k j
i
r ( t ) f ( t ) g ( t ) h ( t )
If and are integrable, then the indefinite and definite integrals of a vector function are defined, respectively, by
g
f , h
i j k
r
( t ) dt f ( t ) dt g ( t ) dt h ( t ) dt
k j
i
r
abb a b
a b
a
( t ) dt f ( t ) dt g ( t ) dt h ( t ) dt
) ( )
( ,
)
( t dt R c where R t r t
r
Vector Functions
Example 8
Integral of a Vector Function
If r(t)=6t
2i+4e
-2tj+8cos4tk then
, 4
sin 2 2
2
4 sin 2 2
2
4 cos 8 4
6 )
(
2 3
3 2
2 1
3
2 2
c k j
i
k j
i
k j
i r
t e
t
c t c
e c
t
dt t dt
e dt
t dt
t
t
t t
where c=c
1i+c
2j+c
3k.
Vector Functions
Example 9
Consider the helix of Example 1.
Find the length of the curve from r(0) to an arbitrary point r(t).
Find the parametric equations of the helix.
9.1 Example 1.
. 0 ,
sin 2 cos
2 )
( t t i t j t k t r
k j
i
r ( t ) 2 cos t 2 sin t t k j i
r ( t ) 2 sin t 2 cos t
2
2 22 2
( ) 2sin 2 cos 1
4sin 4 cos 1 5
t t t
t t
r
5 )
(
t r
t du
du u
s
t( )
t5 5
0
0
r
k j
i
r 2 sin 5 5
cos 5 2
)
( s s s
s
sin 5 2 ) (
cos 5 2 ) (
s s g
s s f
Solution)
Partial Derivatives
Function of two variables
z f ( x , y )
x
y
z
) ,
( x y f
z
Domain of
) ,
( x y
) ,
( x y f
z
) ,
( )
, ,
( x y z where z f x y
y x,
z
: independent variables
: dependent variable
Partial Derivatives
) (x f y
x
x f x
x f dx
dy
x
) ( )
lim (
0
) , ( x y f
z
Partial Derivatives
0
( , ) ( , )
lim
xx x
z f x x y f x y
x x
f z f
x
0
( , ) ( , )
lim
xy y
z f x y y f x y
y y
f z f
y
partial derivative with respect to
x
partial derivative with respect toy
treating as a constant
x
Ordinary Derivatives
Partial Derivatives
Example 3
Partial Derivatives
If z=4x
3y
2-4x
2+y
6+1, find ∂z/∂x and ∂z/∂y.
x y
x x
z 12
2 2 8
5 2
3
6
8 x y y y
z
Partial Derivatives
Example 4
Partial Derivatives
If F(x,y,t)=e
-3πtcos4x sin6y, then the
partial derivatives with respect to x, y, and t are, in turn,
. 6 sin 4
cos 3
) , , (
6 cos 4
cos 6
) , , (
6 sin 4
sin 4
) , , (
3 3
3
y x
e t
y x F
y x
e t
y x F
y x
e t
y x F
t t
t y
t x
Partial Derivatives
Chain Rule
y f (u )
dx du u y dx
dy
) ( , u g x
Chain Rule
If is differentiable and and have continuous first partial derivatives, then
Theorem 9.5
) , ( x y g
u )
, ( v u f
z v h ( x , y )
x v v z x
u u
z x
z
y v v z y
u u
z y
z
( ( )) y f g x
( , ) ( ( , ), ( , ))
z f u v f g x y h x y
Partial Derivatives
Example 5 Chain Rule
If z=u
2-v
3and u=e
2x-3y, v=sin(x
2-y
2), find ∂z/∂x and ∂z/∂y.
Solution)
) cos(
6 4
) cos(
2 3 2
2
2 2
2 3
2
2 2
2 3
2
y x
xv ue
y x
x v e
u
x v v z x
u u
z x
z
y x
y x
( 2 ) cos( )
3 ) 3
(
2 u e
2 3v
2y x
2y
2y v v z y
u u
z y
z
y
x
Partial Derivatives
special case
z f ( v u , )
dt dv v z dt
du u z dt
dz
) ( ,
) (
, u g t v h t x
v v z x
u u
z x
z
y v v z y
u u
z y
z
Chain Rule
z f ( v u , ) , u g ( x , y ) , v h ( x , y ) ( , ) ( ( , ), ( , ))
z f u v f g x y h x y
( , ) ( ( ), ( ))
z f u v f g t h t
Partial Derivatives
If r(x,y,z) = x 2 +y 5 z 3
and x(u,v)=uve 2s , y(u,v)=u 2 -v 2 s, z(u,v)=sin(uvs 2 ), find ∂r/∂s.
2 4 3 2 5 2 2
2 (2
s) 5
z( ) 3 2 cos( )
r r x r y r z
s x s y s z s
x uve y v y z uvs uvs
Chain Rule example)
Partial Derivatives
generalization
z f ( u
1, u
2,..., u
n)
dt du u
z dt
du u
z dt
du u
z dt
dz
n
n
2
2 1
1
) (
) (
) (
2 2
1 1
t g u
t g u
t g u
n n
special case
z f ( v u , )
dt dv v z dt
du u z dt
dz
) ( ,
) (
, u g t v h t
( , ) ( ( ), ( ))
z f u v f g t h t
Partial Derivatives
x v v z x
u u
z x
z
y v v z y
u u
z y
z
Chain Rule
z f ( v u , ) , u g ( x , y ) , v h ( x , y )
generalization
z f ( u
1, u
2,..., u
n)
n
x u u
z x
u u
z x
u u
z dx
dz
1 2
) ,..., ,
(
) ,..., ,
(
) ,..., ,
(
2 1
2 1 2 2
2 1 1 1
k n
n
k k
x x
x g u
x x
x g u
x x
x g u
( , ) ( ( ), ( ))
z f u v f g t h t
Gradient of a Function
Apply Vector differential operator
z y
x
y x
k j
i
j i
k j
i j i
z F y
F x
z F y x F
y f x
y f x f
) , , (
) , ( )
, , (
) , (
z y x F w
y x f z
Differentiable function
Gradients of function Introduce a new vector based on partial differentiation
: “del” or “nabla” f
: “grad “f
Directional Derivatives
x
y
z
i j
) , ( x y f
z
:Rate of change of in the -direction
f i f
x
i
j
Directional Derivatives
x
y
z
i j
) , ( x y f
z
:Rate of change of in the -direction
f j f
y
i
j
u
j
Directional Derivatives
x
y
z
i j
) , ( x y f
z
u
f y
The rate of change of
f
in the direction of :y
f cos
x
fx
The component of in the direction of :
u
cos( ) 2 sin f
y f y
f
y
The component of in the direction of :
u
f x
The rate of change of
f
in the direction of :x
= f cos
x
sinf
y
+
The rate of change of
f
in the direction given by the vector :u D f
u f
f y
j i y
f x
y f x
f
( , )
i
j
+
The rate of change of
f
in the direction given by the vector :u D f
u=
Directional Derivatives
x
y
z
i
j
) , ( x y f
z
u
P
secant tangent
y x f h
y h
x f y
x f y
y x x
f ( , ) ( , ) ( cos , sin ) ( , )
h f ( x x , y y ) f ( x , y )
x
y
) 0 , , ( x y
) 0 , ,
( x x y y
u v h
Slope of indicated secant line is
Directional Derivatives
x
y
z
i
j
) , ( x y f
z
u
P
secant tangent
h f ( x x , y y ) f ( x , y )
x
y
) 0 , , ( x y
) 0 , ,
( x x y y
u v h
Slope of indicated secant line is
Directional Derivatives
x
y
z
i
j
) , ( x y f
z
u
P
secant tangent
y x f h
y h
x f y
x f y
y x x
f ( , ) ( , ) ( cos , sin ) ( , )
h f ( x x , y y ) f ( x , y )
x
y
) 0 , , ( x y
) 0 , ,
( x x y y
As we expect the slope of secant line would be tangent line u
v h
0
h
Directional Derivatives
h
y x f h
y h
x f h
y x f y y
x x
f
( , ) ( , ) (
cos ,
sin )
( , )
Directional Derivative
The directional derivative of in the direction of a unit vector is
provided the limit exists.
Definition 9.5
) , ( x y f
z
0
( cos , sin ) ( , ) ( , ) lim
h
f x h y h f x y
D f x y
h
u
j i
u cos sin
Slope of secant line
x
y
z
i
j
) , (x y f z
u
P secant
tangent
h f(xx,yy) f(x,y)
x
y ) 0 , , (xy
) 0 , , (xx yy
u vh
x z h
y x f y h x y f
x f D
h
) , ( )
, lim (
) , ( implies
0
i 0
( , ) ( , )
f x y h f x y z
Directional Derivatives
) , ) (
, ( )
sin ,
cos lim (
) 0 ( )
0 lim ( )
0
(
0 0D f x y
h
y x f h
y h
x f h
g h
g g
h
h
u
Computing a Directional Derivative If is differentiable function of x and y and then,
Theorem 9.6
) , ( x y f
z
u
f ( x , y ) f ( x , y ) u D
j i
u cos sin
Proof)
) sin ,
cos (
)
( t f x t y t
g
by definition of a derivative
by the Chain Rule,
1 2
1 2
( ) ( cos , sin ) ( cos ) ( cos , sin ) ( sin )
( cos , sin cos ( cos , sin ) sin
d d
g t f x t y t x t f x t y t y t
dt dt
f x t y t f x t y t
(0) ( , cos ( , )sin
g f x y f x y
Directional Derivatives
Computing a Directional Derivative If is differentiable function of x and y and then,
Theorem 9.6
) , ( x y f
z
u
f ( x , y ) f ( x , y ) u D
j i
u cos sin
Proof)
) sin ,
cos (
)
( t f x t y t
g
( , ) (0) ( , cos ( , ) sin
[ ( , ) ( , ) ] (cos sin ) ( , )
x y
x y
D f x y g f x y f x y
f x y f x y f x y
u
i j i j
u
j i y
f x
y f x
f
( , )
Directional Derivatives
Example 2
Gradient at a Point
If F(x,y,z)=xy
2+3x
2-z
3, find ∇F(x,y,z) at (2,-1,4).
Solution)
k j
i
k j i
2 2
3 2
2
3 2
2
3 2
2
3 2
) 6 (
) 3
(
) 3
(
) 3
( )
, , (
z xy
x y
z x
dz xy
z x
dy xy
z x
x xy z
y x F
k j
i 4 48 13
) 4 , 1 , 2
(
F
Directional Derivatives
Computing a Directional Derivative If is differentiable function of x and y and then,
Theorem 9.6
) , ( x y f
z
u
f ( x , y ) f ( x , y ) u D
j i
u cos sin
( , ) [ ( , ) ( , ) ] (cos sin ) ( , )
x y
D f x y f x y f x y f x y
u
i j i j
u
j i y
f x
y f x
f
( , )
y
z
i j
) , ( x y f
z
j
Directional Derivatives
Example 3
Directional Derivative
Find the directional derivative of
f(x,y)=2x
2y
3+6xy at (1,1) in the direction of a unit vector whose angle with the positive x-axis is π/6.
Solution)
j i
j i
) 6 6
( ) 6 4
( ) , (
2 2
3
y x y x
xy y f x
y f x f
j i 12 10
) 1 , 1
(
f
j i
j i
u sin 6
cos 6 sin
cos
(1,1) (1,1)
3 1
(10 12 )
2 2
5 3 6 D f f
u
u
i j i j
Directional Derivatives
Example 4
Directional Derivative
Consider the plane that is
perpendicular to the xy-plane and passes through the points P(2,1) and Q(3,2). What is the slope of the tangent line to the curve on intersection of this plane with the surface f(x,y)=4x
2+y
2at (2,1,17) in the direction of Q?
Solution)
j i
u 2
1 2
1
PQ
2
4
2) ,
( x y x y
f
j i y x
y x
f ( , ) 8 2
j i 2 16 )
1 , 2
(
f
(2,1) (2,1)
1 1
(16 2 )
2 2
9 2
D f f
u
u
i j i j
Directional Derivatives
Example 5
Directional Derivative
Find the directional derivative of F(x,y,z)=xy
2-4x
2y+z
2at (1,-1,2) in the direction of 6i+2j+3k.
Solution)
2 2
2
4
) , ,
( x y z xy x y z
f
k j
i
k j i
z x
xy xy
y
z y x z xy
z y x y xy
z y x x xy
z y x f
2 ) 4 2
( ) 8 (
) 4
(
) 4
(
) 4
( )
, , (
2 2
2 2
2 2
2 2
k j
i 6 4 9
) 2 , 1 , 1
(
f
7 3
2
6 i j k
k j
i k
j i
u 7
3 7
2 7
) 6 3 2 6 7 (
1
6 2 3
(1, 1, 2) (9 6 4 )
7 7 7
54 7
D F
u
i j k i j k
Directional Derivatives
Points in the direction of maximum increase of at P
f f
The rate of change of f in the direction given by the vector : u
( , ) [
x( , )
y( , ) ] (cos sin ) ( , ) D f x y
u f x y i f x y j i j f x y u
cos cos , :angle between and
1 cos 1
D f f f f
u
u u
The maximum value of
D f
uD f
u f
, Whencos 1, 0
has the same direction of
f u
is the direction of maximum increase of at P
f f
is the direction of maximum decrease of at P
f f
Directional Derivatives
Example 6
Max/Min of Directional Derivative
In Example 5 the maximum value of the directional derivative at F at (1,-1,2) is .
133 )
2 , 1 , 1
(
F
The minimum value of D
uF(1,-1,2) is then 133 .
Directional Derivatives
Example 7
Direction of Steepest Ascent
Each year in Los Angeles there is a
bicycle race up to the top of a hill by a road known to be the steepest in the city. To understand why a bicyclist with a modicum of sanity will zigzag up the road, let us suppose the graph of
shown in Figure (a) is a mathematical model of the hill. The gradient of f is
where r=-xi-yj is a vector pointing to
Thus the steepest ascent up the hill is a straight road whose projection in the xy-plane is a radius of the circular base.
Since D
uf=comp
u∇f, a bicyclist will
zigzag, or seek a direction u other than
∇f, in order to reduce this component.
r j
i 2 2 2 2
2 2
3 / 2 3
) 2 , (
y x y
x y y
x y x
x
f
, 4 0
3 , 4 2 ) ,
(x y x2 y2 z f
Directional Derivatives
Example 8
Direction to Cool Off Fastest
Solution)
The temperature in a rectangular box is approximated by
If a mosquito is located at (½,1,1), in which direction should it fly to cool off as rapidly as possible?
. 3 0
, 2 0
, 1 0
) 3 )(
2 )(
1 ( )
, , (
z y
x
z y
x xyz
z y x T
j i
k j
i
) 2 2 )(
3 )(
1 (
) 2 1 )(
3 )(
2 (
) , , ( )
, , ( )
, , ) (
, , (
y z
x xz
x z
y yz
x z y x T y
z y x T x
z y x z T
y x T
4 k 1 1 , 1 2 ,
1
T
To cool off most rapidly, the mosquito should fly in the direction of -1/4k;
that is, it should dive for the floor of
the box, where the temperature is
T(x,y,0)=0.
Directional Derivatives
x
y
z
) , ( x y f
z
surface
c z
plane
c y
x
f ( , )
Surface
x y
Level Curves
increasing value of
f
Level Curves
c 10
11 c
12
c
Directional Derivatives
x
y
z
) , ( x y f
z
surface
c z
plane
c y
x
f ( , )
Surface
x
y
Level Curves
increasing value of
f
Level Curves
c 10
11 c
12 c
P
0u
1P
1u
2
P
2u
3P
3Level Curves
x y
Directional Derivatives
) , ( x y f
z
x
y
z
surface
c z
plane
c y x f
( , )
Surface
increasing value of
f
Level Curves
c 10
11 c
12 c
( , ) D f x y
uThe rate of change of in the direction given by the vector : f u
1
( , )
D f x y
u 0
1 0
( ) ( ) 10 10 f P f P 0
P
0u
3u
1u
2
P
1P
2P
3Directional Derivatives
) , ( x y f
z
x
y
z
surface
c z
plane
c y x f
( , )
Surface
x y
Level Curves
increasing value of
f
Level Curves
c 10
11 c
12 c
P
0u
3u
1u
2
( , ) D f x y
uThe rate of change of in the direction given by the vector : f u
2
( , ) Vs.
3( , ) D f x y
uD f x y
u2 0 3 0
2 0 3 0