ME 2304: 3D Geometry & Vector Calculus

Dr. Faraz Junejo

Gradient of a Scalar field & Directional Derivative

Partial Derivatives

Let f(x,y) be a function with two variables.

If we keep y constant and differentiate f (assuming f is differentiable)

with respect to the variable x, we obtain what is called the partial

derivative of f with respect to x which is denoted by:

xforx

f

Similarly If we keep x constant and differentiate f (assuming f is differentiable) with respect to the variable y, we obtain what is called the partial derivative of f with respect to y which is denoted by

yfory

f

Ex 1. 2( , ) 3 lnf x y x y x y

6 lnf

xy yx

2 13

fx x

y y

Ex 2.

2

( , ) xy yg x y e

2

2 1 xy ygxy e

y

Partial Derivatives: Examples

Ex 3. 4 3( , , ) 2f x y z xy z xy

4 3 2f

y z yx

3 34 2f

xy z xy

4 23f

xy zz

Partial Derivatives: Examples

Example 1: Find the partial derivatives fx and fy if f(x , y) is given by f(x , y) = x2 y + 2x + y

Example 2: Find fx and fy if f(x , y) is given by

f(x , y) = sin(x y) + cos x

Example 3: Find fx and fy if f(x , y) is given by

f(x , y) = x ex y

Example 4: Find fx and fy if f(x , y) is given by

f(x , y) = ln ( x2 + 2 y)

• If f(x, y) = x3 + x2y3 – 2y2

find fx(2, 1) and fy(2, 1)

Example 5

• Holding y constant and differentiating with respect to x, we get:

fx(x, y) = 3x2 + 2xy3

– Thus, fx(2, 1) = 3 . 22 + 2 . 2 . 13

= 16

Example 5 (contd.)

• Holding x constant and differentiating with respect to y, we get:

fy(x, y) = 3x2y2 – 4y

– Thus, fy(2, 1) = 3 . 22 . 12 – 4 . 1

= 8

Example 5 (contd.)

• If

• calculate

( , ) sin1

and

xf x y

y

f f

x y

Exercise: 1

• Using the Chain Rule for functions of one variable, we have:

2

1cos cos

1 1 1 1

cos cos1 1 1 1

f x x x

x y x y y y

f x x x x

y y y y y y

Exercise: 1(contd.)

• Find fx, fy, and fz if f(x, y, z) = exy ln z

– Holding y and z constant and differentiating with respect to x, we have:

– fx = yexy ln z

– Similarly,

– fy = xexy ln z

– fz = exy/z

Exercise: 2

• If f is a function of two variables, then

its partial derivatives fx and fy are also

functions of two variables.

HIGHER DERIVATIVES

• So, we can consider their partial derivatives

(fx)x , (fx)y , (fy)x , (fy)y

• These are called the second partial derivatives of f.

SECOND PARTIAL DERIVATIVES

• If z = f(x, y), we use the following notation:2 2

11 2 2

2 2

12

2 2

21

2 2

22 2 2

( )

( )

( )

( )

x x xx

x y xy

y x yx

y y yy

f f zf f f

x x x x

f f zf f f

y x y x y x

f f zf f f

x y x y x y

f f zf f f

y y y y

NOTATION

• Thus, the notation fxy (or ∂2f/∂y∂x) means

that we first differentiate with respect to x

and then with respect to y.

• In computing fyx , the order is reversed.

SECOND PARTIAL DERIVATIVES

• Find the second partial derivatives of

f(x, y) = x3 + x2y3 – 2y2

– We know that

fx(x, y) = 3x2 + 2xy3 fy(x, y) = 3x2y2 – 4y

Example 6

–Hence,

2 3 3

2 3 2

2 2 2

2 2 2

3 2 6 2

3 2 6

3 4 6

3 4 6 4

xx

xy

yx

yy

f x xy x yx

f x xy xyy

f x y y xyx

f x y y x yy

Example: 6 (contd.)

Ex 3.2 3 5( , ) lnf x y x y x x y

Second-Order Partial Derivatives (fxx, fyy)

23 3

22 20

fy x

x

2 22 1

6f f

xyy x x y y

22

2 26

f xx y

y y

Exercise: 3

Notation for Partial Derivatives

means xf

fx

means yf

fy

xy

ff xy

2

means

yx

ff yx

2

means

• Partial derivatives of order 3 or higher can also be defined.

2 3

2xyy xy y

f ff f

y y x y x

HIGHER DERIVATIVES

• Calculate fxxyz if f(x, y, z) = sin(3x + yz)

– fx = 3 cos(3x + yz)

– fxx = –9 sin(3x + yz)

– fxxy = –9z cos(3x + yz)

– fxxyz = –9 cos(3x + yz) + 9yz sin(3x + yz)

Example 7

Example: 8

Example: 8 (contd.)

Interpretations of Partial Derivatives

• As with functions of single variables partial derivatives represent the rates of change of the functions as the variables change.

• As we saw in the previous section, fx(x , y) represents the rate of change of the function f ( x, y) as we change x and hold y fixed while, fy(x , y) represents the rate of change of f ( x, y) as we change y and hold x fixed.

Scalar FieldEvery point in a region of space is assigned a

scalar value obtained from a scalar function f(x,

y, z), then a scalar field f(x, y, z) is defined in the

region, such as the pressure or temperature in

atmosphere, etc.

Examples of scalar quantitiesAltitude: Temperature:

Electric potential:Pressure:

Scalar Field

Scalar Field :

A scalar quantity, smoothly assigned to each point of a certain region of space is called a scalar field

Examples :

i) Temperature and pressure distribution in the atmosphere

ii) Gravitational potential around the earth

iii) Assignment to each point, its distance from a fixed point

222 zyxr

O

Scalar Field (contd.)

O

),,( zyx

),,( zyxf

x

z

y

Once a coordinate system is set up, a scalar field is mathematically represented by a function : )(),,( rfzyxf

is the value of the scalar assigned to the point (x,y,z)

),,( zyxf

A smooth scalar field implies that the function

, ,is a smooth or differentiable

function of its arguments, x,y,z.

),,( zyxf

Scalar Field (contd.)

Since the scalar field has a definite value at each point, we must have

),,(),,( zyxfzyxf

O),,( zyxf

Consider two coordinate systems.

x

y

z

O’

z

y

x

),,( zyxf

GradientThe gradient of a function, f(x, y), in two dimensions is defined as:

• The gradient of a function is a vector field.

• It is obtained by applying the vector operator to

the scalar function f(x, y)• Such a vector field is called a gradient (or conservative) vector field.

Gradient (contd.)

Del operator

x y z

i j k

Gradient

grad f f f

f fx y z

i j k

Gradient characterizes maximum increase. If at a point

P the gradient of f is not the zero vector, it represents

the direction of maximum space rate of increase in f at

P.

Example: 1

For the scalar field (x,y) = x∅ 2sin5y, calculate gradient of∅

• For the scalar field (x,y) = 3x + 5y∅ , calculate gradient of f.

Solution: Given scalar field (x,y) = 3x + 5y∅

• For the scalar field (x,y) = x∅ 4yz,calculate gradient of ∅.

Example: 2

Gradient of a Scalar field

• In vector calculus, the gradient of a scalar field

is a vector field that points in the direction of

the greatest rate of increase of the scalar field,

and whose magnitude is the greatest rate of

change.

Interpretation

• Consider a room in which the temperature is given by

a scalar field, T, so at each point (x,y,z) the

temperature is T(x,y,z).

• At each point in the room, the gradient of T at that

point will show the direction the temperature rises

most quickly.

• The magnitude of the gradient will determine how

fast the temperature rises in that direction.

• Since temperature T depends on those three variables

we can ask the question: how does T change when we

change one or more of those variables?

• And as always, the answer is found by differentiating

the function. In this case, because the function

depends on more than one variable, we're talking

partial differentiation.

Gradient of temperature field

• Now if we differentiate T with respect to x, that tells us

the change of T in the x-direction. That is therefore the i-

component of the gradient of T.

• You can see that there is going to be three components

of the gradient of T, in the i, j and k directions, which we

find by differentiating with respect to x, y and z

respectively. So this quantity "the gradient of T" must be

a vector quantity. Indeed it is a vector field.

Gradient of temperature field (contd.)

• This vector field is called "grad T" and written like and it is given as:

Gradient of temperature field (contd.)

T

kz

Tj

y

Ti

x

TT

Gradient of temperature field : Summary

• In three dimensions, a scalar field is simply a field that takes on a single scalar value at each point in space. For example, the temperature of all points in a room at a particular time t is a scalar field.

• The gradient of this field would then be a vector that pointed in the direction of greatest temperature increase.

• Its magnitude represents the magnitude of that increase.

Example: 3

If T(x,y,z) is given by:Determine

)sin(2),,( 3 zexzyxT yT

kzexjzexizexT

kzexz

jzexy

izexx

T

kz

Tj

y

Ti

x

TT

yyy

y

yy

)cos(2)sin(2)sin(6

)sin(2

)sin(2)sin(2

332

3

33

Example: 4 Given potential function V = x2y + xy2 + xz2, (a) find the gradient of V, and (b) evaluate it at (1, -1, 3).

Solution:(a)

2 2 2(2 ) ( 2 ) 2

V V VV

x y z

xy y z x xy xz

i j k

i j k

(b) (1, 1,3)

( 2 1 9) (1 2) 6

8 6

V

i j k

= i j k

2 2 2

8 6 1ˆ (8 6 )

1018 ( 1) 6

i j ka i j k Direction of

maximum increase

Summary The gradient of a scalar field is a vector field,

whose:

• Magnitude is the rate of change, and

• which points in the direction of the greatest

rate of increase of the scalar field.

• If the vector is resolved, its components

represent the rate of change of the scalar field

with respect to each directional component.

• Hence for a two-dimensional scalar field ∅ (x,y).

• And for a three-dimensional scalar field ∅ (x, y, z)

• Note that the gradient of a scalar field is the derivative of f in each direction

Summary (contd.)

• The gradient of any scalar field is a vector,

whose direction is the direction in which the

scalar increases most rapidly, and whose

magnitude is the maximum rate of change

Summary (contd.)

Directional Derivative

Directional Derivative: Example

Maximum and minimum value of Directional Derivative

• Since, the directional derivative of f in the direction of n is just the scalar projection of grad f along the direction of n i.e.

fDf

hence

Becausef

Dwhere

fD

fn

fn

fn

ˆ

ˆ

ˆ

,1cos1 have weand 0

, .n̂ and between angle theis

derivative ldirectionarepresent ,

cos

Maximum and minimum value of Directional Derivative

In other words,

• The maximum value of directional derivative

is and it occurs when has the same

direction as

• The minimum value of directional derivative is

a and it occurs when

has the opposite direction i.e.

f n̂

) or θθen f (i.e. wh 01cos

f f and n̂ ) or θθwhen 1801cos

Maximum and minimum value of Directional Derivative

In other words,

• The maximum value of directional derivative

is and it occurs when has the same

direction as

f n̂

) or θθen f (i.e. wh 01cos

Exercise: 1• Calculate the directional derivative of the following

function in the given direction and at the stated point.

(1,2,3)at direction in the 33),( 22 jyxyxf

Exercise: 2

• Calculate the directional derivative of the following

function in the given direction and at the stated point.

(0,-1,2)at 22direction in the ),( 22 kjiyxyxf

Exercise: 2 (contd.)

Summary

• The directional derivative in any direction is given by

the dot product of a unit vector in that direction with

the gradient vector. So in effect, a directional

derivative tells the slope of a surface in a given

direction.

• The directional derivative of f in the direction of n is

just the scalar projection of grad f along the

direction of n.