UNIT 3

PARTIAL DIFFERENTIATION 2

3.1 Jacobin’s

Jacobians

If u and v be continuous and differentiable functions of two other independent variables x and y such as

, then we define the determine

asJacobian of u, v with respect to x, y

Similarly ,

JJ’ = 1

Example 1:

Let x(u,v) = u2 – v2, y(u,v)=uv.find the jacobain j(u,v).

Solution:

Given that x(u,v) =u2-v2 and y(u,v)= 2uv

We know that,

J(u,v)=

Therefore, J(u,v) = 4u2+4v2

Example 2:

Find if u= 2xy, v= x2 – y2 and x = r.cos , y = r.sin

Solution:

=

=

Actually Jacobins are functional determines

Ex.

  Calculate

If

If

ST

4.      find

5.     If and , find

6.    

7.     If  

8.     If , ,

JJ1 = 1

If ,

JJ1=1

Jacobian of composite function (chain rule)

Then

Ex.

If

Where

2.     If   and

Find

3.     If

Find

Jacobian of Implicit function

Let u1, u2 be implicit functions of x1, x2 connected by f1, f2 such there

,

Then

Similarly,

Ex.

If

If

Find

3.2 Taylor’s and Maclaurin’s series for functions of two variables

Taylor’s series

Let be   a function of two variables x  and y  then

Which is known as Taylor’s expansion  of in power of  and .

Malclaurian’s Series

It is particular case of Taylor’s theorem by putting  a=0 and b=0

Example1:Expandin power of up to second  terms.

Given function

Here  

Now,  

and

The Taylor’s expansion in power is

Example2: Expand the  in power of x and y up to third term?

Given function 

Now,

By Maclaurins expansion

+…

3.3 Maxima and Minima of functions of two variables

Let z = f(x, y)

Now for stationary point dz = 0

&

This gives the set of values of x and y for which maxima or minim occurs

Now find

We called it as r, s, t resp.

Thus function has maximum or minimum

ifrt – s2>0

i.e.

further if

; function is minimum at (x, y) &

; Function is maximum at (x, y)

Note that

If

; then function will not have either maxima or minima such point is called saddle point.

If

; then more details are required to justly maxima or minima

Ex. Discuss the stationary values of

Ex. Find the values of x and y for which x2 + y2 + 6x = 12 has a minimum values and find its minimum value.

Divide 120 into three parts so that the sum of their product. Taken two at a times shall be maximum.

Using Lagrange’s method divide 24 into three parts. Such that continued product of the first, square of second, cube of third may be maximum.

Find the maximum and minimum value of x2 + y2 when 3x2 + 4xy + 6y2 = 140

is satisfied.

3.4 Lagrange‟s method of undetermined multipliers

Let be a function of x, y, z which to be discussed for stationary value.

Let be a relation in x, y, z

for stationary values we have,

i.e.    … (1)

Also from we have

    … (2)

Let ‘’ be undetermined multiplier then multiplying equation (2) by and adding in equation (1) we get,

     … (3)

     … (4)

      … (5)

Solving equation (3), (4) (5) & we get values of x, y, z and .

Solution:

Let x, y, z be the three parts of ‘a’ then we get.

    … (1)

Here we have to maximize the product

i.e.

By Lagrange’s undetermined multiplier, we get,

       … (2)

       … (3)

        … (4)

i.e.

        … (2)’

        … (3)’

        … (4)

And

From (1)

Thus .

Hence their maximum product is  .

2.     Find the point on plane nearest to the point (1, 1, 1) using Lagrange’s method of multipliers.

Solution:

Let be the point on sphere which is nearest to the point . Then shortest distance.

Let

Under the condition    … (1)

By method of Lagrange’s undetermined multipliers we have

       … (2)

       … (3)

i.e. &

       … (4)

From (2) we get

From (3) we get

From (4) we get

Equation (1) becomes

i.e.

y = 2

Exercise:

If where x + y + z = 1.

Prove that the stationary value of u is given by,

Reference Books:

1. A text book of Applied Mathematics Volume I and II by J.N. Wartikar and P.N. Wartikar

2. Higher Engineering Mathematics by Dr. B. S. Grewal

3. Advanced Engineering Mathematics by H. K. Dass

4. Advanced Engineering Mathematics by Erwins Kreyszig