Unit - 3

Applications of Partial differentiation

3.1 Jacobians – properties

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

, then we define the determine

as Jacobian of u, v with respect to x, y

Similarly ,

JJ’ = 1

Actually Jacobins are functional determines

Ex.

1.   Calculate
2. If
3. 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.

1. 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

Partial derivative of implicit functions

Consider four variables u, v, x, y related by implicit function.

,

Then

Ex.

If and

Find

If and

Find

Find

If

Find

3.2 Taylor’s and Maclaurin’s theorems (without proofs) for functions of two variables

Expand by Maclaurin’s theorem,

Log sec x

Solution:

Let f(x) = log sec x

By Maclaurin’s Expansion’s,

   (1)

By equation (1)

Prove that

Solution:

Here f(x)  = x cosec x

=

Now we know that

Expand upto x6

Solution:

Here

Now we know that

    … (1)

    … (2)

Adding (1) and (2) we get

Show that

Solution:

Here

Thus

Taylor’s Series Expansion:-

a)      The expansion of f(x+h) in ascending power of x is

b)     The expansion of f(x+h) in ascending power of h is

c)      The expansion of f(x) in ascending powers of (x-a) is,

Using the above series expansion we get series expansion of f(x+h) or f(x).

Expansion of functions using standard expansions

Expand in power of (x – 3)

Solution:

Let

Here a = 3

Now by Taylor’s series expansion,

 … (1)

equation (1) becomes.

Using Taylors series method expand

in powers of (x + 2)

Solution:

Here

a = -2

By Taylors series,

   … (1)

Since

,  , …..

Thus equation (1) becomes

Expand in ascending powers of x.

Solution:

Here

i.e.

Here h = -2

By Taylors series,

    … (1)

equation (1) becomes,

Thus

Expand in powers of x using Taylor’s theorem,

Solution:

Here

i.e.

Here

h = 2

By Taylors series

  … (1)

By equation (1)

Exercise

a)      Expand in powers of (x – 2)

b)     Expand in powers of (x + 2)

c)      Expand in powers of (x – 1)

d)     Using Taylors series, express in ascending powers of x.

e)     Expand in powers of x, using Taylor’s theorem.

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

If rt – s2 >0

i.e.

Further if

1. ; function is minimum at (x, y) &
2. ; Function is maximum at (x, y)

Note that

1. If ; then function will not have either maxima or minima such point is called saddle point.
2. 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 .

1. Decampere a positive number ‘a’ in to three parts, so their product is maximum

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

If where x + y + z = 1.

Prove that the stationary value of u is given by,

Reference Books

1) Higher Engineering Mathematics by B. V. Ramana, Tata McGraw-Hill Publications, New Delhi.

2) A Text Book of Engineering Mathematics by Peter O’ Neil, Thomson Asia Pte Ltd. Singapore.

3) Advanced Engineering Mathematics by C. R. Wylie & L. C. Barrett, Tata Mcgraw-Hill Publishing Company Ltd., New Delhi.