UNIT 6

Jacobian

If u and v are functions of the two independent variables x and y , then the determinant,

Is known as the jacobian of u and v with respect to x and y, and it can be written as,

Suppose there are three functions u , v and w of three independent variables x , y and z then,

The Jacobian can be defined as,

Important properties of the Jacobians-

Property-1-

If u and v are the functions of x and y , then

Proof- Suppose u = u(x,y) and v = v(x,y) , so that u and v are the functions of x and y,

Now,

Interchange the rows and columns of the second determinant, we get

=

= ………….(1)

Differentiate u = u(x,y) and v= v(x,y) partially w.r.t. u and v, we get

Putting these values in eq.(1) , we get

hence proved.

Property-2:

Suppose u and v are the functions of r and s, where r,s are the fuctions of x , y, then,

Proof: =

Interchange the rows and columns in second determinant

We get,

=

=

= =

Similarly we can prove for three variables.

Property-3

If u,v,w are the functions of three independent variables x,y,z are not independent , then,

Proof: here u,v,w are independent , then f(u,v,w) = 0 ……………….(1)

Differentiate (1), w.r.t. x, y and z , we get

…………(2)

………………(3)

………………..(4)

Eliminate from 2,3,4 , we get

Interchanging rows and columns , we get

= 0

So that,

Example-1: If x = r sin , y = r sin , z = r cos, then show that

sin also find

Sol. We know that,

=

=

= ( on solving the determinant)

=

Now using first propert of Jacobians, we get

Example-2: If u = x + y + z , uv = y + z , uvw = z , find

Sol. Here we have,

x = u – uv = u(1-v)

y = uv – uvw = uv( 1- w)

And z = uvw

So that,

=

Apply

=

Now we get,

= u²v(1-w) + u²vw

= u²v

Example-3: If u = xyz , v = x² + y² + z² and w = x + y + z, then find J =

Sol. Here u ,v and w are explicitly given , so that first we calculate

J’ =

J’ = =

= yz(2y-2z) – zx(2x – 2z) + xy (2x – 2y) = 2[yz(y-z)-zx(x-z)+xy(x-y)]

= 2[x²y - x²z - xy² + xz² + y²z - yz²]

= 2[x²(y-z) - x(y² - z²) + yz (y – z)]

= 2(y – z)(z – x)(y – x)

= -2(x – y)(y – z)(z – x)

By the property,

JJ’ = 1

J =

Suppose w = f(x,y) has continuous partial variables fx and fy and if x = x(t) and y = y(t) are differentiable function of t , then the composite function w = f(x(t), y(t)) is a differentiable function of t. we get,

Or,

Suppose w = f(x,y,z) has continuous partial variables fx , fy and fz and if x = x(t) and y = y(t) and z = z(t) are differentiable function of t , then the composite function w = f(x(t), y(t), z(t)) is a differentiable function of t. we get,

Or

Example-1: If w = x² + y – z + sin t and x + y = t then find

Sol. With x, y and z independent , we have

Therefore , we get

= 2x + cos ( x+ y)

Example-2: If w = x² + y – z + sin t and x + y = t then find

Sol. With x, t and z dependent , we have

y = t – x , w = x² + ( t – x ) + sin t

So that,

Chain rule (Jacobians)-

Suppose u and v are the functions of r and s where r,s are the functions of x, y.

Then, according chain rule-

Similarly for three variables-

Which is also called the second property of Jacobians.

As we know that the value of a function at maximum point is called maximum value of a function. Similarly the value of a function at minimum point is called minimum value of a function.

The maxima and minima of a function is an extreme biggest and extreme smallest point of a function in a given range (interval) or entire region. Pierre de Fermat was the first mathematician to discover general method for calculating maxima and minima of a function. The maxima and minima are complement of each other.

Maxima and Minima of a function of one variables

If f(x) is a single valued function defined in a region R then

Maxima is a maximum point if and only if

Minima is a minimum point if and only if

Maxima and Minima of a function of two independent variables

Let be a defined function of two independent variables.

Then the point is said to be a maximum point of if

Or =

For all positive and negative values of h and k.

Similarly the point is said to be a minimum point of if

Or =

For all positive and negative values of h and k.

Saddle point:Critical points of a function of two variables are those points at which both partial derivatives of the function are zero. A critical point of a function of a single variable is either a local maximum, a local minimum, or neither. With functions of two variables there is a fourth possibility - a saddle point.

A point is a saddle point of a function of two variables if

At the point.

Stationary Value

The value is said to be a stationary value of if

i.e. the function is a stationary at (a , b).

Rule to find the maximum and minimum values of

- Calculate.
- Form and solve , we get the value of x and y let it be pairs of values
- Calculate the following values :

4. (a) If

(b) If

(c) If

(d) If

Example1 Find out the maxima and minima of the function

Given …(i)

Partially differentiating (i) with respect to x we get

….(ii)

Partially differentiating (i) with respect to y we get

….(iii)

Now, form the equations

Using (ii) and (iii) we get

using above two equations

Squaring both side we get

Or

This show that

Also we get

Thus we get the pair of value as

Now, we calculate

Putting above values in

At point (0,0) we get

So, the point (0,0) is a saddle point.

At point we get

So the point is the minimum point where

In case

So the point is the maximum point where

Example2 Find the maximum and minimum point of the function

Partially differentiating given equation with respect to and x and y then equate them to zero

On solving above we get

Also

Thus we get the pair of values (0,0), (,0) and (0,

Now, we calculate

At the point (0,0)

So function has saddle point at (0,0).

At the point (

So the function has maxima at this point (.

At the point (0,

So the function has minima at this point (0,.

At the point (

So the function has an saddle point at (

Example3 Find the maximum and minimum value of

Let

Partially differentiating given function with respect to x and y and equate it to zero

..(i)

..(ii)

On solving (i) and (ii) we get

Thus pair of values are

Now, we calculate

At the point (0,0)

So further investigation is required

On the x axis y = 0 , f(x,0)=0

On the line y=x,

At the point

So that the given function has maximum value at

Therefore maximum value of given function

At the point

So that the given function has minimum value at

Therefore minimum value of the given function

Generally to calculate the stationary value of a function with some relation by converting the given function into the least possible independent variables and then solve them. When this method fail we use Lagrange’s method.

This method is used to calculate the stationary value of a function of several variables which are all not independent but are connected by some relation.

Let be the function in the variable x, y and z which is connected by the relation

Rule: a) Form the equation

Where is a parameter.

b) Form the equation using partial differentiation is

(We always try to eliminate)

c) Solve the all above equation with the given relation

These give the value of

These value obtained when substituted in the given function will give the stationary value of the function.

Example1 Divide 24 into three parts such that the continued product of the first, square of second and cube of third may be maximum.

Let first number be x, second be y and third be z.

According to the question

Let the given function be f

And the relation

By Lagrange’s Method

….(i)

Partially differentiating (i) with respect to x,y and z and equate them to zero

….(ii)

….(iii)

….(iv)

From (ii),(iii) and (iv) we get

On solving

Putting it in given relation we get

Or

Or

Thus the first number is 4 second is 8 and third is 12

Example2 The temperature T at any point in space is .Find the highest temperature on the surface of the unit sphere.

Given function is

On the surface of unit sphere given [ is an equation of unit sphere in 3 dimensional space]

By Lagrange’s Method

….(i)

Partially differentiating (i) with respect to x, y and z and equate them to zero

or …(ii)

or …(iii)

…(iv)

Dividing (ii) and (ii) by (iv) we get

Using given relation

Or Or

So that

Or

Thus points are

The maximum temperature is

Example3 If ,Find the value of x and y for which is maximum.

Given function is

And relation is

By Lagrange’s Method

[] ..(i)

Partially differentiating (i) with respect to x, y and z and equate them tozero

Or …(ii)

Or …(iii)

Or …(iv)

On solving (ii),(iii) and (iv) we get

Using the given relation we get

So that

Thus the point for the maximum value of the given function is

Example4 Find the points on the surface nearest to the origin.

Let be any point on the surface, then its distance from the origin is

Thus the given equation will be

And relation is

By Lagrange’s Method

….(i)

Partially differentiatig (i) with respect to x, y and z and equate them to zero

Or …(ii)

Or …(iii)

Or

Or

On solving equation (ii) by (iii) we get

And

On subtracting we get

Putting in above

Or

Thus

Using the given relation we get

= 0.0 +1=1

Or

Thus point on the surface nearest to the origin is