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