Unit 5

Applications of Multivariable Differential Calculus

5.1      Jacobians, Properties of Jacobians, Jacobians of composite functions

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

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

= 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

Jacobian of Implicit functions-

If the variables u,v, x, and y are connected by the following equations-

   ………………… (1)

   ………………….. (2)

u,v are the implicit functions of x and y.

Now differentiate equation (1) and (2) with respect to x and y-

   ………… (3)

……………… (4)

………………. (5)

  ………………. (6)

Now,

  =

Using equation (3), (4), (5), (6), we get-

=

So that-

We can find this for three variables as well.

Example-1: If u = 2axy, v = then prove that-

Sol. Here we have,

u = 2axy, v =

Then

Here - 

So that

Now,

Hence-

Hence proved.

Example-2: If , then prove that-

Sol. Suppose 

And                 

Now

And

So that,

Hence proved.

Example-3: If x + y + z = u,  y + z = uv , z = uvw, then prove that-

Sol. Suppose

And

Hence,

Hence proved.

Jacobians for functional dependence functions-

Note- two functions u and v said to be functionally dependent if their jacobian is equals to zero. That means J (u , v) = 0

Suppose u and v are functionally dependent functions, then

f( u, v) = 0

Differentiate this equation with respect to x and y-

= 0

= 0

There will be a non-trivial solution for ,   to this system exists.

So that-

On transposing, we get-

Example-1: Show that and are functionally dependent.

Sol. Here we have-

and

Now we will find out the Jacobian to check the functional dependence.

=

Here Jacobian is zero, so we can conclude that these functions are functionally dependent.

Example-2: Prove that u, v, w are functionally dependent, where-

Sol. Here we have 

Now we will find out the Jacobian of the given functions-

=

=

= 0

Therefore, u, v, w are functionally dependent.

Key takeaways-

1. 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,

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

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

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

5.2      Errors and approximations

Let f(x, y) be a continuous function of x & y. Let be the increments x & y resp. Then the value of f(x, y) will be

Expanding above expression by taylor’s theorem & since are very small Hence their product and higher powers are negligible and hence neglected we get,

i.e.

Similarly,

If f be a function of variables x, y, z, t, ………. Then we have

Note that

1. Dx, dy, dz, ……. May be taken as actual errors or interments in x, y, z ……… resp. While df is the approximate error in f.
2. ……….. Denotes relative errors in x, y, z,………….. Resp.
3. ………….
4. Denotes percentage errors in, x, y, z, …………. Resp.

Ex.

1. Find the percentage error in the area of an ellipse where error of ly is made in measuring it’s major and minor axes.

Solution:

Let A be the area of an ellipse and ‘a’ & ‘b’ are its semi major and minor axes

Taking log on both sides.

Differentiating we get,

Percentage error ion the area of an ellipse = 2%

2.     The density of a body is calculated from its’s weight W in air and win water if error are made in , find the error in .

Solution:

We know that,

Taking log on both sides we get,

Diff. We get,

Is the required error in density.

3.     Find the percentage error in computing the parallel resistance r of three resistances r1, r2, r3 from the formula.

Where

r1, r2, r3 are each in error of 1 & 2y.

4.     Find the approximate value of

Solution:

Let

    … (1)

Take x = 1, y = 2, z = 2

From equation (1)

approximate value is

In calculating the volume of right circular cone, error of 2y & 1.5y are made in measuring height and radius of base resp. Find they error in the calculated volume.

If , find the approximate value of f if x = 1.99, y = 3.01, z = 0.98.

Key takeaways-

1. If f be a function of variables x, y, z, t, ………. Then we have

5.     dx, dy, dz, ……. May be taken as actual errors or interments in x, y, z ……… resp. While df is the approximate error in f.

6.     ……….. Denotes relative errors in x, y, z,………….. Resp.

5.3      Maxima and minima of functions of two 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

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

4.     (a) If

(b) If

(c) If

(d) If

Example-1: Find out the maxima and minima of the function

Sol.

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 (

Example-3: There is rectangular box which is open at the top, is to have a volume of 32 c.c.

Find the dimensions of the box requiring least (minimum) material to construct it.

Sol. Here it is given that-

Volume (V) = 32 c.c.

Suppose ‘l’ , ‘b’, ‘h’ are the length, breadth, height of the rectangular box respectively and its surface area is ‘S’.

As we know that-

Volume (V) = l b h = 32

b (breadth) = 32/lh

And the surface area of the rectangular box which is open at the top is-

S = 2 (l + b) h + l b   …………… (1)

On putting the value of ‘b’ in (1), we get-

S = 2

S =         ……………… (2)

Differentiate partially equation (2) with respect to l and h respectively, we get-

  …………… (3)  and …………….. (4)

For Max. And Min. S, we get-

And     

The values of ‘l’, ‘h’ and ‘b’ will be-

L = 4, h = 2 and b = 4

Now-

And

So that-

, then S is minimum for l = 4, b = 4, and h = 2

Key takeaways-

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

At the point.

2.     The value is said to be a stationary value of if

5.4      Lagrange’s method of Undetermined multipliers (one condition)

Suppose is a function of three independent variables, where x,y,z are related to a constraint

(x, y, z) = 0   …………………… (1)

So that for stationary values-

,     , 

  ………………….. (2)

By total differentiation of (1), we get-

   …………………… (3)

Now multiply by  and adding to (2), we get-

That equation holds good if-

We can find the values of x, y, z and by solving these equations.

Example-1: Determine the maxima and minima of when 1

Sol. Suppose    ……….. (1)

And =0  …………… (2)

Differentiate partially equation (1) and (2) w.r.t. x and y respectively, we get-

And                            

Now using Lagrange’s equations, we get-

Which gives-

or    ……… (3)

Which gives-

…… (4)

Which gives-

……. (5)

Multiply these equations by x, y, z respectively and adding, we get-

(

Hence we get-

f + 0   then 

Put in (3) , (4) and (5), we get-

x(1 – fa) = 0   ,   y(1 – fb) = 0 and z(1 – fc) = 0

We get-

These gives the maximum and minimum values of f.

Example: 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  .

Example: 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

Maxima and minima of function of two variables and three variables using method of Lagrange’s multiplier-

Generally we 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 fails 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

Example-4: 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

References

1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.

2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.

3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers

4. HK dass, engineering mathematics.