UNIT 3

Differential Calculus-II

3.1. Expansion of functions of several variables (Theorems for two variables)

Taylor’s theorem for functions of two variables-

Suppose f(x , y) be a function of two independent variables x and y. Then,

+ ……………

Maclaurin’s theorem for functions of two variables-

When we put a = 0 and b = 0 (about origin) in Taylor’s series, we get-

+ ……………

Example-1:  Expand f(x , y) = in powers of x and y about origin.

Sol. Here we have the function-

f(x , y) =

Here , a = 0 and b = 0 then

f(0 , 0) =

Now we will find partial derivatives of the function-

Now using Taylor’s theorem-

+………

Suppose h = x and k = y , we get

+…….

= +……….

Example-2: Find the Taylor’s expansion of about (1 , 1) up to second degree term.

Sol. We have,

At (1 , 1)

Now by using Taylor’s theorem-

……

Suppose 1 + h = x  then h =  x – 1

1 + k = y   then k = y - 1

……

  =

……..

Example-3: Calculate log [ ]approximately by using Taylor’s expansion.

Sol.  Suppose-   f(x , y) = log [ ]

f(1 , 1) = log 1 = 0

Now we will find out the derivatives- take a = 1 and b = 1

Using Taylor’s theorem-

+ ……………….

But here,

So that-

Put h = 0.03 and k = -0.02 , we get

Example-4: Expand near the point (1 , π/4) by using Taylor’s theorem.

Sol.                                Let f(x , y) = 

We know that,

+ ……………

Then,

f(x , y) = 

Where h = x – 1 and k =

So that,

f(x , y) =  

Put the above values in Taylor’s expansion-

+……

3.2. Taylor’s and Maclaurin’s Theorem

Taylor’s Theorem-

If f(x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-

+ …….+ + ……..

Which is called Taylor’s theorem.

If we put x = a, we get-

+ …….+ + ……..     (1)

Maclaurin’s Theorem-

If we put a = 0 and h = x then equation(1) becomes-

+ …….

Which is called Maclaurin’s theorem.

Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)  

We get-

+ ……. 

Example-1: Express the polynomial in powers of (x-2).

Sol.  Here we have,

f(x) =

Differentiating the function w.r.t.x-

f’(x) =

f’’(x) = 12x + 14

f’’’(x) = 12

f’’’’(x)=0

Now using Taylor’s theorem-

+ …….     (1)

Here we have, a = 2,

Put x = 2 in the derivatives of f(x), we get-

f(2) = 

f’(2) =

f’’(2) = 12(2)+14 = 38

f’’’(2) = 12   and f’’’’(2) = 0

Now put a = 2 and substitute the above values in equation(1), we get-

Example-2: Expand sin x in powers of

Sol. Let f(x) = sin x

Then,

=

By using Taylor’s theorem-

+ …….     (1)

Here f(x) = sin x and a = π/2

f’(x) = cos x  , f’’(x) = - sin x    ,     f’’’(x) = - cos x  and so on.

Putting x = π/2 , we get

f(x) = sin x = = 1

f’(x) = cos x = = 0

f’’(x) = -sin x = = -1

f’’’(x) = -cos x = = 0

From equation (1) put a = and substitute these values, we get-

+ …….    

= ………………………..

3.3. Jacobian & Properties of 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 =

3.4. 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.

3.5. Functional Dependence

Let be two functions, and u and v are connected by the relation f(u , v) = 0, where f is differentiable, then u and v are said to be functionally dependent on each other.

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.

3.6. Maxima and minima of functions of two variables

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

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

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 (

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

3.7. Lagrange’s Methods of Multipliers

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 Lagranges’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-2: Find the minimum distance from the point (1, 2, 0) to the cone                                                            

Sol. Let there are three points (x, y, z) on the cone then the distance from the point (1, 2, 0) by using distance formula will be-

Suppose                                     ………….. (1)

Subject to                                  ………………. (2)

From equation (1) and (2), for minimum, we get-

………….. (3)

   ……………. (4)

Multiply (4) by and adding in (3),  we get-

  ,         ,     

We get-

We get the values of x and y-

x = 1/2        ,         y = 1

Put the values of x and y in equation (2), we get-

So that the min. Distance form the point (1, 2, 0) is-

Therefore,

                      Ans.

Example-3: Find the volume of the largest rectangular parallelepiped which can be inscribed in the ellipsoid by using method of Lagranges’s multipliers.

Sol. We have-

Let                                            ……………. (1)

Suppose 2x , 2y and 2z are the l , b and h of the rectangular parallelepiped inscribed in the ellipsoid.

Then the volume will be-

V = (2x)(2y)(2z) = 8 xyz  ……………… (2)

Differentiate equation (1) and (2) partially with respect to x and y respectively, we get-

And

Now, Lagrange’s equations are-

So that,

  …………… (3)

………………. (4)

……………………… (5)

Multiply (3), (4) and (5) by x, y, z and adding, we get-

As 

Then,

2

Putting the value of in equation (3), (4), and (5), we get-

  gives        

Which gives-

Now put in (4)-

Similarly,

So that the volume of the largest rectangular parallelepiped = 8xyz

= 8                     Ans.