Limits-

The function f(x,y) is said to tend to limit ‘l’ , as x →a and  y→bIff the limit is dependent on point (x,y) as  x →a and  y→b

We can write this as,

Example-1: evaluate the

Sol. We can simply find the solution as follows,

Example-2:  evaluate

Sol.

-6.

Example-3: evaluate

Sol.

Conitinuity –

At point (a,b) , a function f(x,y) is said to be continuous if,

Working rule for continuity-

Step-1: f(a,b) should be well defined.

Step-2: should exist.

Step-3:

Example-1: Test the continuity of the following function-

Sol. (1) the function is well defined at (0,0)

(2) check for the second step,

That means the limit exists at (0,0)

Now check step-3:

So that the function is continuous at origin.

 Example-2: check for the continuity of the following function at origin,

Sol. (1) Here the function is well defined at (0,0)

(2) check for second step-

Limit f is not unique for different values of m

So that the limit does not exists.

Therefore the function is not continuous at origin.

Steps to check for existence of limit-

Step-1: find the value of f(x,y) along x →a and  y→b

Step-2: find the value of f(x,y) along x →b and  y→a

Note- if the values in step -1 and step-2 are same then we can say that the limits exist otherwise not.

Step-3: if a →0 and  b→0 then find the limit along y =mx , if the value does not contain m then limit exist,  If it contains m then the limit does not exist.

Note-1- put x = 0 and y = 0 in f , then find f1

2 - Put y = 0 and x = 0 In f  then find x2

If f1 and f2 are equal then limit exist otherwise not.

3- put y = mx  then find f3

If f1 = f2 ≠f3, limit does not exist.

4- put y = mx² and find f4,

If f1 = f2 = f3 ≠f4 , limit does not exist

If f1 = f2 = f3 = f4 , limit exist.

Example-1:  Evaluate 

Sol . 1.     

2.

Here f1 = f2

3. now put y = mx, we get

Here f1 = f2 = f3

Now put y = mx²

4.

Therefore ,

                      F1 = f2 = f3 =f4

We can say that the limit exists with 0.

Example-2: evaluate the following-

Sol.  First we will calculate f1 –

Here we see that f1 = 0

Now find f2,

Here , f1 = f2

Therefore the limit exists with value 0.

Partial derivatives-

First order partial differentiation-

Let f(x , y) be a function of two variables. Then the partial derivative of this function with respect to x can be written as and defined as follows:

Now the partial derivative of f with respect to f can be written as and defined as follows:

Note: a. while calculating partial derivatives treat all independent variables, other than the variable with respect to which we are differentiating , as constant.

b. we apply all differentiation rules.

Higher order partial differentiation-

Let f(x , y) be a function of two variables. Then its second-order partial derivatives, third order partial derivatives and so on are referred as higher order partial derivatives.

These are second order four partial derivatives:

(a)     =

(b)  =

(c)   =

(d)     =

b and c are known as mixed partial derivatives.

Similarly we can find the other higher order derivatives.

Example-1: -Calculate    and   for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

sol. To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

= [3x³-5y²+2xy-8x+4y-20]

              = 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

              = 9x² - 0 + 2y – 8 + 0 – 0

              = 9x² + 2y – 8

Similarly partial derivative of f(x,y) with respect to y is:

= [3x³-5y²+2xy-8x+4y-20]

                           = 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

                           = 0 – 10y + 2x – 0 + 4 – 0

                           = 2x – 10y +4.

Example-2: Calculate    and   for the following function

f( x, y) = sin(y²x + 5x – 8)

sol. To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

[sin(y²x + 5x – 8)]

                           = cos(y²x + 5x – 8)(y²x + 5x – 8)

                           = (y² + 50)cos(y²x + 5x – 8)

Similarly partial derivative of f(x,y) with respect to y is,

[sin(y²x + 5x – 8)]

                           = cos(y²x + 5x – 8)(y²x + 5x – 8)

                           = 2xycos(y²x + 5x – 8)

Example-3: Obtain all the second order partial derivative of the function:

f( x, y) = ( x³y² - xy⁵)

sol. 3x²y² - y⁵,      2x³y – 5xy⁴,

  = = 6xy²

  = 2x³ - 20xy³

= = 6x²y – 5y⁴

= = 6x²y -   5y⁴

Example-4: Find

Sol.  First we will differentiate partially with repsect to r,

Now differentiate partially with respect to θ, we get

Example-5:  if,

then find.

Sol-

Example-6: if , then show that-

Sol. Here we have,

u =     …………………..(1)

now partially differentiate eq.(1) w.r to x and y , we get

                                               =

Or

                 ………………..(2)

And now,

                                                =

         ………………….(3)

Adding eq. (1) and (3) , we get

=  0

Hence proved.

Total derivative-

When we measure the rate of change of the dependent variable owing to any change in a variable on which it depends, when none of the variable is assumed to be constant.

Let the function,     u = f( x, y),  such that    x = g(t)  ,     y = h(t)

ᵡ  Then we can write,

This is the total derivative of u with respect to t.

Change of variable-

If w = f (x, y) has continuous partial variables fx and fyand if x = x (t), y = y (t) are

Differentiable functions of t, then the composite function w = f (x (t), y (t)) is a

Differentiable function of t.

In this case, we get,

fx(x (t), y (t)) x’(t)+ fy(x(t), y (t)) y’(t).

Example-:1 let q = 4x + 3y      and    x = t³ + t² + 1    , y = t³ - t² - t

                     Then find  .

Sol:

. =

Where,

f1 = ,  f2 = 

In this example f1 = 4    ,      f2 = 3

Also,              3t² + 2t    ,               

4(3t² + 2t) + 3(

=  21t² + 2t – 3

Example-2: Find      if u = x³y⁴    where   x = t³     and    y = t².

Sol. as we know that by definition,     =

3x²y⁴3t² + 4x³y³2t = 17t¹⁶.

Example-3: if w = x² + y – z + sintand  x + y = t, find

 (a)    y,z

(b)   t, z

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

t = x + y, w = x²+ y - z + sin (x + y).

Therefore,

y,z = 2x + cos(x+y)(x+y)

=  2x + cos (x + y)

With x, t, z independent, we have

                                                          Y = t-x,   w=  x² + (t-x) + sin t

Thus t, z  = 2x - 1

Example-4: If u = u( y – z , z - x , x – y)  then prove that  = 0

Sol. Let,

Then,

By adding all these equations we get,

= 0

hence proved.

Example-5:   if φ( cx – az ,  cy – bz) = 0 then show that ap + bq = c

                          Where p =      q = 

Sol.  We have,

φ( cx – az ,  cy – bz) = 0

φ( r , s) = 0

where,

We know that,

Again we do,

By adding the two results, we get

Example-6: If z is the function of x and y , and x =   , y = , then prove that,

Sol. Here , it is given that,  z is the function of x and y & x , y are the functions of u and v.

So that,

  ……………….(1)

And,

   ………………..(2)

Also there is,

x =   and  y = ,

now,

   ,       ,      ,    

From equation(1) , we get

  ……………….(3)

And from eq. (2) , we get

    …………..(4)

Subtracting eq. (4) from (3), we get

= ) – (

= x

Hence proved.

1. Limits-

The function f(x,y) is said to tend to limit ‘l’ , as x →a and  y→bIff the limit is dependent on point (x,y) as  x →a and  y→b

We can write this as,

2.     Conitinuity –

At point (a,b) , a function f(x,y) is said to be continuous if,

(a)     =

(b)  =

(c)   =

(d)     =

Equation of a tangent plane-

And

Equation of the normal to the plane-

Example: Find the equation of the tangent plane and normal line to the surface

Sol.

Here,

At point (1, 2, -1)-

Therefore the equation of the tangent plane at (1, 2, -1) is-

Equation of normal line is-

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 (

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

1. 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.
2. 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

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

at the point.

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

Suppose f(x, y, z) be the scalar function and it is continuously differentiable then the vector-

Is called gradient of f and we can write is as grad f.

So that-

Here is a vector which has three components

Properties of gradient-

Property-1:

Proof:

First we will take left hand side

                                               L.H.S =

                                                         =

                                                         =

                                                          =

Now taking R.H.S,

                                       R.H.S. =

                                                   =

                                                      =

Here-                     L.H.S. = R.H.S.

         Hence proved.

Property-2: Gradient of a constant (

Proof:

Suppose

                                              Then

We know that the gradient-

                                                    = 0

Property-3: Gradient of the sum and difference of two functions-

If f and g are two scalar point functions, then

Proof:   

L.H.S                                

        Hence proved

Property-4: Gradient of the product of two functions

If f and g are two scalar point functions, then

Proof:

So that-

        Hence proved.

Property-5: Gradient of the quotient of two functions-

If f and g are two scalar point functions, then-

Proof:

So that-

Example-1: If , then show that

1.

2.

Sol.

Suppose   and

Now taking L.H.S,

Which is

                                                                                               Hence proved.

2.

So that

Example: If then find grad f at the point (1,-2,-1).

Sol.

Now grad f at (1 , -2, -1) will be-       

Example: If then prove that grad u , grad v and grad w are coplanar.

Sol.

Here-                     

Now-

Apply

Which becomes zero.

So that we can say that grad u, grad v and grad w are coplanar vectors.

Divergence (Definition)-

Suppose is a given continuous differentiable vector function then the divergence of this function can be defined as-

Curl (Definition)-

Curl of a vector function can be defined as-

Note- Irrotational vector-

If then the vector is said to be irrotational.

Vector identities:

Identity-1:  grad uv = u grad v + v grad u

Proof: 

So that

graduv = u grad v + v grad u

Identity-2: 

Proof:

Interchanging , we get-

We get by using above equations-

Identity-3

Proof:   

So that-

Identity-4

Proof:

So that,

Identity-5                  curl (u

Proof:

So that

curl (u

Identity-6:

Proof:

So that-

Identity-7:

Proof:

So that-

Example-1: Show that-

1.

2.

Sol. We know that-

2. We know that-

= 0

Example-2: If then find the divergence and curl of .

Sol.  we know that-

Now-

Example-3: Prove that

Note- here is a constant vector and

Sol. here    and

So that

Now-

So that-

1. 
2. 
3. 
4. Gradient of a constant (

If f and g are two scalar point functions, then

5.     If f and g are two scalar point functions, then

6.     If f and g are two scalar point functions, then-

7.    

8.    

9.     If then the vector is said to be irrotational.

10. grad uv = u grad v + v grad u

11.

12. curl (u

Directional derivative

Let ϕ be a scalar point function and let ϕ(P) and ϕ(Q) be the values of ϕ at two neighbouring points P and Q in the field. Then,

,are the directional derivative of ϕ in the direction of the coordinate axes at P.

The directional derivative of ϕ in the direction l, m, n=l + m+

The directional derivative of ϕ in the direction of =

Example: Find the directional derivative of 1/r in the direction where

Sol. Here

Now,

And

We know that-

So that-

Now,

            Directional derivative =

Example: Find the directional derivative of

At the points (3, 1, 2) in the direction of the vector .

Sol. Here it is given that-

Now at the point (3, 1, 2)-

Let be the unit vector in the given direction, then

at (3, 1, 2)

Now,

Example: Find the directional derivatives of at the point P(1, 1, 1) in the direction of the line

Sol. Here

Direction ratio of the line are 2, -2, 1

Now directions cosines of the line are-

Which are

Directional derivative in the direction of the line-

1. The directional derivative of ϕ in the direction l, m, n= l + m+
2. The directional derivative of ϕ in the direction of =