UNIT 3

Question Bank

Question-1: evaluate

Sol.

Question-2: 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.

Question-3: : 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.

Question-4: 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.

Question-5: if,

Then find.

Sol-

Question-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.

Question-7: 3: if w = x² + y – z + sint and  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

Question-8: 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.

Question-9: If where then find the value of ?

Given  

Where 

By chain rule

                          Now substituting the value of x ,y,z we get

                          -6

                          8

Question-10: If then calculate

                         Given

                 By Chain Rule

              Putting the value of u =

               Again partially differentiating z with respect to y

             By Chain Rule

               by substituting value

Question-11: : If u = u(), then prove that .

Sol. We are given that,

                                     u = u () = u( r , s)

Where r = and s =

Or we can write as-

           r =     and s =

Differentiate them partially with respect to x , y and z. , we get

As we know that-

           Then,

Adding all these results, we get

                                   .

Question-12: 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

Question-13: Find the value of n so that the equation

Satisfies the relation 

Given   

  Partially differentiating V with respect to r keeping as constant

Again partially differentiating given V with respect to keeping r as constant

Now, we are taking the given relation

Substituting values using eq(i) and eq(ii)

On solving we get 

Question-14: If

Then show that 

      Given 

Partially differentiating u with respect to x keeping y and z as constant

Similarly partially differentiating u with respect to y keeping x and z as constant

   …….(ii)

   Similarly partially differentiating u with respect to z keeping x and y as constant

   …….(iii)

LHS: 

Hence proved

Question-15: 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

+…….

             = +……….

Question-16: 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

……

   =

……..

Question-17: 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 (

Question-18: 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