Question Bank ( unit – 4 GTU)

Question-1: evaluate the

Sol. We can simply find the solution as follows,

Example-2:  evaluate

Sol.

               -6.

Question-2: 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.

Question-3: 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-4: 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-5: 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-6: 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

Question-7: :  if,

Then find.

Sol-

Question-8: verify mixed derivative theorem for

Sol. We will find first derivatives,

Then we find second derivative ,

We see that,

                                         verified.

Question-9: suppose that the function,

                      f(z) = 4x + y + i( -x + 4y)

Discuss df/dz.

Sol. Here,

Question-10: 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)

Question-11: determine the gradient of the function ,

Sol. We know that, the gradient for three vairables,

                           grad f =

So that, we get the gradient as below,

Question-12: 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.

Question-13: 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-14:Find  the points on the surface nearest to the origin.

Sol.

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 differentiating (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