Question Bank (unit-5)

Question-1: 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 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-8: If u = x²(y-x) + y²(x-y), then show that    -2 (x – y)²

Solution - here,  u = x²(y-x) + y²(x-y)

                                u = x²y - x³ + xy² - y³,

         now differentiate u partially with respect to x and y respectively,

                                         =  2xy – 3x² + y²        --------- (1)

                                       = x² + 2xy – 3y²          ---------- (2)

Now adding equation (1) and (2), we get

                                 = -2x² - 2y² + 4xy

                                               = -2 (x² + y² - 2xy)

                                               = -2 (x – y)²   

Question-9: If u(x,y,z) = log( tan x + tan y + tan z) , then prove that ,

Sol. Here we have,

                    u(x,y,z) = log( tan x + tan y + tan z)  ………………..(1)

Diff. Eq.(1) w.r.t. x , partially , we get

                     ……………..(2)

Diff. Eq.(1) w.r.t. y , partially , we get

                    ………………(3)

Diff. Eq.(1) w.r.t. z , partially , we get

     ……………………(4)

Now multiply eq. 2 , 3 , 4 by sin 2x , sin 2y , sin 2z respectively and adding , in order to get the final result,

We get,

                                                    =

So that, 

                                                       hence proved.

Question-10: 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

Question-11: 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-12: :   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

Question-13: 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.

Question-14: 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

Question-15: 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 =

Question-16: Prove that the following functions are functionally dependent.

                   f(x , y) = log x – log y  and  g(x , y) =

Sol.   We have,

                           f(x , y) = log x – log y  …………………(1)

                          g(x , y) =    …………………………(2)

Here we will find x from the second equation and substitute in (1) , which will show the relation between the two.

Write f(x,y) = f and g(x,y) = g for convenience,

From equation(1) , we get

2g.xy = 

Lets consider,

x = y [ g +

Put this value in eq.(1), we get

f = log x – log y

= log y [ g + – log y

= log y + log y ( g + – log y

f =  log (g +

The functional relationship F(u,v) = 0 is,

 F(u,v) = u -  log (v +

Now we can verify that,

F(f(x,y) , g(x,y)) = 0

The jacobian will also be zero.

Question-17: The temperature T at any point   in space is  .Find  the  highest temperature on the surface of the unit sphere.

Sol. 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

Question-18: 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 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