Unit-4

Multivariable Calculus (Differentiation)

Question-1: (i) Evaluate

Ans 1)

-6.

(ii) Evaluate

Sol.

Question-2: Calculate    and   for the following function

                           f(x , y) = 3x³-5y²+2xy-8x+4y-20

Ans 2) 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-3: if , then show that-

Ans 3) 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-4: Find the directional derivative of

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

Ans 4) 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,

Question-5: let q = 4x + 3y      and    x = t³ + t² + 1    , y = t³ - t² - t

Then find .

Ans 5)          

. =

                      Where, f1 =         ,f2 = 

In this example f1 = 4    ,      f2 = 3

Also,         

    3t² + 2t    ,               

4(3t² + 2t) + 3(

=  21t² + 2t – 3

Question-6: if w = x² + y – z + sint and  x + y = t, find

                   (a)    y,z

                     (b)   t, z  

Ans 6) 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-7: If z is the function of x and y , and x =   , y = , then prove that,

Ans 7) 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-8: Find the equation of the tangent plane and normal line to the surface

Ans 8)

Here,

At point (1, 2, -1)-

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

Equation of normal line is-

Question-9: Find out the maxima and minima of the function

  Ans 9)

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

Question-10: If then prove that grad u, grad v and grad w are coplanar.

Ans 10)

Here-                     

Now-

Apply

Which becomes zero.

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

Question-11: If then find the divergence and curl of .

Ans 11)  We know that-

Now-

Question-12: Prove that

Note- here is a constant vector and

Ans 12) here    and

So that

Now-

So that-