Unit-1

Partial differentiation and its Applications

Question Bank

Question-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.

Question-2: 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-3: 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-4: 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-5: if ∅ is a differentiable function such that y = ∅(x) satisfies the equation

x³ + y³ +sin xy = 0 then find .

Sol. Suppose      f(x,y) =  x³ + y³ +sin xy

Then,

fᵡ=  3x² + y cosxy

Fy = 2y + x cosxy

So,

Question-6: 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-7: 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

Thust, z   =  2x - 1

Question-8: 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-9: 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-10: 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-11: Calculate the percentage increase in the pressure p corresponding to a reduction

Of 1/2 % in the volume V, if the p and V are related by = C, where C is a constant.

Sol.

Here we have-

= C

Taking log on both sides,

Hence increase in the pressure p = 0.7 percent

Question-12: 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-13: 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

Question-14: Solve the differential equation-

Given the boundary condition that-

At x = 0,

Sol.

Here we have-

On integrating partially with respect to x, we get-

Here f(y) is an arbitrary constant.

Now form the boundary condition-

When x = 0,

Hence-

On integrating partially w.r.t.x, we get-

Question-15: Solve-

Sol.

Here we have-

Integrate w.r.t. x, we get-

Integrate w.r.t. x, we get-

Integrate w.r.t. y, we get-

Question-16: Solve-

Sol.

We have-

Then the auxiliary equations are-

Consider first two equations only-

On integrating

…….. (2)

Now consider last two equations-

On integrating we get-

…………… (3)

From equation (2) and (3)-

Question-17: Solve-

Sol.

Let u = x + by

So that-

Put these values of p and q in the given equation, we get-

Question-18: Solve-

Sol.

Let
 

Charpit’s subsidiary equations are-

So that- dq = 0 or q = a

On putting q = a in (1) we get-

Such that-

Integrating

Or

Which is the required solution.