Question Bank

UNIT-2

Question Bank

UNIT-2

FUNCTIONS OF SEVERAL VARIABLE

Q 1: Find the limit of the following points

Solution:

We apply the value of each x by finding the respective value through appliying limits

f(x)=

f(x)=

Finding limit at x= 1.9

f(x)=

f(1.9)=

=

= 0.3448

Limit at x=1.99

f(1.99)=

=

= 0.33444

Limit at x=1.999

f(1.999)=

=

= 0.33344

Limit at x=2.001

f(2.001)=

=

= 0.33322

Limit at x=2.01

f(2.01)=

=

= 0.3225

From the above table we have to estimate the limit when x tends to 2

Here answer is 0.333....

Q 2:

Determine whether the following is dis-continuous at x=-1,0,

Solution :

Given 

Verifying for continuity at x=-1

Therefore here = f(-1)

f(x)=f(-1)

Hence the function is continuous at x=-1

Verifying at x=0

Therefore here = f(0)

f(x)=f(0)

Hence the function is continuous at x=0

Q 3:

                        =

                        =(3)(

                         =

Q 4:

Solution:

Partial derivative for the given equation is,

Q 5:

Find the partial derivative of the following

Z= ,x=st , y=

Solution:

.

Q 6:

Solve the following D.E

Solution :

Approximating over all change in y

  = 0.02w          =        =

Substituting into equation we get,

(-0.03s) + (0.01d) =0.11

Therefore y decreases by approximately 11 percent.

Q 7:

Suppose   : is given by

( = B +,

Where B is an mn –matrix and .For example if B= and =

f(x,y,z) = + =

We claim that D. = B for all .to check this note

So indeed, D  = B

Q 8:

Find the total derivative of f(x,y)=2x+3y with respect to x,

Given that y= sin-1(x)

Solution:

=

2x+3y+3x.x +

Q 9:

The radius and height of a cylinder are both 2cm.the radius is decreased at 1 cms and the height is increasing at 2 cms. What is the change in volume at this instant

The volume of a right circular cylinder is,

V=

We are taking total derivative of this whole to get,

2(2)(2)(-1)+ = -8 +8 =0

At this moment, the volume of the  cylinder is not changing.

Q 10:

Determine the jacobian matrix of the function

f: given by f(x,y,z)=(xy+2yz+2xy2z).

SOLUTION:

We first write f = f( where are given by the formulas we know compute the gradients of these functions. We have that,

                             = [y,x+2z,2y]

        = [2y2z,4xyz,2xy2]

The jacobian matrix f is therefore the 2 matrix whose first row is and the second row is  so

Df ( x ,y ,z) =

Q 11:

Consider

Put

.

Thus degree of f(x, y) is

Note that

If be a homogeneous function of degree n then z can be written as

Q 12:

f(B) =      Solve the given function.

     =

       =             [Recursive function for the gamma function]

       =       [Recursive formula for the gamma function]

       =

        =     [By the definition of Beta function]

       =

 Q 13:

B  =

               = 

               =

              =

         =    [because

Q 14:

Evaluate       . DA

Solution :

Here s is the region bounded by the lines on the real lines

X+ y=0; x+ y=-1

x- y=0  ; x-y = 8

Let

u=x+ y , 0 u 1

v=x-y ,  0 v 8

Now we have two choices

*solve for x  and y if possible

*use the inverse transmission

• Solving for x and y

X=1/2(u)+1/2(v)  and y =1/2(u-1/2(v)

Using of inverse method:

  and

Q 15:

Evaluate    .dA

Solution:

Here s is the region bounded by the lines on the real lines

X - 2y=0; x  - 2y= 4

3x-y=1  ;3 x-y = 8

Let

u=x – 2y

v= 3x- y  

Now we have two choices

*solve for x  and y if possible

*use the inverse transmission

Using of inverse method:

  and

where  ,   ,       

   = -1+6 =5

=

Q 16:

Given a plane z= 3x+4y+2 that lies above the rectangle [0,5] [1,4]. Find the surface area

Solution:

The area of the surface with equation z=f(x,y),(x,y) D ,where are continuous,

Is A(S)= dA

We have z=2+3x+4y.

Then,and =4

A(S) =

From the dimensions of the rectangle, we get the limits of x as (0,5) and the limits of y as(1,4)

On substituting the known values in the expression for area ,we get

A(S) =

Evaluate the iterated integral.

A(S) =

         =

          =15

Q 17:

As an example, let us consider the following integral in two dimensions:

I=

Where C is a straight line from the origin to (1,1), as shown the figure ,Let s be the arc length measured from the origin .We then have

x =s=

y=s sin =

The endpoint (1,1) corresponds to s=.Thus , the line integral becomes

I=  

Q 18 :

          =      by substituting x (2n+1)

                       = .

                      =

                      =

Q 19:

Consider the Dirichlet problem for the wave equation which describes a string attached between walls with one end attached permanently and with the other moving with the constant velocity ie. The d’Almbert equation on the triangular region of the Cartesian product of the space and the time.

u(x,t)-

u(x,t)= 0 and u(

We can check by substituting that the solution fulfilling first condition is

U(x,t) = f(t-x)-f(x+t)

Additionally we want

f(t-

Substituting

We get the condition of sef-semetry

f(

It is fulfilled by considering a simple example sin[log

With

Thus in general,

f(

And we get the general solution as,