Unit - 2

Extrema of functions of two variables

Q1) Define relative maximum and relative minimum.

A1)

Relative maximum

Relative maximum: f (x, y) is said to have a relative maximum at a point (a, b) if

f (a, b) > f (a + h, b + k)

For small positive or negative values of h and k i.e., f (a, b) the value of the function f at (a, b) is greater than the value of the function f at all points in some small neighbourhood of (a, b).

Relative minimum-

f (x, y) has a relative minimum at (a, b) if

f (a, b) < f (a + h, b + k).

Q2) What are the maxima and minima of a function of two independent variables.

A2)

Let be a defined function of two independent variables.

Then the point is said to be a maximum point of if

Or   =

For all positive and negative values of h and k.

Similarly the point is said to be a minimum point of if

Or   =

For all positive and negative values of h and k.

Q3) What is the saddle point?

A3)

Critical points of a function of two variables are those points at which both partial derivatives of the function are zero. A critical point of a function of a single variable is either a local maximum, a local minimum, or neither. With functions of two variables there is a fourth possibility - a saddle point.

A point is a saddle point of a function of two variables if

At the point.

Q4) Explain the procedure to find maximum and minimum values of f(x, y).

Step by step method:

1. Calculate.
2. Form and solve , we get the value of x and y let it be pairs of values
3. Calculate the following values :

4.     (a) If

(b) If

(c) If

(d) If

Q5) Find out the maxima and minima of the function

A5)

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

Q6) Divide 24 into three parts such that the continued product of the first, square of second and cube of third may be maximum.     

A6)

Let first number be x, second be y and third be z.

According to the question

Let the given function be  f

And the relation

By Lagrange’s Method

   ….(i)

Partially differentiating (i) with respect to x,y and z and  equate  them  to zero

   ….(ii)

   ….(iii)

   ….(iv)

From (ii),(iii)  and (iv) we get

On solving

Putting it in given relation we get

Or 

Or

Thus the first number is 4 second is 8 and third is 12

Q7) If , Find the value of x and y for  which is maximum.

A7)

Given function is

And relation is

By Lagrange’s Method

[]     ..(i)

Partially differentiating (i) with respect to  x, y and z and equate them tozero

Or       …(ii)

Or       …(iii)

Or       …(iv)

On solving (ii),(iii) and (iv) we  get

Using the given relation  we get

So that

Thus the point for the maximum value of the given function is

Q8) Define the vector function and differentiation of vector.

A8)

Vector function-

Suppose be a function of a scalar variable t, then-

Here vector varies corresponding to the variation of a scalar variable t that its length and direction be known as value of t is given.

Any vector  can be expressed as-

Here , , are the scalar functions of t.

Differentiation of a vector-

We can denote it as-

Similarly is the second order derivative of

Note- gives the velocity and gives acceleration.

Q9) A particle moves along the curve , here ‘t’ is the time. Find its velocity and acceleration at t = 2.

A9)

Here we have-

Then, velocity

Velocity at t = 2,

=

Acceleration =

Acceleration at t = 2,

Q10) A particle is moving along the curve x = 4 cos t, y = 4 sin t, z = 6t. Then find the velocity and acceleration at time t = 0 and t = π/2.

And find the magnitudes of the velocity and acceleration at time t.

A10)

Suppose

Now,

At t = 0	4
At t = π/2	-4
At t = 0	|v|=
At t = π/2	|v|=

Again acceleration-

Now-

At t = 0
At t = π/2
At t = 0	|a|=
At t = π/2	|a|=

Q11) Show that  where

A11)

Here it is given-

=

Therefore-

      [Note-

Hence proved

Q12) Prove that

A12)

Interchanging , we get-

We get by using above equations-

Q13) Show that-

1.

2.

A13)

We know that-

2. We know that-

= 0

Q14) If then find the divergence and curl of .

A14)

We know that-

Now-

Q15) Evaluate 

A15)

Given

Here limits of inner integral are functions of y therefore integrate w.r.t y,

Q16) Evaluate

A16)

Let, 

Here limits for both x and y are constants, the integral can be evaluated first w.r.t any of the variables x or y.

Q17) Evaluate the following by changing to polar coordinates,

A17)

In this problem, the limits for y are 0 to and the limits for are 0 to 2.

Suppose,

y =

Squaring both sides,

y² = 2x - x²

x² + y² = 2x

But in polar coordinates,

We have,

r² = 2r cosθ

r = 2 cosθ

From the region of integration, r lies from 0 to 2 cosθ  and θ varies from 0  to π / 2.

As we know in case of polar coordinates,

Replace x by r cosθ and y by r sinθ, dy dx by r drdθ,

We get,

Q18) Evaluate the following integral by converting into polar coordinates.

A18)

Here limits of y,

y =

y² = 2x - x²

x² + y² = 2x

x² + y² - 2x = 0  ………………(1)

Eq. (1) represent a circle whose radius is 1 and centre is ( 1, 0)

Lower limit of y is zero.

Region of integration in upper half circle,

First we will covert into polar coordinates,

By putting

x by r cos θ and y by r sinθ ,  dy dx by r drdθ,

Limits of r are 0 to 2 cosθ  and limits of  θ  are from 0  to π / 2.

Q19) Evaluate

A19)

Let the integral,

I =

=

Put x = sinθ

= π / 24   ans.