Unit - 2

Extrema of functions of two variables

Let z = f (x, y) be a function of two independent variables x and y.

Definition of relative maximum and relative minimum-

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).

Note-

- f has a maximum at (a, b) if has the same negative sign for all small values of h, k; i.e., < 0 a minimum at (a, b) if has the same positive sign, which means, > 0. Where

= f (a + h, b + k) − f (a, b)

2. Extremum is a point which is either a maximum or minimum. The value of the function f at an extremum (maximum or minimum) point is known as the extremum (maximum or minimum) value of the function f.

3. Saddle point or minimax is a point where function is neither maximum nor minimum. At such point f is maximum in one direction while minimum in another direction.

Maxima and minima of function of two variables-

As we know that the value of a function at maximum point is called maximum value of a function. Similarly the value of a function at minimum point is called minimum value of a function.

The maxima and minima of a function is an extreme biggest and extreme smallest point of a function in a given range (interval) or entire region. Pierre de Fermat was the first mathematician to discover general method for calculating maxima and minima of a function. The maxima and minima are complement of each other.

Maxima and Minima of a function of one variables

If f(x) is a single valued function defined in a region R then

Maxima is a maximum point if and only if

Minima is a minimum point if and only if

Maxima and Minima of a function of two independent variables

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.

Saddle point:

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.

Stationary Value

The value is said to be a stationary value of if

i.e. the function is a stationary at (a, b).

Rule to find the maximum and minimum values of

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

4. (a) If

(b) If

(c) If

(d) If

Example1 Find out the maxima and minima of the function

Sol:

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

Example2 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 (

Key takeaways-

- The maxima and minima of a function is an extreme biggest and extreme smallest point of a function in a given range (interval) or entire region.
- If f(x) is a single valued function defined in a region R then

Maxima is a maximum point if and only if

Minima is a minimum point if and only if

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

At the point.

Constrained optimization problems are problems for which a function f(x) is to be minimized or maximized subject to constraints.

Generally to calculate the stationary value of a function with some relation by converting the given function into the least possible independent variables and then solve them. When this method fail we use Lagrange’s method.

This method is used to calculate the stationary value of a function of several variables which are all not independent but are connected by some relation.

Let be the function in the variable x, y and z which is connected by the relation

Rule:

a) Form the equation

Where is a parameter.

b) Form the equation using partial differentiation is

(We always try to eliminate)

c) Solve the all above equation with the given relation

These give the value of

These value obtained when substituted in the given function will give the stationary value of the function.

Example1 Divide 24 into three parts such that the continued product of the first, square of second and cube of third may be maximum.

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

According to the question

Let the given function be

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

Example 2 The temperature T at any point in space is . Find the highest temperature on the surface of the unit sphere.

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

Example3 If , Find the value of x and y for which is maximum.

Given function is

And relation is

By Lagrange’s Method

[] ..(i)

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

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

Example4 Find the points on the surface nearest to the origin.

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 differentiating (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

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.

Note-

1. Velocity =

2. Acceleration =

The derivative of the velocity vector V(t) is called acceleration A(t), and it is given as-

Rules for differentiation-

1.

2.

3.

4.

5.

Example-1: A particle moves along the curve , here ‘t’ is the time. Find its velocity and acceleration at t = 2.

Sol. Here we have-

Then, velocity

Velocity at t = 2,

=

Acceleration =

Acceleration at t = 2,

Example-2: If and then find-

1.

2.

Sol. 1. We know that-

2.

Example-3: 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.

Sol. 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|= |

Scalar point function-

If for each point P of a region R, there corresponds a scalar denoted by f(P), in that case f is called scalar point function of the region R.

Note-

Scalar field- this is a region in space such that for every point P in this region, the scalar function ‘f’ associates a scalar f(P).

Vector point function-

If for each point P of a region R, then there corresponds a vector then is called a vector point function for the region R.

Vector field-

Vector filed is a reason in space such that with every point P in the region, the vector function associates a vector (P).

Note-

Del operator-

The del operated is defined as-

Example: Show that where

Sol. Here it is given-

=

Therefore-

[Note-

Hence proved

Divergence (Definition)-

Suppose is a given continuous differentiable vector function then the divergence of this function can be defined as-

Curl (Definition)-

Curl of a vector function can be defined as-

Note- Irrotational vector-

If then the vector is said to be irrotational.

Vector identities:

Identity-1: grad uv = u grad v + v grad u

Proof:

So that

Graduv = u grad v + v grad u

Identity-2:

Proof:

Interchanging , we get-

We get by using above equations-

Identity-3

Proof:

So that-

Identity-4

Proof:

So that,

Identity-5: curl (u

Proof:

So that

Curl (u

Identity-6:

Proof:

So that-

Identity-7:

Proof:

So that-

Example-1: Show that-

1.

2.

Sol. We know that-

2. We know that-

= 0

Example-2: If then find the divergence and curl of .

Sol. We know that-

Now-

Example-3: Prove that

Note- here is a constant vector and

Sol. Here and

So that

Now-

So that-

Key takeaways-

- Gradient of a constant (

If f and g are two scalar point functions, then

5. If f and g are two scalar point functions, then

6. If f and g are two scalar point functions, then-

7.

8.

9. If then the vector is said to be irrotational.

10. grad uv = u grad v + v grad u

11.

12. curl (u

13. Any vector can be expressed as-

Here , , are the scalar functions of t.

1. Velocity =

2. Acceleration =

3. Scalar field- this is a region in space such that for every point P in this region, the scalar function ‘f’ associates a scalar f(P).

4. Vector field-

Vector filed is a reason in space such that with every point P in the region, the vector function associates a vector (P).

Double Integral over Rectangular and general regions

Consider a function f (x, y) defined in the finite region R of the x-y plane. Divide R into n elementary areas A1, A2,…,An. Let (xr, yr) be any point within the rth elementary are Ar

f (x, y) dA = f (xr, yr) A

Evaluation of Double Integral when limits of Integration are given(Cartesian Form).

Ex-1: Evaluate

Sol:

Given

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

Ex-2: Evaluate

Sol:

Given:

Here the limits of inner integration are functions of y therefore first integrate w.r.t y.

Put 1 – x = a (constant for inner integral)

Put y = at dy = a dt

y | 0 | a |

t | 0 | 1 |

Ex-3: Evaluate

Sol:

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.

Ex-4: Evaluate

Sol:

Let

In polar coordinates, we need to evaluate

Over the region bounded by θ1 and θ2.

And the curves r1(θ) and r2(θ)

Example-1: Evaluate the following by changing to polar coordinates,

Sol. 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,

Example-2: Evaluate the following integral by converting into polar coordinates.

Sol.

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 convert 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.

Example-3: Evaluate

Sol. Let the integral,

I =

=

Put x = sinθ

= π / 24 ans.

Key takeaways:

In polar coordinates, we need to evaluate

Over the region bounded by θ1 and θ2.

References:

1. S C Mallik and S Arora: Mathematical Analysis, New Age International.

2. Publications S Ghorpade, B V Limaye, Multivariable calculus, Springer international edition.

3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers.

4. Advanced engineering mathematics, by HK Dass.

5. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.