Unit 3

Multivariable calculus

First order partial differentiation-

Let f(x , y) be a function of two variables. Then the partial derivative of this function with respect to x can be written as and defined as follows:

=

Now the partial derivative of f with respect to f can be written as and defined as follows:

=

Note: a. While calculating partial derivatives treat all independent variables, other than the variable with respect to which we are differentiating, as constant.

b. We apply all differentiation rules.

Higher order partial differentiation-

Let f(x , y) be a function of two variables. Then its second-order partial derivatives, third order partial derivatives and so on are referred as higher order partial derivatives.

These are second order four partial derivatives:

(a) =

(b) =

(c) =

(d) =

b and c are known as mixed partial derivatives.

Similarly, we can find the other higher order derivatives.

Example-1: -Calculate and for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

Solution: 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.

Example-2: Calculate and for the following function

f( x, y) = sin(y²x + 5x – 8)

Solution: To calculate treat the variable y as a constant, then differentiate

f(x, y) with respect to x by using differentiation rules,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= (y² + 50)cos(y²x + 5x – 8)

Similarly partial derivative of f(x,y) with respect to y is,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= 2xycos(y²x + 5x – 8)

Example-3: Obtain all the second order partial derivative of the function:

f(x, y) = ( x³y² - x y⁵)

Solution: 3x²y² - y⁵, 2x³y – 5xy⁴,

= = 6xy²

= 2x³ - 20xy³

= = 6x²y – 5y⁴

= = 6x²y - 5y⁴

Example-4: Find

Solution: First we will differentiate partially with respect to r,

Now differentiate partially with respect to θ, we get

Example-5: if, then find

Solution:

Example-6: if , then show that-

Solution: 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.

When we measure the rate of change of the dependent variable owing to any change in a variable on which it depends, when none of the variable is assumed to be constant.

Let the function, u = f( x, y), such that x = g(t) , y = h(t)

Then we can write,

=

=

This is the total derivative of u with respect to t.

Change of variable-

If w = f (x, y) has continuous partial variables fx and fy and if x = x (t), y = y (t) are

Differentiable functions of t, then the composite function w = f (x (t), y (t)) is a

Differentiable function of t.

In this case, we get,

fx (x (t), y (t)) x’(t)+ fy (x(t), y (t)) y’(t).

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

Then find .

Solution: . =

Where, f1 = , f2 =

In this example f1 = 4, f2 = 3

Also,

3t² + 2t ,

4(3t² + 2t) + 3(

= 21t² + 2t – 3

Example-2: Find if u = x³y⁴ where x = t³ and y = t².

Solution: As we know that by definition, =

3x²y⁴3t² + 4x³y³2t = 17t¹⁶.

Example-3: if w = x² + y – z + sintan x + y = t, find

(a) y, z

(b) t, z

Solution: 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

Example-4: If u = u (y – z, z - x, x – y) then prove that = 0

Solution: Let, r = y - z, s = z - x, t = x – y, u = u(r, s, t)

Then,

By adding all these equations, we get,

= 0 hence proved.

Example-5: if φ (cx – az, cy – bz) = 0 then show that ap + bq = c

Where p = q =

Solution: We have,

Φ(cx – az , cy – bz) = 0

φ(r , s) = 0

Where,

We know that,

Again, we do

By adding the two results, we get

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

Solution: 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.

Differentiability-

Let f(x) be a single valued function of the variable x,then,

f’(x) =

Provided that the limit exists and is dependent of the path along which

Example 7: suppose that the function,

f(z) = 4x + y + i( -x + 4y)

Discuss df/dz.

Solution: Here,

Example-2: if

, Then df/dz , z = 0.

Solution:

In calculus, the general Leibniz rule named after Gottfried Wilhelm Leibniz, generalizes the product rule (which is also known as "Leibniz's rule"). It states that if f and g are n-times differentiable functions, then the product fg is also n-times differentiable and its nth derivative is given by

(fg)n = (nk) f(n-k).g(k)

Alternatively, by letting F = f ∘ g (equiv., F(x) = f(g(x)) for all x), one can also write the chain rule in Lagrange's notation, as follows:

F’(X)= f’(g(x)).g’(x)

The chain rule may also be rewritten in Leibniz's notation in the following way. If a variable z depends on the variable y, which itself depends on the variable x (i.e., y and z are dependent variables), then z, via the intermediate variable of y, depends on x as well. In which case, the chain rule states that:

=

The Chain Rule Formula is as follows –

= .

Example 1: Differentiate y = cos x2

Solution:

Given,

y = cos x2

Let u = x2, so that y = cos u

Therefore: =2x

= -sin u

And so, the chain rule says:

= .

= -sin u × 2x

= -2x sin x2

Example 2:

Differentiate f(x)=(1+x2)5.

Solution:

Using the Chain rule,

=

Let us take y = u5 and u = 1+x2

Then = (u5) = 5u4

= (1 + x2 )= 2x.

= 5u4⋅2x = 5(1+x2)4⋅2x

= 10x(1+x4)

Determine the image of a region under a given transformation of variables.

Compute the Jacobian of a given transformation.

Evaluate a double integral using a change of variables.

Evaluate a triple integral using a change of variables.

Example 1:

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

4 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

Example 2:

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

=

Example 3:

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

Example 4:

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

I=

Solution: 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=

Jacobians, Errors and Approximations, maxima and minima

Jacobians

If u and v be continuous and differentiable functions of two other independent variables x and y such as

, then we define the determine

as Jacobian of u, v with respect to x, y

Similarly,

JJ’ = 1

Actually, Jacobins are functional determines

Ex.

- Calculate
- If
- If

ST

4. find

5. If and , find

6.

7. If

8. If , ,

JJ1 = 1

If ,

JJ1=1

Jacobian of composite function (chain rule)

Then

Ex.

- If

Where

2. If and

Find

3. If

Find

Jacobian of Implicit function

Let u1, u2 be implicit functions of x1, x2 connected by f1, f2 such there

,

Then

Similarly,

Ex.

If

If

Find

Partial derivative of implicit functions

Consider four variables u, v, x, y related by implicit function.

,

Then

Ex.

If and

Find

If and

Find

Find

If

Find

Example 1:

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) =

Example 2:

Solution: 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

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 variable

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

Example 1: Find out the maxima and minima of the function

Solution: 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

Example 2: 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 (

Example 3: Find the maximum and minimum value of

Let

Solution: Partially differentiating given function with respect to x and y and equate it to zero

..(i)

..(ii)

On solving (i) and (ii) we get

Thus, pair of values are

Now, we calculate

At the point (0,0)

So further investigation is required

On the x axis y = 0, f(x,0) =0

On the line y=x,

At the point

So that the given function has maximum value at

Therefore, maximum value of given function

At the point

So that the given function has minimum value at

Therefore, minimum value of the given function

Let be a function of x, y, z which to be discussed for stationary value.

Let be a relation in x, y, z

for stationary values we have,

i.e. … (1)

Also, from we have

… (2)

Let ‘’ be undetermined multiplier then multiplying equation (2) by and adding in equation (1) we get,

… (3)

… (4)

… (5)

Solving equation (3), (4) (5) & we get values of x, y, z and .

Example 1: Decampere a positive number ‘a’ in to three parts, so their product is maximum

Solution: Let x, y, z be the three parts of ‘a’ then we get.

… (1)

Here we have to maximize the product

i.e.

By Lagrange’s undetermined multiplier, we get,

… (2)

… (3)

… (4)

i.e.

… (2)’

… (3)’

… (4)

And

From (1)

Thus .

Hence their maximum product is .

Example 2: 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

Textbooks:

1. Erwin Kreyszig, Advanced Engineering Mathematics, 10/e, John Wiley & Sons, 2011.

2. B. S. Grewal, Higher Engineering Mathematics, 44/e, Khanna Publishers, 2017.

References:

1. R. K. Jain and S. R. K. Iyengar, Advanced Engineering Mathematics, 3/e, Alpha Science

International Ltd., 2002.

2. George B. Thomas, Maurice D. Weir and Joel Hass, Thomas Calculus, 13/e, Pearson

Publishers, 2013.

3. Glyn James, Advanced Modern Engineering Mathematics, 4/e, Pearson publishers, 201.