Unit III

Multivariable calculus

Q1) Calculate and for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

A1) 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.

Q 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)

Q 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

Q 4: Find

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

Now differentiate partially with respect to θ, we get

Q 5: if, then find

Solution:

Q 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.

Q 7: 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

Q 8: Find if u = x³y⁴ where x = t³ and y = t².

Solution: As we know that by definition, =

3x²y⁴3t² + 4x³y³2t = 17t¹⁶.

Q 9: 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

Q 10: 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.

Q 11: 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

Q 12: 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.

Q 13: suppose that the function,

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

Discuss df/dz.

Solution: Here,

Q 14: if

, Then df/dz , z = 0.

Solution:

Q 15: 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

Q 15:

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)

Q 16:

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

3 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 17:

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 18: 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 19: 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=

Q 20: 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 21:

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

Q 22: 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

Q 23: 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 (

Q 24: 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

Q 25: 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 .

Q 26: 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