Solution

Assignment No: 1 1. In the form of interval, real line can be denoted by (-∞ , +∞)

2. To each point on a co-ordinate line or real line, there is associated

i. A real number

ii. An integer

iii. A natural number

iv. A rational number.

3. How many real numbers lie in the interval [1, 5]

i. 5

ii. 3

iii. 10

iv. Infinite

4. All the three axes are positive in

i. Third octant

ii. First octant

iii. Second octant

iv. Eigth octant

5. Let w = f(x, y, z) such that 2 2

2 2

( 1) ( 2) ( 4)x y zwa b c− − −

= + +2

2 where a,

b and c are real numbers, not equal to zero. Is the function defined at

origin? If yes, what is its value and if not, give the reason.

Solution:

Origin means (x, y, z) = (0, 0, 0). So putting these values in given equation 2 2

2 2

2 2 2

2 2 2

2 2 2

( 1) ( 2) ( 4)

(0 1) (0 2) (0 4)

1 4 16

x y zwa b c

a b c

a b c

− − −= + +

− − −= + +

= + +

2

2

Thus the given function is defined at origin and its value is

2 2

1 4 1a b c

+ + 2

6 .

6. Find the domain of the following functions.

i. 2 2( , ) 4 2f x y x y= − −

Domain of f consists of region in xy-plane where 2 24 2x y− − ≥ 0

2 2

2 2

4 22 4

x yx y≥ +

+ ≤

Now is an equation of surface of elliptic cylinder in

three dimensional space which has infinite length along z-axis.

2 22x y+ = 4

Thus domain of f is inside and on the surface of elliptic

cylinder.

ii. 2 2

1( , )( )

f x y 2x y=

+

Domain of f is entire xy-plane except the origin because at origin

function is not defined.

iii. 2 3( , ) cos ( )f x y x y xy xy= + − +

Domain of f is entire xy-plane.

7. Which of the following statement is true? Correct the false statement.

i. A straight line is a special kind of curve.

TRUE

Generally, a curve is considered to be any one-dimensional

collection of points. For your convenience, think a curve as a

thread which we use in our daily life. The straight line is a special

kind of curve.

ii. Plane is an example of surface.

TRUE

Surface is a two-dimensional geometric figure (a collection of

points) in three-dimensional space. The simplest example is a

plane—a flat surface. Some other common surfaces are spheres,

cylinders, and cones.

iii. The intersection of two surfaces in three dimensional space gives

a surface.

FALSE

Correct statement is “Intersection of two surfaces in three

dimensional space gives a curve”.

8. Identify the curve or surface in three dimensional space for each of the

following equation.

i. y2 + z2 = 100

This is the equation of surface of circular cylinder which has

infinite length along x-axis.

ii. x = 0

This is the equation of yz-plane.

iii. 2 24 1 ,y z x+ = = 0

1

Since intersection of two surfaces is a curve in three dimensional

space. So a curve in three dimensional space is represented by two

equations representing the intersecting surfaces. 2 24y z+ = is the equation of surface of an elliptic cylinder

which has infinite length along x-axis and 0x = is the equation of

yz-plane. The intersection of these two surfaces is an elliptic curve.

Thus these two equations shows elliptic curve in yz-plane.

iv. 2y x=

This is the equation of surface of half cylinder in three dimensional

space.

v. z = z0 , if ( , , )r zθ represent spherical co-ordinates, and z0 is any

real number.

z = z0 is a plane at point z0 on z-axis. It is parallel to xy-plane.

9. Following are the co-ordinates of point in different co-ordinate systems.

Complete the table.

Rectangular Co-ordinate

( , , )x y z

Spherical Co-ordinate

( , , )ρ θ φ

Cylindrical Co-ordinate

( , , )r zθ

3, ,

3 2π π⎛ ⎞

⎜ ⎟⎝ ⎠

sin

3 sin2

3(1)

3

3cos

3 cos2

3(0)0

3, ,03

r

r

r

r

z

z

zz

ρ φπ

θ θπθ

ρ φπ

π

=

=

=

==

=

=

=

==

⎛ ⎞⇒ ⎜ ⎟⎝ ⎠

sin cos

3 sin cos2 313(1)2

32sin sin

3 sin sin2 3

33(1)2

32

cos

3 cos2

3(0)0

3 3, ,02 2

x

x

x

x

y

y

y

y

z

z

zz

ρ φ θπ π

ρ φ θπ π

ρ φπ

=

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

=

=

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

=

=

=

==

⎛ ⎞⇒ ⎜ ⎟⎜ ⎟

⎝ ⎠

( )

cos(0)cos(0)( 1)0

sin(0)sin(0)(0)0

2

0,0, 2

x rxxxy ryyyz z

z

θπ

θπ

=== −======

= −

⇒ −

( )

2 2

2 2

1

(0) ( 2)

2

tan

0tan2

tan 0tan 00

2, ,0

r z

rz

ρ

ρ

ρθ θθ π

φ

φ

φφφ

π

−

= +

= + −

===

=

=−

=

==

⇒

( )0, , 2π −

10. Show that whether the limit of the following function exist or not at the

origin. 2 2

4 4( , )3

x yf x yx y

=+

.

Solution:

Let us recall Rule for Non-Existence of a Limit If in

( , ) ( , )

lim ( , )x y a b

f x y→

we get two or more different values as we approach (a, b) along different

paths, then ( , ) ( , )

lim ( , )x y a b

f x y→

does not exist.

The paths along which (a, b) is approached may be straight lines or plane

curves .

Two of the more common paths to check are the x and y-axis so let’s try

those.

If we approach (0, 0) along the x-axis, their y = 0 . This means that along

the x-axis, we will plug-in y = 0 into the function and then take the limit as

x approaches zero. 2 2

4 4( , ) (0,0) ( , ) (0,0)

2 2

4 4( , ) (0,0)

( , ) (0,0)

lim ( , ) lim3

(0)lim3(0)

lim 0

0

x y x y

x y

x y

x yf x yx yx

x

→ →

→

→

=+

=+

=

=

So, along the x-axis the function will approach zero as we move towards the origin. Now, let’s try the y-axis. Along this axis we have x = 0 and so the limit becomes,

2 2

4 4( , ) (0,0) ( , ) (0,0)lim ( , ) lim

3x y x y

x yf x yx y→ →

=+

2 2

4 4( , ) (0,0) ( , ) (0,0)

(0)lim lim 0 0(0) 3x y x y

yy→ →

= =+

=

The same limit along two paths

Let’s take another path. Move towards the origin along the path y = x.

To do this we will replace all the y’s with x’s and then let x approach zero.

( )( )

2 2

4 4( , ) (0,0) ( , ) (0,0)

2 2

4 4( , ) (0,0)

4

4( , ) (0,0)

( , ) (0,0)

lim ( , ) lim3

lim3

lim41lim4

14

x y x y

x x

x y

x y

x yf x yx y

x x

x xxx

→ →

→

→

→

=+

=+

=

=

=

A different value from the previous two paths. This means that the limit

DOES NOT exist.

11. In which region the following function is continuous? 2 2( , ) ln ( 1)f r s r s= + −

Solution:

The function is continuous in the region where 2 2 1 0r s+ − > because

the natural logarithm (ln) is defined for non-zero positive real numbers

only.

This implies 2 2 1r s+ >

Since is the equation of a circle, center at origin and unit

radius.

2 2 1r s+ =

Thus f is continuous in the region outside a unit circle center at origin.

12. Let r is a function and s, t, u, v are variables. Partial derivative is a

derivative of a function r with respect to u

i. If r is a function of s and s is a function of u. FALSE

( ) , ( )Say r f s s g u then we havedr dr dsdu ds du

= =

=

ii. If r is a function of s and s is a function of two variables u and v.

TRUE

( ) , ( , )Say r f s s g u v then we haver dr su ds u

= =∂ ∂

=∂ ∂

iii. If r is a function of s and t and each of s and t is a function of u.

FALSE

( , ) , ( ) , ( )Say r f s t s g u t h u then we havedr r ds r dtdu s du t du

= = =∂ ∂

= +∂ ∂

iv. If r is a function of s and t and each of s and t is a function of both

u and v. TRUE

( , ) , ( , ) , ( , )Say r f s t s g u v t h u v then we haver r s r tu s u t u

= = =∂ ∂ ∂ ∂ ∂

= +∂ ∂ ∂ ∂ ∂

13. Let 3 3

2 2( , ) x y xyf x yx y

−=

+ and f (0, 0) = 0

Find ( , )xf x y and . ( , )yf x y

Solution: 3 3

2 2( , ) x y xyf x yx y

−=

+

2 3 2 2 3 3

2 2 2

(3 )( ) ( )(2 )( , )( )x

x y y x y x y xy xf x yx y

− + − −=

+

4 2 3 2 3 5 4 2

2 2 2

3 3 2 2( )

3x y x y x y y x y x yx y

+ − − − +=

+

4 2 3

2 2 2

4( )

5x y x y yx y+ −

=+

3 2 2 2 3 3

2 2 2

5 3 2 3 2 4 3 2

2 2 2

5 3 2 4

2 2 2

( 3 )( ) ( )(2 )( , )( )

3 3 2 2( )

4( )

y

4

x xy x y x y xy yf x yx y

x x y x y xy x y xyx y

x x y xyx y

− + − −=

+

+ − − − +=

+

− −=

+

14. Show whether the function ( , ) sin2xz x y y⎛= −⎜

⎝ ⎠⎞⎟ satisfy Laplace’s

Equation or not?

Solution:

As we know for a function z(x, y) ,Laplace Equation is 2 2

2 2 0z zx y∂ ∂

+ =∂ ∂

So for ( , ) sin2xz x y y⎛ ⎞= −⎜ ⎟

⎝ ⎠

2

2

2

2

2 2

2 2

1cos2 2

1sin2 4

cos2

sin2

1 sin sin4 2

5 sin 04 2

z xyxz xy

xz xyyz xy

yz z x xy y

x yxy

∂ −⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠∂ ⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠∂ ⎛ ⎞= −⎜ ⎟∂ ⎝ ⎠∂ ⎛ ⎞= − −⎜ ⎟∂ ⎝ ⎠∂ ∂ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = − − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞= − − ≠⎜ ⎟⎝ ⎠

2

Since for given function z, 2 2

2 2 0z zx y∂ ∂

+ ≠∂ ∂

so the given function does not

satisfy Laplace Equation.

15. For a function 2 4( , ) 2xf x y y x e= + ,

5

3 2 0fy x∂

=∂ ∂

Show the calculation steps.

Solution:

2 4( , ) 2xf x y y x e= +

5

3 2

fy x∂

∂ ∂ means differentiating f with respect to x, two times and then with

respect to y three times. If we differentiate this function in this order it will

be difficult because due to the presence of 4 xx e , number of terms increase

after each differentiation.

But if we first differentiate this function three times with respect to y and

then two times with respect to x, this fifth derivative can be calculated in

few steps.

As we know that this function and all of its partial derivatives are defined

and continuous everywhere, so we use Euler’s Theorem for mixed

derivatives by which 5 5

3 2 2 3

f fy x x y∂ ∂

=∂ ∂ ∂ ∂

So 2 4( , ) 2xf x y y x e= +

42 xf yx ey∂

=∂

24

2 2 xf x ey

∂=

∂

3

3 0fy

∂=

∂

4

3 0fx y∂

=∂ ∂

5

2 3 0fx y∂

=∂ ∂

5

3 2 0

Thusf

y x∂

=∂ ∂

16. Let , ltxw where x e yy

= = n t= . Find the derivative of w with respect

to t.

Solution:

Since w is a function of two variables x and y , and each x and y are

function of one variable t. So by chain rule

(1)dw w dx w dydt x dt y dt

∂ ∂= + − − − − − − − − − −∂ ∂

Since , ,txw x e yy

= = = ln t so

1tdx dye anddt dt t

= =

2

2

1

1ln (ln )

t

w w xandx y y y

Put values of x and yw wandx t y t

∂ ∂ −= =

∂ ∂

∂ ∂= =

∂ ∂e−

Put values in equation (1)

( ) 2

2

2

1 1ln (ln )

ln(ln )

( ln 1)(ln )

tt

t t

t

dw eedt t tt

te t et t

e t tt t

− ⎛ ⎞= + ⎜ ⎟⎝ ⎠

−=

−=

Observe that dwdt

is a function of t only.