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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

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ii. 3

en
iii. 10

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iv. Infinite
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4. VU
All the three axes are positive in
H .co

i. Third octant
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ii. First octant

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si at

iii. Second octant

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iv. Eigth octant

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( x − 1) 2 ( y − 2) 2 ( z − 4) 2
st //vu

5. Let w = f(x, y, z) such that w = + + where a,

a2 b2 c2
Be tp:

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

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
( x − 1) 2 ( y − 2) 2 ( z − 4) 2
w= + +
a2 b2 c2
(0 − 1) 2 (0 − 2) 2 (0 − 4) 2
= + +
a2 b2 c2
1 4 16
= 2 + 2 + 2
a b c

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

1 4 16
2
+ 2 + 2.
a b c
6. Find the domain of the following functions.

i. f ( x , y) = 4 − x2 − 2 y 2

Domain of f consists of region in xy-plane where 4 − x 2 − 2 y 2 ≥ 0

4 ≥ x2 + 2 y 2
x2 + 2 y 2 ≤ 4

Now x 2 + 2 y 2 = 4 is an equation of surface of elliptic cylinder in

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

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

ud
cylinder.

St
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VU
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p
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te .n
si at
eb nn
W ja
st //vu
Be tp:
ht

1
ii. f ( x, y ) =
( x + y 2 )2
2

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

function is not defined.
iii. f ( x, y ) = x 2 + y 3 − xy + cos ( xy )
Domain of f is entire xy-plane.

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

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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

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

ud
cylinders, and cones.

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

p
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a surface.
te .n

FALSE
si at

Correct statement is “Intersection of two surfaces in three

eb nn

dimensional space gives a curve”.

W ja
st //vu

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

following equation.
ht

i. y2 + z2 = 100
This is the equation of surface of circular cylinder which has
infinite length along x-axis.

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ii. x = 0
This is the equation of yz-plane.

ts
en
ud
St
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VU
H .co
p
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te .n
si at
eb nn
W ja
st //vu
Be tp:

iii. y 2 + 4 z 2 = 1 , x = 0
ht

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.
y 2 + 4 z 2 = 1 is the equation of surface of an elliptic cylinder
which has infinite length along x-axis and x = 0 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.

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ts
en
ud
iv. y = x 2

St
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VU
This is the equation of surface of half cylinder in three dimensional
H .co

space.
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te .n
si at
eb nn
W ja
st //vu
Be tp:
ht

v. z = z0 , if (r , θ , z ) 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.

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

Complete the table.

ts
en
ud
St
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Rectangular Co-ordinate Spherical Co-ordinate Cylindrical Co-ordinate
VU
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( x, y , z ) (ρ, θ , φ ) (r , θ , z )
p
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⎛ π π⎞
⎜ 3, , ⎟
te .n

⎝ 3 2⎠ r = ρ sin φ
si at

π
r = 3 sin
eb nn

2
W ja

r = 3(1)
st //vu

r= 3
Be tp:

θ =θ
ht

π
θ=
3
z = ρ cos φ
π
z = 3 cos
2
z = 3(0)
z=0
⎛ π ⎞
⇒ ⎜ 3, , 0 ⎟
⎝ 3 ⎠

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x = ρ sin φ cos θ
π π
x = 3 sin cos
2 3
⎛1⎞
x = 3(1) ⎜ ⎟
⎝2⎠
3
x=
2
y = ρ sin φ sin θ
π π
y = 3 sin sin
2 3
⎛ 3⎞
y = 3(1) ⎜⎜ ⎟⎟
⎝ 2 ⎠

ts
3
y=

en
2

ud
z = ρ cos φ
π

St
z = 3 cos
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2 VU
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z = 3(0)
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z=0
te .n

⎛ 3 3 ⎞
⇒ ⎜⎜
si at

, , 0 ⎟⎟
⎝ 2 2 ⎠
eb nn

x = r cos θ
( 0, π , − 2 )
W ja

ρ = r2 + z2
st //vu

x = (0) cos π
ρ = (0)2 + (− 2)2
x = (0)(−1)
Be tp:

x=0 ρ= 2
ht

y = r sin θ θ =θ
y = (0) sin π θ =π
y = (0)(0) r
tan φ =
y=0 z
0
z=z tan φ =
− 2
z=− 2 tan φ = 0
(
⇒ 0, 0, − 2 ) φ = tan −1 0
φ =0
⇒ ( 2, π , 0 )

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10. Show that whether the limit of the following function exist or not at the
origin.
x2 y 2
f ( x, y ) = .
x4 + 3 y 4
Solution:
Let us recall Rule for Non-Existence of a Limit
If in
lim f ( x, y )
( x , y ) → ( a ,b )

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

ts
paths, then lim f ( x, y )

en
( x , y ) → ( a ,b )

does not exist.

ud
St
The paths along which (a, b) is approached may be straight lines or plane
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curves . VU
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Two of the more common paths to check are the x and y-axis so let’s try
te .n
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those.
eb nn
W ja

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

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

x approaches zero.
ht

x2 y2
lim f ( x, y ) = lim
( x , y ) →(0,0) ( x , y ) → (0,0) x 4 + 3 y 4

x 2 (0) 2
= lim
( x , y ) →(0,0) x 4 + 3(0) 4

= lim 0
( x , y ) → (0,0)

=0
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,

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x2 y2
lim f ( x, y ) = lim
( x , y ) → (0,0) ( x , y ) → (0,0) x 4 + 3 y 4

(0) 2 y 2
= lim = lim 0 =0
( x , y ) → (0,0) (0) 4 + 3 y 4 ( x , y ) → (0,0)

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.
x2 y2
lim f ( x, y ) = lim
( x , y ) → (0,0) ( x , y ) → (0,0) x 4 + 3 y 4

= lim
( x )( x )
2 2

( x , x ) → (0,0) x 4 + 3x 4

ts
en
x4
= lim
4 x4

ud
( x , y ) → (0,0)

St
= lim
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( x , y ) →(0,0) 4
VU
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1
=
4
p
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A different value from the previous two paths. This means that the limit
te .n
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DOES NOT exist.

eb nn
W ja

11. In which region the following function is continuous?

st //vu

f (r , s ) = ln (r 2 + s 2 − 1)
Be tp:

Solution:
ht

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

the natural logarithm (ln) is defined for non-zero positive real numbers
only.
This implies r 2 + s 2 > 1

Since r 2 + s 2 = 1 is the equation of a circle, center at origin and unit

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

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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 have
dr dr ds
=
du 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 have
∂r dr ∂s
=
∂u ds ∂u

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

en
FALSE

ud
Say r = f ( s, t ) , s = g (u ) , t = h(u ) then we have

St
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dr ∂r ds ∂r dt
= VU +
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du ∂s du ∂t du
p
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iv. If r is a function of s and t and each of s and t is a function of both

te .n

u and v. TRUE
si at

Say r = f ( s, t ) , s = g (u , v ) , t = h(u , v ) then we have

eb nn

∂r ∂r ∂s ∂r ∂t
= +
W ja

∂u ∂s ∂u ∂t ∂u
st //vu

x 3 y − xy 3
13. Let f ( x, y ) = and f (0, 0) = 0
Be tp:

x2 + y 2
ht

Find f x ( x, y ) and f y ( x, y ) .

Solution:
x 3 y − xy 3
f ( x, y ) = 2
x + y2

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

3x 4 y + 3x 2 y 3 − x 2 y 3 − y 5 − 2 x 4 y + 2 x 2 y 3
=
( x 2 + y 2 )2

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x4 y + 4 x2 y3 − y5
=
( x2 + y 2 )2

( x 3 − 3xy 2 )( x 2 + y 2 ) − ( x 3 y − xy 3 )(2 y )
f y ( x, y ) =
( x 2 + y 2 )2
x 5 + x 3 y 2 − 3x 3 y 2 − 3 xy 4 − 2 x 3 y 2 + 2 xy 4
=
( x 2 + y 2 )2
x 5 − 4 x 3 y 2 − xy 4
=
( x 2 + y 2 )2

⎛ x⎞
14. Show whether the function z ( x, y ) = sin ⎜ y − ⎟ satisfy Laplace’s
⎝ 2⎠

ts
Equation or not?

en
ud
Solution:
As we know for a function z(x, y) ,Laplace Equation is

St
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∂2 z ∂2 z VU
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+ =0
∂x 2 ∂y 2
p
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⎛ x⎞
te .n

So for z ( x, y ) = sin ⎜ y − ⎟
⎝ 2⎠
si at
eb nn

∂z ⎛ x⎞ ⎛ −1 ⎞
= cos ⎜ y − ⎟ ⎜ ⎟
W ja

∂x ⎝ 2⎠ ⎝ 2 ⎠
st //vu

∂ z
2
⎛ x⎞ ⎛1⎞
= − sin ⎜ y − ⎟ ⎜ ⎟
∂x 2
⎝ 2⎠ ⎝4⎠
Be tp:

∂z ⎛ x⎞
ht

= cos ⎜ y − ⎟
∂y ⎝ 2⎠
∂2 z ⎛ x⎞
= − sin ⎜ y − ⎟
∂y 2
⎝ 2⎠
∂2 z ∂2 z ⎛1⎞ ⎛ x⎞ ⎛ x⎞
+ 2 = − ⎜ ⎟ sin ⎜ y − ⎟ − sin ⎜ y − ⎟
∂x 2
∂y ⎝4⎠ ⎝ 2⎠ ⎝ 2⎠
5 ⎛ x⎞
= − sin ⎜ y − ⎟ ≠ 0
4 ⎝ 2⎠

∂2 z ∂2 z
Since for given function z, + ≠ 0 so the given function does not
∂x 2 ∂y 2
satisfy Laplace Equation.

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

∂5 f
=0
∂y 3 ∂x 2
Show the calculation steps.
Solution:
f ( x, y ) = y 2 x 4 e x + 2

∂5 f
means differentiating f with respect to x, two times and then with
∂y 3 ∂x 2
respect to y three times. If we differentiate this function in this order it will
be difficult because due to the presence of x 4 e x , number of terms increase

ts
after each differentiation.

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

ud
St
then two times with respect to x, this fifth derivative can be calculated in
el m
few steps. VU
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As we know that this function and all of its partial derivatives are defined
p
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and continuous everywhere, so we use Euler’s Theorem for mixed

te .n
si at

derivatives by which
eb nn

∂5 f ∂5 f
=
W ja

∂y 3 ∂x 2 ∂x 2 ∂y 3
st //vu

So f ( x, y ) = y 2 x 4 e x + 2
Be tp:

∂f
ht

= 2 yx 4 e x
∂y

∂2 f
= 2 x4 ex
∂y 2

∂3 f
=0
∂y 3

∂4 f
=0
∂x∂y 3

∂5 f
=0
∂x 2 ∂y 3

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Thus
∂5 f
=0
∂y 3 ∂x 2

x
16. Let w = where x = et , y = ln t . Find the derivative of w with respect
y
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

ts
dw ∂w dx ∂w dy

en
= + − − − − − − − − − −(1)
dt ∂x dt ∂y dt

ud
St
x
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Since w = , x = et , y = ln t so
y VU
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dx dy 1
p
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= et and =
dt dt t
te .n

∂w 1 ∂w − x
si at

= and = 2
∂x ∂y
eb nn

y y
W ja

Put values of x and y

st //vu

∂w 1 ∂w −et
= and =
∂x ln t ∂y (ln t ) 2
Be tp:

Put values in equation (1)

ht

dw 1 t −et ⎛ 1 ⎞
=
dt ln t
( ) (ln t )2 ⎜⎝ t ⎟⎠
e +

tet ln t − et
=
t (ln t ) 2
et (t ln t − 1)
=
t (ln t ) 2
dw
Observe that is a function of t only.
dt

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