Unit - 2
Partial differentiation and its applications
Function of two variables-
R denotes the set of real numbers. Let’s D is a collection of pairs of real numbers (x, y), which means D is a subset of R × R.
Then a real valued function of two variables of f is a rule that assign to each point (x, y) in D a unique real number denoted by f (x, y).
The set D is called the domain of f.
The set [ f (x, y): (x, y) belongs to D], which is the set of values the function f takes, is called range of f.
Here we generally use the letter z to denote the value that a function of two variables takes,
Then we will have,
z = f (x, y)
here we will call that z is the dependent variable and x and y are independent variables.
Example-1: The area of a rectangular figure whose length is l and breadth are b, given by l × b.
Here independent variables are l and b, but dependent variable is area.
Example-2: the volume of a cylinder is given by πr²h, where r is the radius and h are the height of the cylinder.
In which radius and height are independent variables and volume is dependent.
Example-3: the volume of cuboid is given by l × b× h. where l, b and h are the length, breadth and height respectively.
L, b and h are independent variable and volume of cuboid is dependent variable.
Limits-
The function f (x, y) is said to tend to limit ‘l’, as x →a and y→b Iff the limit is dependent on point (x, y) as x →a and y→b
We can write this as,
Example-1: Evaluate the 
Sol. We can simply find the solution as follows,
Example-2: Evaluate 
Sol.
-6.
Example-3: evaluate 
Sol.
Conitinuity –
At point (a,b) , a function f(x,y) is said to be continuous if,
Working rule for continuity-
Step-1: f(a,b) should be well defined.
Step-2: 
should exist.
Step-3: 
Example-1: Test the continuity of the following function-
Sol.
(1) the function is well defined at (0,0)
(2) check for the second step,
That means the limit exists at (0,0)
Now check step-3:
So that the function is continuous at origin.
Example-2: check for the continuity of the following function at origin,
Sol. (1) Here the function is well defined at (0,0)
(2) check for second step-
Limit f is not unique for different values of m
So that the limit does not exists.
Therefore, the function is not continuous at origin.
Steps to check for existence of limit-
Step-1: find the value of f(x,y) along x →a and y→b
Step-2: find the value of f (x, y) along x →b and y→a
Note- if the values in step -1 and step-2 are same then we can say that the limits exist otherwise not.
Step-3: if a →0 and b→0 then find the limit along y =mx, if the value does not contain m, then limit exist, if it contains m then the limit does not exist.
Note-1- put x = 0 and y = 0 in f, then find f1
2 - Put y = 0 and x = 0 In f then find x2
If f1 and f2 are equal then limit exist otherwise not.
3- put y = mx then find f3
If f1 = f2 ≠f3, limit does not exist.
4- put y = mx² and find f4,
If f1 = f2 = f3 ≠ f4, limit does not exist
If f1 = f2 = f3 = f4, limit exist.
Example-1: Evaluate 
Sol . 1
. 
2. 
Here f1 = f2
3. now put y = mx, we get
Here f1 = f2 = f3
Now put y = mx²
4. 
Therefore,
F1 = f2 = f3 =f4
We can say that the limit exists with 0.
Example-2: evaluate the following-
Sol. First, we will calculate f1 –
Here we see that f1 = 0
Now find f2,
Here , f1 = f2
Therefore the limit exists with value 0.
Partial derivatives
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
Sol. 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)
Sol. 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)
= 2xy cos (y²x + 5x – 8)
Example-3: Obtain all the second order partial derivative of the function:
f (x, y) = (x³y² - xy⁵)
Sol.
3x²y² - y⁵, 
2x³y – 5xy⁴,
= 
= 6xy²
= 
2x³ - 20xy³
= 
= 6x²y – 5y⁴
= 
= 6x²y - 5y⁴
Example-4: Find
Sol. First, we will differentiate partially with respect to r,
Now differentiate partially with respect to θ, we get
Example-5: if,
then find.
Sol-
We have
Example-6: if 
, then show that- 
Sol. Here we have,
u = 
…………………. (1)
now partially differentiate eq. (1) w.r.t to x and y, we get
= 
Or
………………. (2)
And now,
= 
…………………. (3)
Adding eq. (1) and (3), we get
= 0
Hence proved.
Key takeaways-
3. While calculating partial derivatives treat all independent variables, other than the variable with respect to which we are differentiating, as constant.
(a) 
= 
(b) 
= 
(c) 
= 
(d) 
= 
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)
ᵡ 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) is
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 
.
Sol:
. = 
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.
Sol. as we know that by definition, 
= 
3x²y⁴3t² + 4x³y³2t = 17t¹⁶
Example-3: if w = x² + y – z + sin t and x + y = t, find
(a) 
y, z
(b) 
t, z
Sol. 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
Thus
t, z = 2x - 1
Example-4: If u = u (y – z, z - x, x – y) then prove that 
= 0
Sol. Let,
Then
By adding all these equations, we get,
hence proved.
Example-5: if φ (cx – az, cy – bz) = 0 then show that ap + bq = c
Where p = 
q = 
Sol. 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,
Sol. 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.
Key takeaways-
A composite function is a composition / combination of the functions. In this value of one function depends on the value of another function. A composite function is created when one function is put in another.
Let 
i.e., 
To differentiate composite function chain rule is used:
Chain rule:
2. If 
be an implicit relation between x and y.
Differentiating with respect to x we get
We get 
Example1: If 
where 
then find the value of 
?
Sol.
Given 
Where 
By chain rule
Now substituting the value of x, y, z we get
-6
8
Example2: If 
then calculate 
Sol
Given 
By Chain Rule
Putting the value of u = 
Again, partially differentiating z with respect to y
By Chain Rule
by substituting value 
Example 3: If 
.
Show that 
Given 
Partially differentiating u with respect to x and using chain rule
………(i)
Partially differentiating z with respect to y and using chain rule
= 
………. (ii)
Partially differentiating z with respect to t and using chain rule
Using (i) and (ii) we get
Hence 
Example4: If 
where the relation is 
.
Find the value of 
Let the given relation is denoted by 
We know that 
Differentiating u with respect to x and using chain rule
Example5: If 
and the relation is 
. Find 
Given relation can be rewrite as
.
We know that
Differentiating u with respect to x and using chain rule
Implicit differentiation-
let f (x, y) = 0
where y = ∅(x)
by the chain rule, with x = x and y = ∅(x), we get
Here we assume that y is a differentiable funtion of x.
Example-1: if ∅ is a differentiable function such that y = ∅(x) satisfies the equation
x³ + y³ +sin xy = 0 then find 
.
Sol. Suppose f (x, y) = x³ + y³ +sin xy
Then,
fᵡ = 3x² + y cos xy
fy = 2y + x cos xy
so,
Example-2:
Let 
Find 
Sol. Take partial derivative on both side w.r. t. x, treat y as constant
Example-3: if x²y³ + cos y cos z = x² cos x sin y, then find 
Sol. Differentiate partially w.r.t. x and treat y as constant,
If u and v are functions of the two independent variables x and y, then the determinant,
Is known as the Jacobian of u and v with respect to x and y, and it can be written as,
Suppose there are three functions u, v and w of three independent variables x, y and z then,
The Jacobian can be defined as,
Important properties of the Jacobians-
Property-1-
If u and v are the functions of x and y, then
Proof- Suppose u = u (x, y) and v = v (x, y), so that u and v are the functions of x and y,
Now,
Interchange the rows and columns of the second determinant, we get
Differentiate u = u (x, y) and v= v (x, y) partially w.r.t. u and v, we get
Putting these values in eq. (1), we get
hence proved.
Property-2:
Suppose u and v are the functions of r and s, where r, s are the functions of x, y, then,
Proof: 
= 
Interchange the rows and columns in second determinant
We get,
= 
= 
= 
= 
Similarly, we can prove for three variables.
Property-3
If u, v, w are the functions of three independent variables x, y, z is not independent, then,
Proof: here u, v, w is independent, then f (u, v, w) = 0 ………………. (1)
Differentiate (1), w.r.t. x, y and z, we get
………… (2)
……………… (3)
………………. (4)
Eliminate 
from 2,3,4, we get
Interchanging rows and columns, we get
= 0
So that,
Example-1: If x = r sin
, y = r sin
, z = r cos
, then show that
sin
also find 
Sol. We know that,
= 
= 
= 
( on solving the determinant)
= 
Now using first property of Jacobians, we get
Example-2: If u = x + y + z, uv = y + z, uvw = z, find 
Sol. Here we have,
x = u – uv = u(1-v)
y = uv – uvw = uv (1- w)
And z = uvw
So that,
=
Apply 
=
Now we get,
= u²v(1-w) + u²vw
= u²v
Example-3: If u = xyz, v = x² + y² + z² and w = x + y + z, then find J = 
Sol. Here u, v and w are explicitly given, so that first we calculate
= yz(2y-2z) – zx (2x – 2z) + xy (2x – 2y)
= 2[yz(y-z)-zx(x-z) +xy(x-y)]
= 2[x²y - x²z - xy² + xz² + y²z - yz²]
= 2[x²(y-z) - x (y² - z²) + yz (y – z)]
= 2(y – z) (z – x) (y – x)
= -2(x – y) (y – z) (z – x)
By the property,
JJ’ = 1
Jacobian of Implicit functions-
If the variables u, v, x, and y are connected by the following equations-
………………… (1)
………………….. (2)
u, v are the implicit functions of x and y.
Now differentiate equation (1) and (2) with respect to x and y-
………… (3)
……………… (4)
………………. (5)
………………. (6)
Now,
= 
Using equation (3), (4), (5), (6), we get-
= 
So that-
We can find this for three variables as well.
Example-1: If u = 2axy, v = 
then prove that-
Sol. Here we have,
u = 2axy, v = 
Then
Here - 
So that
Now,
Hence-
Hence proved.
Example-2: If 
, then prove that-
Sol. Suppose 
And 
Now
And
So that,
Hence proved.
Example-3: If x + y + z = u, y + z = uv, z = uvw, then prove that-
Sol. Suppose
And
Hence,
Hence proved.
Jacobians for functional dependence functions-
Note- two functions u and v said to be functionally dependent if their Jacobian is equals to zero. That means J (u, v) = 0
Suppose u and v are functionally dependent functions, then
f (u, v) = 0
Differentiate this equation with respect to x and y-
= 0
= 0
There will be a non-trivial solution for 
, 
to this system exists.
So that-
On transposing, we get-
Example-1: Show that 
and 
are functionally dependent.
Sol. Here we have-
and
Now we will find out the Jacobian to check the functional dependence.
= 
Here Jacobian is zero, so we can conclude that these functions are functionally dependent.
Example-2: Prove that u, v, w is functionally dependent, where-
Sol. Here we have
Now we will find out the Jacobian of the given functions-
= 
= 
= 0
Therefore, u, v, w is functionally dependent.
Key takeaways-
Is known as the Jacobian of u and v with respect to x and y, and it can be written as,
2. If u and v are the functions of x and y, then
3. Suppose u and v are the functions of r and s, where r, s are the functions of x, y, then,
4. If u, v, w are the functions of three independent variables x, y, z is not independent, then,
Homogeneous function - A function f (x, y) is said to be homogeneous of degree n if,
f (kx, ky) = kⁿf (x, y)
Here, the power of k is called the degree of homogeneity.
Or
A function f (x, y) is said to be a homogenous function in which the power of each term is the same.
Example:
1. The function-
Is a homogeneous function of order 3.
Euler’s theorem
Statement – if u = f (x, y) be a homogeneous function in x and y of degree n, then
x
+ y 
= nu
Proof: Here u is a homogeneous function of degree n,
u = xⁿ f(y/x) ----------------(1)
Partially differentiate equation (1) with respect to x,
= n
f(y/x) + xⁿ f’(y/x). (
)
Now multiplying by x on both sides, we get
x
= n
f(y/x) + xⁿ f’(y/x). (
) ---------- (2)
Again, partially differentiate equation (1) with respect to y,
= xⁿ f’(y/x).
Now multiplying by y on both sides,
y 
= xⁿ f’(y/x).
---------------(3)
By adding equation (2) and (3),
x
y 
= n
f(y/x) + + xⁿ f’(y/x). (
) + xⁿ f’(y/x).
x
y 
= n
f(y/x)
Here u = f (x, y) is homogeneous function, then - u = 
f(y/x)
Put the value of u in equation (4),
x
y 
= nu
Which is the Euler’s theorem.
Let’s understand Eulers’s theorem by some examples:
Example1- If u = x²(y-x) + y²(x-y), then show that 
-2 (x – y) ²
Solution - here, u = x²(y-x) + y²(x-y)
u = x²y - x³ + xy² - y³,
now differentiate u partially with respect to x and y respectively,
= 2xy – 3x² + y² --------- (1)
= x² + 2xy – 3y² ---------- (2)
Now adding equation (1) and (2), we get
= -2x² - 2y² + 4xy
= -2 (x² + y² - 2xy)
= -2 (x – y) ²
Example2- If u = xy + sin(xy), show that 
= 
.
Solution – u = xy + sin(xy)
= y+ ycos(xy)
= x+ xcos(xy)
x (- sin(xy). (y)) + cos(xy)
= 1 – xysin(xy) + cos(xy) -------------- (1)
1 + cos(xy) + y(-sin(xy) x)
= 1 – xysin(xy) + cos(xy) -----------------(2)
From equation (1) and (2),
= 
Example-3: If u (x, y, z) = log (tan x + tan y + tan z), then prove that,
Sol. Here we have,
u (x, y, z) = log (tan x + tan y + tan z) ………………. (1)
diff. eq. (1) w.r.t. x, partially, we get
……………. (2)
diff. eq. (1) w.r.t. y, partially, we get
……………… (3)
diff. eq. (1) w.r.t. z, partially, we get
…………………… (4)
Now multiply eq. 2, 3, 4 by sin 2x, sin 2y, sin 2z respectively and adding, in order to get the final result,
We get,
= 
So that,
hence proved.
Key takeaways-
x
+ y 
= nu
Taylor’s Theorem-
If f (x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-
+ …….+ 
+ ……..
Which is called Taylor’s theorem.
If we put x = a, we get-
+ …….+ 
+ …….. (1)
Maclaurin’s Theorem-
If we put a = 0 and h = x then equation (1) becomes-
+ …….
Which is called Maclaurin’s theorem.
Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)
We get-
+ …….
Example-1: Express the polynomial 
in powers of (x-2).
Sol. Here we have,
f(x) = 
differentiating the function w.r.t. x-
f’(x) = 
f’’(x) = 12x + 14
f’’’(x) = 12
f’’’’(x)=0
now using Taylor’s theorem-
+ ……. (1)
Here we have, a = 2,
Put x = 2 in the derivatives of f(x), we get-
f (2) = 
f’ (2) = 
f’’ (2) = 12(2) +14 = 38
f’’’ (2) = 12 and f’’’’ (2) = 0
now put a = 2 and substitute the above values in equation (1), we get-
Taylor’s theorem for functions of two variables-
Suppose f (x, y) be a function of two independent variables x and y. then,
+ ……………
Maclaurin’s series is the special case of Taylor’s series-
When we put a = 0 and b = 0 (about origin) in Taylor’s series, we get-
+ ……………
Example-2: Expand f (x, y) = 
in powers of x and y about origin.
Sol. Here we have the function-
f (x, y) = 
here, a = 0 and b = 0 then
f (0, 0) = 
now we will find partial derivatives of the function-
Now using Taylor’s theorem-
+………
Suppose h = x and k = y, we get
+…….
= 
+……….
Example-3: Find the Taylor’s expansion of 
about (1, 1) up to second degree term.
Sol. We have,
At (1, 1)
Now by using Taylor’s theorem-
……
Suppose 1 + h = x then h = x – 1
1 + k = y then k = y - 1
……
= 
……..
Key takeaways-
+ …….+ 
+ ……..
2. Maclaurin’s Theorem-
+ …….
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 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
![2 2 [ 2 ] 2
@f-= 0, @f--= 0, and @-f- @-f-- -@-f- < 0
@x @y @x2 @y2 @x @y](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1642784202_148292.png)
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 
4. (a) If 
(b) If 
(c) If 
(d) If 
Example1 Find out the maxima and minima of the function
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 sides, 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 a saddle point at (
Key takeaways-
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
![2 2 [ 2 ] 2
@f-= 0, @f--= 0, and @-f- @-f-- -@-f- < 0
@x @y @x2 @y2 @x @y](https://glossaread-contain.s3.ap-south-1.amazonaws.com/epub/1642784205_7955992.png)
at the point.
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 fails, 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 
This 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 f
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
Example2 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 
References:
