Unit - 2
Partial differentiation and its applications
Q1) Evaluate the
A1)
We can simply find the solution as follows,
Q2) Test the continuity of the following function-
A2)
(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.
Q3) Evaluate
A3)
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.
Q4) Calculate and for the following function
f (x, y) = 3x³-5y²+2xy-8x+4y-20
A4)
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.
Q5) Obtain all the second order partial derivative of the function:
f (x, y) = (x³y² - xy⁵)
A5)
3x²y² - y⁵, 2x³y – 5xy⁴,
= = 6xy²
= 2x³ - 20xy³
= = 6x²y – 5y⁴
= = 6x²y - 5y⁴
Q6) If, then find.
A6)
We have
Q7) let q = 4x + 3y and x = t³ + t² + 1, y = t³ - t² - t Then find .
A7)
. =
Where,
f1 = , f2 =
In this example f1 = 4, f2 = 3
Also,
3t² + 2t,
4(3t² + 2t) + 3(
= 21t² + 2t – 3
Q8) If u = u (y – z, z - x, x – y) then prove that = 0
A8)
Let,
Then
By adding all these equations, we get,
Hence proved.
Q9) If z is the function of x and y , and x = , y = , then prove that,
A9) 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,
= and y = ,
now,
, , ,
From equation(1) , we get
……………….(3)
And from eq. (2) , we get
…………..(4)
Subtracting eq. (4) from (3), we get
= ) – (
Hence proved.
Q10) If where then find the value of ?
A10) Given
Where
By chain rule
Now substituting the value of x, y, z we get
-6
8
Q11) If where the relation is .
Find the value of
A11)
Let the given relation is denoted by
We know that
Differentiating u with respect to x and using chain rule
Q12) If and the relation is . Find
A12)
Given relation can be rewrite as
.
We know that
Differentiating u with respect to x and using chain rule
Q13) if ∅ is a differentiable function such that y = ∅(x) satisfies the equation
x³ + y³ +sin xy = 0 then find .
A13)
Suppose f (x, y) = x³ + y³ +sin xy
Then,
fᵡ = 3x² + y cos xy
fy = 2y + x cos xy
so,
Q14) If x = r sin , y = r sin , z = r cos, then show that
sin also find
A14) We know that,
=
=
= ( on solving the determinant)
=
Now using first property of Jacobians, we get
Q15) If u = xyz, v = x² + y² + z² and w = x + y + z, then find J =
A15)
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
Q16) If , then prove that-
A16)
Suppose
And
Now
And
So that,
Hence proved.
Q17) Show that and are functionally dependent.
A17)
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.
Q18) If u = x²(y-x) + y²(x-y), then show that -2 (x – y) ²
A18)
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
Q19) If u (x, y, z) = log (tan x + tan y + tan z), then prove that,
A19)
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.
Q20) Express the polynomial in powers of (x-2).
A20)
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-
Q21) Find the Taylor’s expansion of about (1, 1) up to second degree term.
A21)
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
……
=
……..
Q22) Find the maximum and minimum point of the function
A22)
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 (
Q23) Divide 24 into three parts such that the continued product of the first, square of second and cube of third may be maximum.
A23)
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