Unit-5
Applications of Multivariable Differential Calculus
Question-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 propert of Jacobians, we get
Question-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
Question-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
Question-4: If u = 2axy, v = then prove that-
Sol. Here we have,
u = 2axy, v =
Then
Here -
So that
Now,
Hence-
Hence proved.
Question-5: If x + y + z = u, y + z = uv , z = uvw, then prove that-
Sol. Suppose
And
Hence,
Hence proved.
Question-6: 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.
Question-7: Prove that u, v, w are functionally dependent, where-
Sol. Here we have
Now we will find out the Jacobian of the given functions-
=
=
= 0
Therefore, u, v, w are functionally dependent.
Question-8: Find the percentage error in the area of an ellipse where error of ly is made in measuring it’s major and minor axes.
Solution:
Let A be the area of an ellipse and ‘a’ & ‘b’ are its semi major and minor axes
Taking log on both sides.
Differentiating we get,
Percentage error ion the area of an ellipse = 2%
Question-9: Find out the maxima and minima of the function
Sol.
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
Question-10: Determine the maxima and minima of when 1
Sol. Suppose ……….. (1)
And =0 …………… (2)
Differentiate partially equation (1) and (2) w.r.t. x and y respectively, we get-
And
Now using Lagrange’s equations, we get-
Which gives-
or ……… (3)
Which gives-
…… (4)
Which gives-
……. (5)
Multiply these equations by x, y, z respectively and adding, we get-
(
Hence we get-
f + 0 then
Put in (3) , (4) and (5), we get-
x(1 – fa) = 0 , y(1 – fb) = 0 and z(1 – fc) = 0
We get-
These gives the maximum and minimum values of f.
Question-11: 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
Unit-5
Applications of Multivariable Differential Calculus
Question-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 propert of Jacobians, we get
Question-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
Question-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
Question-4: If u = 2axy, v = then prove that-
Sol. Here we have,
u = 2axy, v =
Then
Here -
So that
Now,
Hence-
Hence proved.
Question-5: If x + y + z = u, y + z = uv , z = uvw, then prove that-
Sol. Suppose
And
Hence,
Hence proved.
Question-6: 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.
Question-7: Prove that u, v, w are functionally dependent, where-
Sol. Here we have
Now we will find out the Jacobian of the given functions-
=
=
= 0
Therefore, u, v, w are functionally dependent.
Question-8: Find the percentage error in the area of an ellipse where error of ly is made in measuring it’s major and minor axes.
Solution:
Let A be the area of an ellipse and ‘a’ & ‘b’ are its semi major and minor axes
Taking log on both sides.
Differentiating we get,
Percentage error ion the area of an ellipse = 2%
Question-9: Find out the maxima and minima of the function
Sol.
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
Question-10: Determine the maxima and minima of when 1
Sol. Suppose ……….. (1)
And =0 …………… (2)
Differentiate partially equation (1) and (2) w.r.t. x and y respectively, we get-
And
Now using Lagrange’s equations, we get-
Which gives-
or ……… (3)
Which gives-
…… (4)
Which gives-
……. (5)
Multiply these equations by x, y, z respectively and adding, we get-
(
Hence we get-
f + 0 then
Put in (3) , (4) and (5), we get-
x(1 – fa) = 0 , y(1 – fb) = 0 and z(1 – fc) = 0
We get-
These gives the maximum and minimum values of f.
Question-11: 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