B.tech semester i-unit-iii_application of partial differentiation
Partial differentiation B tech
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Transcript of Partial differentiation B tech
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Presentation on Partial Differentiation and its Application
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Partial DifferentiationDefinition :-
A partial differential equation is an equation involving a function of two or more variables and some of its partial derivatives. Therefore a partial differential equation contains one dependent variable and more than one independent variable. Here z will be taken as the dependent variable and x and y the independent variable so that .
We will use the following standard notations to denote the partial derivatives.
yxfz , .
,, qyzp
xz
t
yzs
yxzr
xz
2
22
2
2
,,
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Partial Derivatives of First Orderf (x, y) is a function of two variables. The first order partial derivative of f with respect to x at a point (x, y) is
0
( , ) ( , )limh
f f x h y f x yx h
f (x, y) is a function of two variables. The first partial derivative of f with respect to y at a point (x, y) is
0
( , ) ( , )limk
f f x y k f x yy k
∂f/∂x and ∂f/∂y are called First Order Partial
Derivatives of f .
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Example
Partial Derivatives of First Order ExampleCompute the first order partial derivatives
2 3( , ) 3 2 .f x y x y y 22 33 6 yxfxyf yx
Compute the first order partial derivatives Example
432),( yxyxf
3 2 yx ff
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Partial Derivatives of Higher Order
if z(x, y) is a function of two variables. Then ∂z/∂x and ∂z/∂y are also functions of two variables and their partials can be taken. Hence we can differentiate with respect to x and y again and find ,
Definition :-
1. 2zxx
2 fxx
2zx2
2 fx2 fxx
Take the partial withrespect to x, and thenwith respect to x again.
2.2zyx
2 fyx
fxy
Take the partial withrespect to x, and thenwith respect to y.
3.2zxy
2 fxy
fyx
Take the partial withrespect to y, and thenwith respect to x.
4.2zyy
2 fyy
2zy2
2 fy2 fyy
Take the partial with respect to y, and then with respect to y again.
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Example
Partial Derivatives of Higher Order Example
Example
Find the second-order partial derivatives of the function2( , ) 3 lnf x y x y x y
yxf
yxf
yxfyf yxxyyyxx
16166 2
6 lnxf xy y 2 13yf x xy
Find the second-order partial derivatives of the function2a. ( , ) 3 2f x y x y
yff yx 43
0f 0f 4f 0f yxxyyyxx
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Find the higher-order partial derivatives of the function2
( , ) xyf x y eExample
2 22xy xyxf e y e
x
2 2
2xy xyyf e xye
y
24 xyxxf y e 2 22 1xy
yxf ye xy
2xyxy2xyxy xy1ye2xye2yye2f
222
2xyxyxyyy xy21xe2xy2eyex2f
222
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Homogeneous Functions A function f(x,y) is said to be homogeneous function of degree n if it can be expressed as
xyxn
yxynOR
Euler’s Theorem on Homogeneous Functions if f is a homogeneous function of x and y of degree n then
nfyfy
xfx
if f is a homogeneous function of degree n, then
fnny
fyyxfxy
xfx )1(2 2
22
2
2
22
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Euler’s Theorem Example
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The Chain Rule
Suppose z = f(x, y) is a differentiable function of x and y, where x = g(s, t) and y = h(s, t) are differentiable functions of s and t.Then,
z z x z y z z x z ys x s y s t x t y t
If u=f(x,y), where x=g(t) and y=h(t) then we can express u as a function of t alone by substituting the value of x and y in f(x,y).Now to find du/dt without actually substituting the values of x and y in f(x,y), we establish the following Chain rule :-
dtdy
yu
dtdx
xu
dtdu
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The Chain Rule ExampleIf z = ex sin y, where x = st2 and y = s2t, find ∂z/∂s and ∂z/∂t.
Example
Applying the Chain Rule, we get the following results.
2 2
2
2 2 2
( sin )( ) ( cos )(2 )
sin( ) 2 cos( )
x x
st st
z z x z ys x s y s
e y t e y st
t e s t ste s t
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2 2
2
2 2 2
( sin )(2 ) ( cos )( )
2 sin( ) cos( )
x x
st st
z z x z yt x t y t
e y st e y s
ste s t s e s t
And
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JACOBIANSIf x and y are functions of two independent variables u and v ,then the determinant Is called Jacobian of x , y
w.r.t. u , v
and is denoted by
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Properties of Jacobians The Jacobian matrix is the Inverse matrix of i.e.,1.
2. The Multiplication of jacobian matrix and is = 1.
. =1.
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Example of Jacobians
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Taylor’s theorem for a function of two variables
Now let f(x , y ) be the function of two independent variables x and y. If y is kept constant then Taylor’s theorem for a function of a single variable x
Putting x=a and y=b , we have
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Taylor’s theorem for a function of two variables
Putting a+h=x and b+k=y so that h=x-a and k=y-b , we have
Maclaurin’s Theorem of two variablePutting a=0 ,b=0 , we have
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Taylor’s theorem for a function of two variables Example
Example
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Maxima And Minima for a function of two variables
• The Function f(x,y) is maximum at (x,y) if for all small positive or negative values of h and k; we have
• f(x+h , y+k) – f(x,y) < 0
• Similarly f(x,y) is minimum at (x,y) if for all small positive or negative values of h and k, we have
• f(x+h , y+k) – f(x,y) > 0
• Thus ,from the defination of maximum of f(x,y) at (x,y) we note that f(x+h , y+k) – f(x,y) preserves the same sign for a maximum it is negative and for a minimum it is positive
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Working rule for Maxima And Minima for a function of two variables
• Working rule to find maximum and minimum values of a function f(x,y)
• (1) find ∂f/∂x and ∂f/∂y• (2) a necessary condition for maximum or minimum value is
∂f/∂x=0 , ∂f/∂y=0• solve simultaneous equations ∂f/∂x=0 , ∂f/∂y=0• Let (a₁,b₁) , (a₂,b₂) … be the solutions of these equations.• Find ∂²f/∂x²=r , • ∂²f/∂x ∂y=s , • ∂²f/∂y²=t• (3) if rt-s²>0 and r>0 at one or more points then those are
the points of minima.
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Working rule for Maxima And Minima for a function of two variables
• (4) if rt-s²>0 or r<0 at one or more points then those points are the points of maxima.
• (5) if rt-s²<0 ,then there are no maximum or minimum at these points. Such points are called saddle points.
• (6) if rt-s²=0 nothing can be said about the maxima or minima .it requires further investigation.
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Example for Maxima And Minima for a function of two variables
Example
Discuss the maxima and minima of f(x,y)= xy+27(1/x + 1/y)
∂f/∂x=y-(27/x²) , ∂f/∂y=x-(27/y²)For max. or min ,values we have ∂f/∂x=0 , ∂f/∂y=0.y-(27/x²)=0…(1)x-(27/y²)=0…(2)Giving x=y=3
²f/∂x²=r =54/x³ ∂²f/∂x ∂y=s=1 , ∂²f/∂y²=t=27/y³r(3,3)=3s(3,3)=1t(3,3)=3rt-s²=9-1=8>o , since r,t are both >0
We get minimum value at x=y=3 which is 27.
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Lagrange’s MethodA method to find the local minimum and maximum of a function with two variables subject to conditions or constraints on the variables involved.Suppose that, subject to the constraint g(x,y)=0, the function z=f(x,y) has a local maximum or a local minimum at the point
. Form the function
, , , ,F x y f x y g x y Then there is a value of such that is a solution of the system of equations
0 0( , , )x y
0 1
0 2
, 0 3
F f gx x xF f gy y yF g x y
provided all the partial derivatives exists.
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Working Rule for Lagrange’s MethodStep 1: Write the function to be maximized (or minimized) and the
constraint in the form: Find the maximum (or minimum) value of
subject to the constraintStep 2: Construct the function F:
,z f x y
, 0g x y
, , , ,F x y f x y g x y
Step 3: Set up the system of equations
0 1
0 2
, 0 3
FxFyF g x y
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Working Rule for Lagrange’s MethodStep 4: Solve the system of equations for x, y and .Step 5: Test the solution to determine maximum or
minimum point.
0 0( , , )x y
Find D* = Fxx . Fyy - (Fxy)2
If D* 0 Fxx 0 maximum pointFxx 0 minimum point
D* 0 Test is inconclusiveStep 6: Evaluate at each solution
found in Step 5. ,z f x y 0 0( , , )x y
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Example for Lagrange’s Method
Find the minimum off(x,y) = 5x2 + 6y2 - xy
subject to the constraintx+2y = 24
Solution:F(x,y, ) = 5x2 + 6y2 - xy + (x + 2y - 24)Fx = F = 10x - y + ; Fxx = 10
xFy = F = 12y - x + 2 ; Fyy = 12
yF = F = x + 2y - 24 ; Fxy = -1
Example
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The critical point,10x - y + = 012y - x + 2= 0x + 2y - 24= 0
The solution of the system is x = 6, y = 9, = -51
D*=(10)(12)-(-1)2=119>0Fxx = 10>0
We find that f(x,y) has a local minimum at (6,9).
f(x,y) = 5(6)2+6(9)2-6(9)= 720
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Differentiation Under Integral SignDifferentiation under the integral sign is an operation is used to evaluate certain integrals. Under fairly loose conditions on the function being integrated, differentiation under the integral sign allows one to interchange the order of integration and differentiation. In its simplest form, called the Leibniz integral rule, differentiation under the integral sign makes the following equation valid under light assumptions on f :
This simpler statement is known as Leibniz integral rule.Examples
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Example for Differentiation Under Integral Sign
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