partial differentiation
DESCRIPTION
Maths 2 Partial DifferentialTRANSCRIPT
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Chain Rule And Euler’s Theoram
G.H.Patel College of Engineering And Technology
Created By Enrolment number130110120022 To 130110120028
Tanuj Parikh Akash Pansuriya
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The Chain Rule
In this section, we will learn
about:
The Chain Rule
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thE CHAIN RULE
• Recall that the Chain Rule for functions of a single variable gives the following rule for differentiating a composite function.
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• Suppose that z = f(x, y) is a differentiable function of x and y, where x = g(t) and y = h(t) are both differentiable functions of t.
– Then, z is a differentiable function of tand
THE CHAIN RULE (CASE 1)
dz f dx f dy
dt x dt y dt
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• If y = f(x) and x = g(t), where f and g are differentiable functions, then y is indirectly a differentiable function of t, and
THE CHAIN RULE
dy dy dx
dt dx dt
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• For functions of more than one variable, the Chain Rule has several versions.
– Each gives a rule for differentiating a composite function.
THE CHAIN RULE
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-Example(1)
• The pressure P (in kilopascals), volume V(in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by the equation
PV = 8.31T
THE CHAIN RULE (CASE 1)
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• Find the rate at which the pressure is changing when:
– The temperature is 300 K and increasing at a rate of 0.1 K/s.
– The volume is 100 L and increasing at a rate of 0.2 L/s.
THE CHAIN RULE (CASE 1)
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• If t represents the time elapsed in seconds, then, at the given instant, we have:
– T = 300
– dT/dt = 0.1
– V = 100
– dV/dt = 0.2
THE CHAIN RULE (CASE 1)
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• Since , the Chain Rule gives:
– The pressure is decreasing at about 0.042 kPa/s.
THE CHAIN RULE (CASE 1)
8.31T
PV
2
2
8.31 8.13
8.31 8.31(300)(0.1) (0.2)
100 100
0.04155
dP P dT P dV dT T dV
dt T dt V dt V dt V dt
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• We now consider the situation where z = f(x, y), but each of x and y is a function of two variables s and t: x = g(s, t), y = h(s, t).
– Then, z is indirectly a function of s and t,and we wish to find ∂z/∂s and ∂z/∂t.
THE CHAIN RULE (CASE 1)
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• Recall that, in computing ∂z/∂t, we hold s fixed and compute the ordinary derivative of z with respect to t.
– So, we can apply Theorem 2 to obtain:
THE CHAIN RULE (CASE 1)
z z x z y
t x t y t
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• 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,
THE CHAIN RULE (CASE 2)
z z x z y z z x z y
s x s y s t x t y t
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-Example(2)
• If z = ex sin y, where x = st2 and y = s2t, find ∂z/∂s and ∂z/∂t.
– Applying Case 2 of the Chain Rule, we get the following results.
THE CHAIN RULE (CASE 2)
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THE CHAIN RULE (CASE 2)
2 2
2
2 2 2
( sin )( ) ( cos )(2 )
sin( ) 2 cos( )
x x
st st
z z x z y
s x s y s
e y t e y st
t e s t ste s t
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THE CHAIN RULE (CASE 2)
2 2
2
2 2 2
( sin )(2 ) ( cos )( )
2 sin( ) cos( )
x x
st st
z z x z y
t x t y t
e y st e y s
ste s t s e s t
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• Case 2 of the Chain Rule contains three types of variables:
– s and t are independent variables.
– x and y are called intermediate variables.
– z is the dependent variable.
THE CHAIN RULE
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• To remember the Chain Rule, it’s helpful to draw a tree diagram, as follows.
THE CHAIN RULE
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• We draw branches from the dependent variable z to the intermediate variables x and y to indicate that z is a function of x and y.
TREE DIAGRAM
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• Then, we draw branches from x and yto the independent variables s and t.
– On each branch, we write the corresponding partial derivative.
TREE DIAGRAM
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• To find ∂z/∂s, we find the product of the partial derivatives along each path from z to s and then add these products:
TREE DIAGRAM
z z x z y
s x s y s
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• Similarly, we find ∂z/∂t by using the paths from z to t.
TREE DIAGRAM
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• Suppose u is a differentiable function of the n variables x1, x2, …, xn and each xj
is a differentiable function of the m variables t1, t2 . . . , tm.
THE CHAIN RULE (GEN. VERSION)
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• Then, u is a function of t1, t2, . . . , tm
and
• for each i = 1, 2, . . . , m.
THE CHAIN RULE (GEN. VERSION)
1 2
1 2
n
i i i n i
xx xu u u u
t x t x t x t
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-Example(3)
• If u = x4y + y2z3, where x = rset, y = rs2e–t, z = r2s sin t
find the value of ∂u/∂s when r = 2, s = 1, t = 0
THE CHAIN RULE (GEN. VERSION)
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• With the help of this tree diagram, we have:
THE CHAIN RULE (GEN. VERSION)
3 4 3 2 2 2(4 )( ) ( 2 )(2 ) (3 )( sin )t t
u
s
u x u y u z
x s y s z s
x y re x yz rse y z r t
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• When r = 2, s = 1, and t = 0, we have:
x = 2, y = 2, z = 0
• Thus,
THE CHAIN RULE (GEN. VERSION)
(64)(2) (16)(4) (0)(0)
192
u
s
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29
Homogeneous Functions
• A function f(x1,x2,…xn) is said to be homogeneous of degree k if
f(tx1,tx2,…txn) = tk f(x1,x2,…xn)
– when a function is homogeneous of degree one, a doubling of all of its arguments doubles the value of the function itself
– when a function is homogeneous of degree zero, a doubling of all of its arguments leaves the value of the function unchanged
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30
Homogeneous Functions
• If a function is homogeneous of degree k, the partial derivatives of the function will be homogeneous of degree k-1
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31
Euler’s Theorem
• If we differentiate the definition for homogeneity with respect to the proportionality factor t, we get
ktk-1f(x1,…,xn) = x1f1(tx1,…,txn) + … + xnfn(x1,…,xn)
• This relationship is called Euler’s theorem
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32
Euler’s Theorem
• Euler’s theorem shows that, for homogeneous functions, there is a definite relationship between the values of the function and the values of its partial derivatives
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Fall 2002
This definition can be further enlarged to include transcendental functions also as follows. A function f(x,y) is said to be homogeneous function of degree n if it can be expressed as
OR
x
yxn
y
xyn
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Fall 2002
Theorem: if f is a homogeneous function of x and y of degree n then
Cor: if f is a homogeneous function of degree n, then
nfy
fy
x
fx
fnny
fy
yx
fxy
x
fx )1(2
2
22
2
2
22
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Fall 2002
Euler’s theorem for three variables: If f is a homogeneous function of three independent variables x, y, z of order n, then
nfzfyfxf zyx
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Modified EULER’s theorem
)('
)(
uf
ufn
y
uy
x
ux
If Z is a Homogeneous function of degree n in
the variables x and y and z=f(u)
If z is a homogeneous function of degree n in
the variables x and y and z=f(u)
],1)(')[(22
2
22
2
22
ugug
yx
uxy
y
uy
x
ux
)('
)()(
uf
ufnug
where
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