Problem - Implicit function theorem

In summary: That condition is that ##A_x \, dy \, dz + A_y \, dx \, dz + A_z \, dx \, dy = 0## for any choice of ##dx, dy, dz##; that is, that ##A_x, A_y, A_z## are linearly dependent. In that case, one can solve for one of the three variables in terms of the other two, and then the implicit function theorem applies to the other two variables.
  • #1
ORF
170
18

Homework Statement



I have 3 functions which define 3 constraints:
A(x,y,z)=0 (nonlinear)
B(y,z)=y-z=0
C(w,x,y,z)=w-(x+y+z)=0

I'm asked to calculate the partial of a function h, with respect to the variable w; with h=h(x,y,z).

Homework Equations


A(x,y,z)=0 (nonlinear)
B(y,z)=y-z=0
C(w,x,y,z)=w-(x+y+z)=0

The Attempt at a Solution



a. With B, make y=z in all of the equations:
A(x,y)=0 (no lineal)
C(w,x,y)=w-(x+2y)=0
h=h(x,y,y)=h(x,y)

b. The subindex point out the variable of the partial; apply the chain rule
h_w = h_x * x_w + h_y * y_w,

c. For the partial derivatives x_w and y_w we can use the implicit function theorem:

x_w = - C_w / C_x = 1,
y_w = - C_w / C_y = 1/2
with C_w = \frac{\partial C}{\partial w}, etc

Doubt: if I make y=z at first the result if different from the case I make y=z at the end. What case is the correct?

Simplied example:
h(x,y)=f(x)+g(y); the constraint is x=y; and w=x+y.

a. Making x=y at the end
h_w = f_x * x_w + g_y * y_w = f(x)_x + g(y)_y
x=y,
h_w = f(x)_x + g(x)_x

b. Making x=y at first. The variable w=2x, h(x,x) = f(x) + g(x)
h_w = f_x * x_w + g_x * x_w
x_w = 1/2
h_w = 0.5 f(x)_x + 0.5 g(x)_x

Which is the correct?

Thank you in advance :)

Greetings!
 
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  • #2
ORF said:

Homework Statement



I have 3 functions which define 3 constraints:
A(x,y,z)=0 (nonlinear)
B(y,z)=y-z=0
C(w,x,y,z)=w-(x+y+z)=0

I'm asked to calculate the partial of a function h, with respect to the variable w; with h=h(x,y,z).

Homework Equations


A(x,y,z)=0 (nonlinear)
B(y,z)=y-z=0
C(w,x,y,z)=w-(x+y+z)=0

The Attempt at a Solution



a. With B, make y=z in all of the equations:
A(x,y)=0 (no lineal)
C(w,x,y)=w-(x+2y)=0
h=h(x,y,y)=h(x,y)

b. The subindex point out the variable of the partial; apply the chain rule
h_w = h_x * x_w + h_y * y_w,

c. For the partial derivatives x_w and y_w we can use the implicit function theorem:

x_w = - C_w / C_x = 1,
y_w = - C_w / C_y = 1/2
with C_w = \frac{\partial C}{\partial w}, etc

Doubt: if I make y=z at first the result if different from the case I make y=z at the end. What case is the correct?

Simplied example:
h(x,y)=f(x)+g(y); the constraint is x=y; and w=x+y.

a. Making x=y at the end
h_w = f_x * x_w + g_y * y_w = f(x)_x + g(y)_y
x=y,
h_w = f(x)_x + g(x)_x

b. Making x=y at first. The variable w=2x, h(x,x) = f(x) + g(x)
h_w = f_x * x_w + g_x * x_w
x_w = 1/2
h_w = 0.5 f(x)_x + 0.5 g(x)_x

Which is the correct?

Thank you in advance :)

Greetings!

You can write three equations:
[tex] A_x dx + A_y dy + A_z dz = 0\\
dy - dz = 0\\
dw - dx - dy - dz = 0
[/tex]
Solving for ##dx, dy, dz## in terms of ##dw## allows you to express ##dh = h_x dx + h_y dy + h_z dz## solely in terms of ##dw##.

If you don't like using ##dv## on a variable ##v##, think instead of a finite, but small, increment ##\Delta v## and first-order Taylor expansions in terms of ##\Delta x, \, \Delta y, \, \Delta z, \, \Delta w##. You want the first-order Taylor expansion of ##h(x,y,z)## in terms of ##\Delta w##.

Solvability of the linear system above requires a condition on ##A(x,y,z)##; that is the same condition needed for applicability of the implicit function theorem to the multivariate linear system ##F_i(x,y,z,w) = 0, i = 1,2,3##.
 
Last edited:

Related to Problem - Implicit function theorem

1. What is the implicit function theorem?

The implicit function theorem is a mathematical theorem that allows for the determination of a function's derivative without having to explicitly solve for the function. It relates to the concept of implicit functions, which are functions that are defined implicitly by an equation rather than explicitly by a formula. The theorem is used in calculus and is a fundamental tool in many areas of mathematics and science.

2. What is the significance of the implicit function theorem?

The implicit function theorem is significant because it allows for the analysis of functions that cannot be solved explicitly, and it provides a way to find derivatives of these functions. It also has many applications in areas such as optimization, differential equations, and physics. The theorem is essential for solving many problems in mathematics and other fields.

3. What are the conditions for the implicit function theorem to hold?

The implicit function theorem holds under certain conditions, including that the function must be continuous and have continuous partial derivatives. The equation defining the function must also have a unique solution for the dependent variable in terms of the independent variables. If these conditions are met, then the theorem can be applied to find the derivative of the implicit function.

4. How is the implicit function theorem applied in real-world problems?

The implicit function theorem is used in many real-world problems, such as in engineering, economics, and biology. For example, in engineering, it can be used to find the rate of change of a variable in a system that is not explicitly defined. In economics, it can be used to analyze the relationship between different variables in an economic model. In biology, it can be used to study the behavior of complex systems, such as biochemical reactions.

5. Are there any limitations or extensions to the implicit function theorem?

While the implicit function theorem is a powerful tool, it does have its limitations. For example, it only applies to functions of several variables, not just one variable. There are also extensions of the theorem, such as the inverse function theorem and the implicit function theorem for systems of equations. These extensions allow for the analysis of more complex functions and systems. Additionally, there are generalizations of the theorem to different contexts, such as the implicit function theorem for Banach spaces.

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