Proving Identity Using Constant Constraint r

In summary: Since we are holding r constant, dr=0 and this simplifies to: (\partial f/\partial x)(\partial x/\partial r) + (\partial f/\partial y)(\partial y/\partial r) + (\partial f/\partial z)(\partial z/\partial r) =0 Now, we can use the fact given in the hint: (\partial x/\partial y)_z *(\partial y/\partial z)_x *(\partial z/\partial x)_y = -1 to rewrite the partial derivatives in terms of each other:(\partial x/\partial r)=-
  • #1
LoopQG
22
0

Homework Statement



Let f(x,y,z)=0 and r=r(x,y,z) be another constraint. show that if r is held constant then

[tex] (\partial x/\partial y)_r *(\partial y/\partial z)_r *(\partial z/\partial x)_r = 1 [/tex]

hint: consider dr and use the fact:

[tex] (\partial x/\partial y)_z *(\partial y/\partial z)_x *(\partial z/\partial x)_y = -1 [/tex]


The Attempt at a Solution



I don't see the connection with r,

[tex] df=(\partial f/\partial x)dx + (\partial f/\partial y)dy + (\partial f/\partial z)dz =0 [/tex]

and similarly

[tex] dr=(\partial r/\partial x)dx + (\partial r/\partial y)dy + (\partial r/\partial z)dz [/tex]

and i know holding r constant means dr=0 but I don't see how to even start this, any help is appreciated, thanks.
 
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  • #2


I believe it is important to fully understand the problem before attempting to solve it. In this case, we are given two constraints: f(x,y,z)=0 and r=r(x,y,z). We are also given the hint to consider dr and use the fact that

(\partial x/\partial y)_z *(\partial y/\partial z)_x *(\partial z/\partial x)_y = -1

This fact suggests that there is a relationship between the partial derivatives of x, y, and z with respect to each other.

To start, we can rewrite the first constraint as f(x,y,z)=0 as f(x,y,z)-0=0. This shows that we can consider the function f as a level surface in a three-dimensional space. Similarly, we can rewrite the second constraint as r(x,y,z)-r=0. This means that we can consider r as another level surface in the same three-dimensional space.

Now, we are asked to show that if r is held constant, then the partial derivatives of x, y, and z with respect to each other, denoted as (\partial x/\partial y)_r, (\partial y/\partial z)_r, and (\partial z/\partial x)_r, will equal 1. This means that the level surface of r is perpendicular to the level surface of f.

To understand this better, let's consider a simpler example. Imagine a level surface of a hill, represented by the function z=f(x,y). Now, let's add another level surface of a road, represented by the function z=r(x,y). If we hold the road constant, meaning we drive along it without changing our elevation, we can see that the hill and the road are perpendicular to each other. This is similar to what we are trying to show in this problem.

To formally prove this, we can use the chain rule for partial derivatives. Starting with the first constraint, we can rewrite it as f(x,y,z(r))=0. This means that f is a function of x, y, and z, which in turn is a function of r. Using the chain rule, we can write:

(\partial f/\partial x)dx + (\partial f/\partial y)dy + (\partial f/\partial z)dz =0

as

(\partial f/\partial x)(\partial x/\partial r)dr + (\partial f/\partial y)(\
 

FAQ: Proving Identity Using Constant Constraint r

What is "Proving Identity Using Constant Constraint r"?

"Proving Identity Using Constant Constraint r" is a method for verifying a person's identity by using a constant constraint, also known as a secret parameter or shared key, which is unique to that individual. This method is commonly used in cryptography and can be implemented in various authentication systems.

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The method works by generating a random value, known as a challenge, and sending it to the individual who wants to prove their identity. The individual then uses the constant constraint r to encrypt the challenge and send it back to the verifier. The verifier can then decrypt the response using the same constant constraint r. If the decrypted value matches the original challenge, the individual's identity is considered verified.

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One of the main benefits of this method is that it provides a high level of security. The constant constraint r is unique to each individual and is not shared with anyone else, making it difficult for unauthorized individuals to impersonate someone else. Additionally, this method is relatively simple and easy to implement in various systems.

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