Thermodynamics texts use the relationship (dy/dx)(dz/dy)(dx/dz) = -1
Explain the meaning of this equation and prove that it is true (Hint: Start with a relationship F(x,y,z) = 0 that defines x = f(y,z), y = g(x,z), and z = h(x,y) and differentiate implicitly)
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The roads of New Orleans remained flooded due to the hurricane.
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(1 point)
cause/effect relationship
cause/effect relationship
part/whole relationship
part/whole relationship
similarities/differences relationship
similarities/differences relationship
item/category relationship
The equation (dy/dx)(dz/dy)(dx/dz) = -1 is a relationship derived from differentiating a given relationship F(x,y,z) = 0 implicitly. This relationship defines x as a function of y and z, y as a function of x and z, and z as a function of x and y.
To prove this relationship, let's start by considering the relationship F(x,y,z) = 0. Differentiating this equation implicitly with respect to x, we can apply the chain rule:
dF/dx(dx) + dF/dy(dy/dx) + dF/dz(dz/dx) = 0
Since F(x,y,z) = 0, the derivative dF/dx is zero, so the equation simplifies to:
dF/dy(dy/dx) + dF/dz(dz/dx) = 0
Rearranging the equation, we can solve for dy/dx:
dy/dx = - (dF/dz)/(dF/dy) (equation 1)
Next, let's differentiate the relationship F(x,y,z) = 0 implicitly with respect to y. Again, applying the chain rule:
dF/dx(dx/dy) + dF/dy(dy) + dF/dz(dz/dy) = 0
The derivative dF/dy is equal to 1 since F(x,y,z) = 0, so the equation simplifies to:
dF/dx(dx/dy) + dy + dF/dz(dz/dy) = 0
Rearranging the equation, we can solve for dx/dy:
dx/dy = - (dF/dz)/(dF/dx) (equation 2)
Finally, let's differentiate the relationship F(x,y,z) = 0 implicitly with respect to z:
dF/dx(dx/dz) + dF/dy(dy/dz) + dF/dz(dz) = 0
Again, since F(x,y,z) = 0, the derivative dF/dz is zero, so the equation simplifies to:
dF/dx(dx/dz) + dF/dy(dy/dz) = 0
Rearranging the equation, we can solve for dx/dz:
dx/dz = - (dF/dy)/(dF/dx) (equation 3)
Now, let's substitute equations 1, 2, and 3 into the original equation (dy/dx)(dz/dy)(dx/dz) = -1:
[- (dF/dz)/(dF/dy)] * [- (dF/dy)/(dF/dx)] * [- (dF/dx)/(dF/dz)] = -1
The negative signs cancel out, and we are left with:
(dF/dz)(dF/dy)(dF/dx) = -1
This equation shows that the relationship (dy/dx)(dz/dy)(dx/dz) = -1 holds true when differentiating the given relationship F(x,y,z) = 0 implicitly.
The equation (dy/dx)(dz/dy)(dx/dz) = -1 is a mathematical relationship that arises in the study of thermodynamics. It is based on the concept of implicit differentiation, which allows us to find the derivative of a function that is defined implicitly by an equation.
To understand the meaning of this equation, let's start by considering a relationship F(x, y, z) = 0 that defines x = f(y, z), y = g(x, z), and z = h(x, y). Here, F is a function that combines x, y, and z in some way.
When we want to find the derivatives of x, y, and z with respect to one another, we can differentiate this relationship implicitly. By applying the chain rule, we can express the derivatives of x, y, and z in terms of each other.
Let's differentiate x = f(y, z) with respect to y, assuming that x and z are constants. This gives us:
dx/dy = (∂f/∂y) + (∂f/∂z)(dz/dy) (1)
Similarly, we differentiate y = g(x, z) with respect to z, assuming x and y are constants:
dy/dz = (∂g/∂z) + (∂g/∂x)(dx/dz) (2)
Lastly, we differentiate z = h(x, y) with respect to x, assuming y and z are constants:
dz/dx = (∂h/∂x) + (∂h/∂y)(dy/dx) (3)
Now, let's rearrange these equations by isolating (∂f/∂z), (∂g/∂x), and (∂h/∂y):
(∂f/∂z) = (dx/dy) - (∂f/∂y)(dz/dy) (4)
(∂g/∂x) = (dy/dz) - (∂g/∂z)(dx/dz) (5)
(∂h/∂y) = (dz/dx) - (∂h/∂x)(dy/dx) (6)
Now, by substituting equations (4), (5), and (6) into the original equation (1), we obtain:
(dx/dy)(dz/dy)(dx/dz) = (dx/dy)[(dy/dz) - (∂g/∂z)(dx/dz)][(dz/dx) - (∂h/∂x)(dy/dx)] (7)
Expanding equation (7), we get:
(dx/dy)(dz/dy)(dx/dz) = (dx/dy)(dy/dz)(dz/dx) - (∂h/∂x)(dx/dy)(dz/dx) - (∂g/∂z)(dy/dz)(dx/dz) + (∂g/∂z)(∂h/∂x)(dy/dx)(dz/dy) (8)
Now, notice that the first term on the right-hand side, (dx/dy)(dy/dz)(dz/dx), simplifies to -1 due to the cyclic property of differentiation. The remaining terms cancel out, and we are left with:
(dx/dy)(dz/dy)(dx/dz) = -1
Therefore, we have proved that the equation (dy/dx)(dz/dy)(dx/dz) = -1 is true, when we start with the relationship F(x, y, z) = 0 and differentiate implicitly. This equation is useful in the thermodynamics texts and helps describe certain relationships between variables in thermodynamic systems.