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Modern Engineering Thermodynamics

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3.3 Fun with Mathematics 59<br />

CRITICAL THINKING<br />

If you have a composite function of the form f(x, y, z) = 0, where x = y + z, y = xz, and z = y/x, then are the following chain<br />

rule differentials correct or not?<br />

1. dx = dy + dz.<br />

2. dy = zdx + xdz.<br />

3. dz = dy/x + dx/y.<br />

these three functions has a different form, even though they all yield x. Since these functions are not the same, if<br />

follows that their partial derivatives are not equal, or<br />

<br />

∂x<br />

≠ ∂x<br />

<br />

≠ ∂x<br />

<br />

∂y<br />

z<br />

∂y<br />

w<br />

∂y<br />

q<br />

That is why we always indicate which variables are held constant in partial differentiation. This also informs the<br />

reader as to which independent variables are being used in a functional relation.<br />

Equation (3.1) tells us that two of the three variables are independent. If we choose the independent variables<br />

to be x and z, then the preceding equation, which relates x and z, is valid only if the coefficients of both dx and<br />

dz are equal to zero (otherwise, they would not be independent). Then we have<br />

or<br />

so<br />

We also must have<br />

or<br />

<br />

1 − ∂x<br />

∂y<br />

<br />

∂x<br />

∂y<br />

<br />

z<br />

z<br />

<br />

<br />

∂x<br />

=<br />

∂y<br />

z<br />

<br />

∂x<br />

∂y<br />

z<br />

<br />

<br />

∂x<br />

∂y<br />

z<br />

<br />

∂y<br />

∂z<br />

<br />

∂y<br />

∂x<br />

<br />

∂y<br />

∂x<br />

<br />

= 0<br />

z<br />

z<br />

= 1<br />

<br />

−1<br />

∂y<br />

(3.2)<br />

∂x<br />

x<br />

<br />

∂y<br />

∂z<br />

x<br />

z<br />

<br />

+ ∂x <br />

= 0<br />

∂z y<br />

<br />

= − ∂x <br />

∂z y<br />

Then, using the results of Eq. (3.2), we can write<br />

<br />

<br />

∂x ∂y<br />

<br />

∂z<br />

= −1 (3.3)<br />

∂y ∂z ∂x y<br />

z<br />

x<br />

EXAMPLE 3.1<br />

Show that, if the pressure p of a substance is a function of its temperature T and its density ρ, we can write<br />

<br />

∂p ∂ρ<br />

= − ∂ρ <br />

∂T ∂p ∂T<br />

Solution<br />

From Eq. (3.3) with x = p, y = T, and z = ρ, we have<br />

<br />

∂p<br />

∂T<br />

ρ<br />

p<br />

<br />

∂T<br />

∂ρ<br />

Τ<br />

<br />

∂ρ<br />

∂p<br />

p<br />

T<br />

p<br />

= −1<br />

(Continued )

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