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hw1_solutions

# hw1_solutions - MS&E 303 Fall 2003 Homework#1 Due Wednesday...

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MS&E 303 – Fall 2003 Homework #1 – Due Wednesday, September 10 Thermodynamics is an abstract science which seeks to explain the behavior of heat and equilibrium around us. As we jump into the abstract, we will also try to relate it to physical problems in everyday life. However, the first step is to understand and work with the mathematics of thermodynamics which is based on total (exact) differentials and the Legendre Transform. 1. For the function z = x 2 y 3 + x 3 y 2 , determine the following differentials: (a) ∂z ∂x y = 2 xy 3 + 3 x 2 y 2 (b) ∂z ∂y x = 3 x 2 y 2 + 2 x 3 y (c) ∂z ∂y z = 0 (d) ∂x ∂y z = - ∂z ∂y x ∂z ∂x y = - 3 x 2 y 2 + 2 x 3 y 2 xy 3 + 3 x 2 y 2 = - 3 xy + 2 x 2 2 y 2 + 3 xy (e) The relative change in the x with respect to motion in y required in order to remain on the surface while traveling along a path where x 2 + z is constant. ∂x ∂y x 2 + z = - " ∂x 2 + z ∂y x ` ∂x 2 + z ∂x y # = - " ∂z ∂y x ` ˆ 2 x + ∂z ∂x y !# = - 3 x 2 y 2 + 2 x 3 y 2 x + 2 xy 3 + 3 x 2 y 2 = - 3 xy 2 + 2 x 2 y 2 + 2 y 3 + 3 xy 2 (f) The rate of change of z with respect to the total distance traveled p x 2 + y 2 under conditions maintaining a constant x + z . This is somewhat more complicated. The first step is converting the English to math – we need to evaluate ˆ ∂z p x 2 + y 2 ! x + z . The square root is particularly ugly, so we eliminate it immediately using the chain rule. There is a bit of subtlety here, which I’ll explain in a moment. ˆ ∂z p x 2 + y 2 ! x + z = ∂z ∂x 2 + y 2 x + z ˆ ∂x 2 + y 2 p x 2 + y 2 ! x + z = ∂z ∂x 2 + y 2 x + z ∂w 2 ∂w x + z = 2 p x 2 + y 2 ∂z ∂x 2 + y 2 x + z where we used the substitution w = p x 2 + y 2 . The second partial derivative can be readily eval- uated even with the ugly constant condition since it involves the same variable in the numerator and denorminator. The other coordinates have to move in strange ways to satisfy the constant condition, but w and w 2 are still related in a trivial way. The next logical step is to invert the remaining differential so that the sum can be separated into two differentials. ∂z ∂x 2 + y 2 x + z = 1 ` ∂x 2 + y 2 ∂z x + z 1

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∂x 2 + y 2 ∂z x + z = ∂x 2 ∂z x + z + ∂y 2 ∂z x + z = 2 x ∂x ∂z x + z + 2 y ∂y ∂z x + z = - 2 x + 2 y ∂y ∂z x + z Just thinking about the differential, the first one had to equal - 1. But you can use the triplet rule to show it as well. The second must be done using the triplet rule - pulling the constant condition up into the numerator.
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