non_ind_variable

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MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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N. Non-independent Variables 1. Partial differentiation with non-independent variables. Up to now in calculating partial derivatives of functions like w = f (x, y) or w = f (x, y, z), we have assumed the variables x, y (or x, y,z) were independent. However in real-world applications this is frequently not so. Computing partial derivatives then becomes confusing, but it is better to face these complications now while you are still in a calculus course, than wait to be hit with them at the same time that you are struggling to cope with the thermodynamics or economics or whatever else is involved. For example, in thermodynamics, three variables that are associated with a contained gas are its p = pressure, v = volume, T = temperature, and you can express other thermodynamic variables like the internal energy U and entropy S in terms of p, v, and T. However, p, v, and T are not independent variables. If the gas is a so-called "ideal gas", they are related by the equation (I) pv = nRT (n, R constants). To see what complications this produces, let's consider first a purely mathematical example. dw Example 1. Let w = x2 + y2 + z2, where z = + y2. Calculate - . dx Discussion. aw (a) If we think of x and y as the independent variables, then we can calculate - ax by two different methods: (i) using z = + Y2 to get rid of z, we get w = + + (x2 + y2)2 = + + x4 + 2x2y2 + y4; (ii) or by using the chain rule, remembering z is a function of x and y, w = x2+y2+z2 so the two methods agree. (b) On the other hand, if we think of x and z as the independent variables, using say method (i) above, we get rid of y by using the relation = z - x2, and get w = + y2 + z2 = z2 + (2 - x2) + = Z + z2;
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N. NON-INDEPENDENT VARIABLES 1 These answers are genuinely different - we cannot convert one into the other by using the relation z = x2 + y2. Will the right dwldx please stand up? The answer is, that there is no one right answer, because the problem was not well-stated. When the variables are not independent, an expression like dwldx does not have a definite meaning. To see why this is so, we interpret the above example geometrically. Saying that x, y, z satisfy the relation z = + y2 means that the point (x, y, z) lies on the paraboloid surface formed by rotating z = y2 about the z-axis. The function I z I w = +y2 +z2 measures the square of the distance from the origin. To be defi- nite, let's suppose we are at the starting point P = Po : (1,0,1) indicated, and we want to calculate dwldx at this point. Case (a) If we take x and y to be the independent variables, then to find dwldx, we hold y fixed and let x vary. So P moves in the xz-plane towards A, along the path shown. As P moves along this path, evidently w, the square of its distance from the dw origin, is steadily increasing: - > 0 and in fact the calculations for (a) on - dx i3w the previous page show that - = 6.
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non_ind_variable - MIT OpenCourseWare http:/ocw.mit.edu...

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