Question 12.11.1: Consider the following system of two linear equations in fou......
Consider the following system of two linear equations in four variables:
5u + 5\nu = 2x − 3y
2u + 4\nu = 3x − 2y
It has two degrees of freedom. In fact, it defines u and \nu as functions of x and y. Differentiate the system and then find the differentials du and d\nu expressed in terms of dx and dy. Derive the partial derivatives of u and \nu w.r.t. x and y. Check the results by solving the system explicitly for u and \nu.
For both equations, take the differential of each side and use the rules in
Section 12.9. The result is
5du + 5d\nu = 2dx − 3dy
2du + 4d\nu = 3dx − 2dy
Note that in a linear system like this, without any constant terms, the differentials satisfy the same equations as the variables. Solving simultaneously for du and d\nu in terms of dx and dy yields
Now we can read off the partial derivatives: u^{\prime}_{x} = −\frac{7}{10} , u^{\prime}_{y} = −\frac{1}{5}, \nu^{\prime}_{x}= \frac{11}{10} , and \nu^{\prime}_{y}= −\frac{2}{5}. Suppose that instead of finding the differential, we solve the given equation system directly for u and \nu as functions of x and y. The result is u = −\frac{ 7}{10} x − \frac{1}{5}y and \nu = \frac{11}{10} x − \frac{2}{5}y. From these expressions we easily confirm the values found for the partial derivatives.