By Francis B. Hildebrand, Mathematics
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In this expression, MT is the transpose of the matrix M; that is it is the matrix obtained by interchanging rows and columns of the matrix MiTj = M ji . 17) our goal of expressing the spherical coordinate components (u r , u θ , u ϕ ) of the vector u in the Cartesian components (u x , u y , u z ). 17) with the inverse matrix (MT )−1 , which gives: ⎛ ⎞ ⎛ ⎞ ux ur ⎝ u θ ⎠ = MT −1 ⎝ u y ⎠ . 18) uϕ uz However, now we have only shifted the problem because we do not know the inverse (MT )−1 . We could of course painstakingly compute this inverse, but this would be a laborious process that we can avoid.
In this example we have solved the problem in little steps. In general we take larger steps in the problems in this book, and you will have to discover how to divide a large step into smaller steps. The next problem is a “large” problem; solve it by dividing it into smaller problems. First formulate the smaller problems as ingredients for the large problem before you actually start working on the smaller problems. Make it a habit whenever you are solving problems to first formulate a strategy for how you are going to attack a problem before you actually start working on the sub-problems.
In that case the vectors ∇ f and δr are parallel, and cos θ = 1. 11) where δr is a step in the direction of increasing values of f . Summarizing this we obtain the following properties of the gradient vector ∇ f : 1 The gradient of a function f is a vector that is perpendicular to the surface f = const. 2 The gradient points in the direction of increasing values of f . 3 The magnitude of the gradient is the change δ f in the function in the direction of the largest increase divided by the distance |δr| in that direction.