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Unformatted text preview: Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution to Assignment 0: Math review/practice 1. Know your recitation ... hopefully, you dont need to see a solution set for this one ... 2. Partial derivatives and coordinate conversions. (a) f x = 1 ( x 2 + y 2 + z 2 ) 2 x q x 2 + y 2 + z 2 = x ( x 2 + y 2 + z 2 ) 3 / 2 . By symmetry, it should be obvious that f y = y ( x 2 + y 2 + z 2 ) 3 / 2 , f z = z ( x 2 + y 2 + z 2 ) 3 / 2 . (b) Note that the magnitude of the radial displacement vector  ~ r  = x 2 + y 2 + z 2 . For simplicity, lets denote this r . Then, ~ f = x x + y y + z z ( x 2 + y 2 + z 2 ) 3 / 2 , = ~ r r 3 , = r r 2 . On the last line, weve defined the unit vector r = ~ r/r . It is the unit vector that points in the radial direction (with respect to a specified origin). You should be aware that flipping back and forth between the forms on the 2nd and 3rd lines is something we do a lot, depending on which is more convenient. (c) When we convert to cylindrical coordinates, x 2 + y 2 + z 2 becomes ( r c ) 2 cos 2 + ( r c ) 2 sin 2 + z 2 = r 2 c + z 2 . The function f thus becomes f = 1 / q r 2 c + z 2 . The partial derivatives we want are therefore given by f r c = r c ( r 2 c + z 2 ) 3 / 2 , f = , f z = z ( r 2 c + z 2 ) 3 / 2 . (d) In these coordinates, ~ f = r c r c + z z ( r 2 c + z 2 ) 3 / 2 . This is equivalent to the answer we worked out in part (b) provided that the radial displacement vector is ~ r = r c r c + z z . (e) When we convert to spherical coordinates, we have x 2 + y 2 + z 2 = r 2 sin 2 cos 2 + r 2 sin 2 sin 2 + r 2 cos 2 = r 2 sin 2 + r 2 cos 2 = r 2 . So, f = 1 /r . [ CAUTION : For some reason, many math classes and textbooks define spherical coordinates in such a way that and are reversed relative to what we use....
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This homework help was uploaded on 04/11/2008 for the course PHYSICS 8.022 taught by Professor Hughes during the Spring '08 term at MIT.
 Spring '08
 Hughes
 Mass

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