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Unformatted text preview: AE 352: Aerospace Dynamics II, Fall 2008 Homework 1 Due Friday, September 5 Problem 1. Show that ε kij ε kpq = δ ip δ jq δ iq δ jp through the following steps: a) Using the facts below, show that [ δ im δ jn ( e m × e n )] · [ δ pr δ qs ( e r × e s )] = ε kij ε kpq (1) Facts: i. e i × e j = ε kij e k ii. e i = δ im e m iii. e i × e j = ( δ im e m ) × ( δ jn e n ) = δ im δ jn ( e m × e n ) = ε kij e k iv. δ pr δ qs ( e r × e s ) = ε lpq e l Solution. Using the Facts, we have: δ im δ jn ( e m × e n )] · [ δ pr δ qs ( e r × e s )] = ( ε kij e k ) · ( ε lpq e l ) (Fact 4) = ε kij ε lpq ( e k · e l ) (just rearranging) = ε kij ε lpq δ kl (Fact 2) = ε kij ε kpq b) You can see that Equation (1) can be written as 3 X m = 1 3 X n = 1 3 X r = 1 3 X s = 1 [ δ im δ jn δ pr δ qs ]( e m × e n ) · ( e r × e s ) = ε kij ε kpq (2) Now, consider all the possible choices of combinations of m , n , r , s , and show that the left hand side becomes δ ip δ jq δ iq δ jp . (Hint: For example, consider the case of m = n and show that LHS = 0 for any value of r , s .) Solution. Let’s look at all the possible choices for m , n , r , s : (i) let m = n , r , s anything: e m × e n = 0 (parallel vectors). Therefore, ( e m × e n ) · ( e r × e s ) = 0....
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This note was uploaded on 09/16/2009 for the course AE ae352 taught by Professor Sri during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
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