Unformatted text preview: ( x , y , z ) = (0,0,0) into T and check if the equation yields any feasible solution of t . First, check x and y : ( ) ( ) ( ) ( )( ) 2 2 2 21 5 1 2 2 5 21 10 21 3 7 t t t t t t t t t t==+ == which means t can be 3 or 7. Next, check if any of these two possible solutions also conforms with yz equation Substitute t = 3, L.H.S. : (213 2 )/2(3) = 2 R.H.S. : 9(33 3 /27)/3 2 = 2 = L.H.S. So t = 3 is a solution. Substitute t = 7: L.H.S. : (217 2 )/2(7) = 2 R.H.S. : 9(37 3 /27)/7 2 = 1.78 ≠ L.H.S. which means t = 7 is not a valid solution. That implies the pilot can hit the target at the origin only when he/she pulls the trigger at t = 3....
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This note was uploaded on 04/15/2010 for the course MATH 20C taught by Professor Lit during the Spring '10 term at UCLA.
 Spring '10
 lit
 Math, Calculus

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