Unformatted text preview: Solution: Let f ( x,y,z ) = ( xj ) 2 +( yk ) 2 +( zl ) 2r 2 . Then we know that ∇ f = 2 h xj,yk,zl i . Letting ( x ,y ,z ) be a general point on the sphere, we now have what we need to compute the normal line: * x y z + = ∇ f ( x ,y ,z ) t + * x y z + Thus, we get the following three equations: x = ( xj ) t + x y = ( yk ) t + y z = ( zl ) t + z Evaluating this line at the point t =1 gives us ( x,y,z ) = ( j,k,l ), the center of the circle. 1...
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 Spring '08
 BOCIU
 Calculus, Derivative, positive real number

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