Engineering Calculus Notes 315

# Engineering Calculus Notes 315 - K/ 2. Now suppose ( s i ,t...

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3.5. SURFACES AND THEIR TANGENT PLANES 303 (Do you see why this is necessary?) Given ( s 0 ,t 0 ) in the domain of −→ p , we can use the continuity of K ( s,t ) to replace it with a positive constant K that works for all ( s,t ) suFciently near ( s 0 ,t 0 ). In particular, we can identify three (overlapping) sets of θ -values, say Θ j ( j = 1 , 2 , 3) such that every θ belongs to at least one of them, and for every θ Θ j the estimate above works at ( s 0 ,t 0 ) using the j th coordinate: | v j ( s 0 ,t 0 ) | > K 2 . But if this estimate works using the j th component for −→ v ( s 0 ,t 0 ), then it also works for all −→ v ( s,t,θ ) with ( s,t ) suFciently near ( s 0 ,t 0 ). Thus by restricting to parameter values close to our “base” ( s 0 ,t 0 ), we can pick an index j ( θ ) independent of ( s,t ) for which the j th component of −→ v ( s,t,θ ) always has absolute value greater that
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Unformatted text preview: K/ 2. Now suppose ( s i ,t i ), i = 1 , 2 are distinct pairs of parameter values near ( s ,t ), and consider the straight line segment joining them in parameter space. This line segment can itself be parametrized as ( s ( ) ,t ( )) = ( s 1 ,t 1 ) + ( s, t ) , 1 where s = s 2 s 1 t = t 2 t 1 . Choose satisfying s = b v b cos t = b v b sin and suppose j as above. Assume without loss of generality that j = 1, so the j th component is x . Then writing x as a function of along our line segment, we have x ( ) = x ( s 1 + s,t 1 + t )...
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## This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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