Engineering Calculus Notes 315

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 )...
View Full Document

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.

Ask a homework question - tutors are online