Engineering Calculus Notes 293

Engineering Calculus Notes 293 - 281 3.5. SURFACES AND...

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

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

Unformatted text preview: 281 3.5. SURFACES AND THEIR TANGENT PLANES → 1 and the linearization of f (x, y ) at − = (1, 2 ) is x T(1, 1 ) f (x, y ) = 2 1 3 + (x − 1) − 8 2 y− 1 2 . If we use the parameters s = △x = x − 1 1 t = △y = y − 2 then the tangent plane is parametrized by x= y= z= 1 +s 1 2 1 8 (3.22) +t + s − 3 t; 2 1 the basepoint of this parametrization is P (1, 2 , 1 ). 8 If we should want to express this tangent plane by an equation, we would need to find a normal vector. To this end, note that the parametrization above has the natural direction vectors → → − =− +− v1 → k ı → → − = − − 3− . k v2 → 2 Thus, we can find a normal vector by taking their cross product →− − → N = →1 × − 2 v v →→→ −−− ı k =10 1 3 0 1 −2 → → 3→ − = −− + − + k . ı 2 It follows that the tangent plane has the equation 0 = −(x − 1) + 3 2 y− 1 2 + z− 1 8 ...
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