Engineering Calculus Notes 293

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

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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 ...
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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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