Unformatted text preview: P = (2 , 1 , 3) lies on the surface z = x 2y 2 . The surface z = x 2y 2 is the level surface x 2y 2z = 0 of the function of three variables f ( x , y , z ) = x 2y 2z . A vector perpendicular to this level surface at P = (2 , 1 , 3) is ~ u = grad f (2 , 1 , 3) = (2 x ~ i2 y ~ j~ k ) ± ± ± ± (2 , 1 , 3) = 4 ~ i2 ~ j~ k Therefore, the line through P ( P has position vector ~ r o = 2 ~ i + ~ j + 3 ~ k ) in the direction ~ u is ~ r ( t ) = ~ r o + t ~ u = (2 + 4 t ) ~ i + (12 t ) ~ j + (3t ) ~ k or, in component form x ( t ) = 2 + 4 t y ( t ) = 12 t z ( t ) = 3t...
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 Spring '08
 Staff
 Multivariable Calculus, Cone, level surface x2

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