Engineering Calculus Notes 307

Engineering Calculus Notes 307 - 3.5. SURFACES AND THEIR...

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3.5. SURFACES AND THEIR TANGENT PLANES 295 We calculate the partials ∂f ∂x = 2 x ∂f ∂y = 6 y ∂f ∂z = 8 z which gives the gradient −→ f (2 , 2 , 1) = 4 −→ ı 12 −→ 8 −→ k . Thus the tangent plane is the plane through P (2 , 2 , 1) perpendicular to 4 −→ ı 12 −→ 8 −→ k , which has equation 4( x 2) 12( y + 2) 8( z + 1) = 0 or 4 x 12 y 8 z = 8 + 24 + 8 = 40 . We note that this is the same as d (2 , 2 , 1) f ( x, y, z ) = 0 with
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