Engineering Calculus Notes 309

# Engineering Calculus Notes 309 - with this as a level...

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3.5. SURFACES AND THEIR TANGENT PLANES 297 while the equation of the tangent plane can be formulated in terms of the normal vector −→ n = −→ v x × −→ v y = ( −→ ı + 1 2 −→ k ) × ( −→ 3 2 −→ k ) = p 1 2 P −→ ı p 3 2 P −→ + −→ k as 1 2 ( x 2) + 3 2 ( y + 2) + ( z + 1) = 0 or 1 2 x + 3 2 y + z = 1 3 1 = 5 which we recognize as our earlier equation, divided by 8. As a second example, we consider the surface x 3 y 2 z + x 2 y 3 z + xyz 3 = 30 near the point P ( 2 , 3 , 1). This time, it is not feasible to solve for any one of the variables in terms of the others; our only choice is to work directly
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Unformatted text preview: with this as a level surface of the function f ( x,y,z ) = x 3 y 2 z + x 2 y 3 z + xyz 3 . The partials of this function are ∂f ∂x = 3 x 2 y 2 z + 2 zy 3 z + yz 3 ∂f ∂y = 2 x 3 yz + 3 x 2 y 2 z + xz 3 ∂f ∂z = x 3 y 2 + x 2 y 3 + 3 xyz 2 . The values of these at our point are ∂f ∂x ( − 2 , 3 , 1) = 3 ∂f ∂y ( − 2 , 3 , 1) = 58 ∂f ∂z ( − 2 , 3 , 1) = 18...
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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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