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Engineering Calculus Notes 303

Engineering Calculus Notes 303 - y 2 − φ x,y 2 = 1 and...

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3.5. SURFACES AND THEIR TANGENT PLANES 291 at the point (1 , 1 , 2), these values are ∂f ∂x (1 , 1 , 2) = 8 ∂f ∂y (1 , 1 , 2) = 2 ∂f ∂z (1 , 1 , 2) = 4 so we see from the Implicit Function Theorem that we can solve for any one of the variables in terms of the other two. For example, near this point we can write z = φ ( x,y ) where 4 x 2
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Unformatted text preview: + y 2 − φ ( x,y ) 2 = 1 and φ (1 , − 1) = 2; the theorem tells us that φ ( x,y ) is di±erentiable at x = 1, y = − 1, with ∂φ ∂x (1 , − 1) = − ∂f/∂x ∂f/∂z = − 8 − 4 = 2 and ∂φ ∂y (1 , − 1) = − ∂f/∂y ∂f/∂z = − − 2 − 4 = 1 2 ....
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