Engineering Calculus Notes 308

# Engineering Calculus Notes 308 - ∂φ ∂x = − − x 4 R...

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296 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION We note in passing that in this case we could also have solved for z in terms of x and y : 4 z 2 = 20 x 2 3 y 2 z 2 = 5 x 2 4 3 y 2 4 z = ± r 5 x 2 4 3 y 2 4 and since at our point z is negative, the nearby solutions are z = r 5 x 2 4 3 y 2 4 . This would have given us an expression for the ellipsoid near (2 , 2 , 1) as the graph z = φ ( x,y ) of the function of x and y φ ( x,y ) = r 5 x 2 4 3 y 2 4 . The partials of this function are
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Unformatted text preview: ∂φ ∂x = − − x/ 4 R 5 − x 2 4 − 3 y 2 4 ∂φ ∂y = − − 3 y/ 4 R 5 − x 2 4 − 3 y 2 4 ; at our point, these have values ∂φ ∂x (2 , − 2) = 1 2 ∂φ ∂y (2 , − 2) = − 3 2 so the parametric form of the tangent plane is x = 2 + s y = − 2 + t z = − 1 + s 2 − 3 t 2...
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