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Math252Ch16Gallery

# Math252Ch16Gallery - 2 f(x,y = x y 2(A Hyperbolic...

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f(x,y) = x 2 - y 2 (A Hyperbolic Paraboloid; “Saddle”) Contour Plot

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f(x,y) = 4 3 x + 2 y The x -slopes are –3 everywhere (i.e., at all points on the plane); the y -slopes are 2 everywhere. If we fix any y -value (for example, y = 0, which corresponds to the x -axis), we get a cross-sectional line with a slope of –3 in the x -direction. If we fix any x -value, (for example, x = 0, which corresponds to the y -axis) we get a cross-sectional line with a slope of +2 in the y -direction.
f(x,y) or z = x 4 + y 2 I've plotted the points (2, 3, 0) and (2, 3, f (2,3) = 25), which lies on the surface. If we fix y to be some value k , we get f ( x , k ) = x 4 + k 2 = x 4 + some number. Then, the corresponding cross-section is a steep quartic (fourth-degree) curve. If we fix x to be some value k , we get f ( k , y ) = k 4 + y 2 = some number + y 2 . Then, the corresponding cross-section is a not-as-steep quadratic (second-degree) curve.

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Here's the corresponding contour diagram: It turns out that f x
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