The problem asks us to describe the graphs of the following functions of two variables: 7. $f(x, y) = 6$ 8. $f(x, y) = 6 - x$ 9. $f(x, y) = 6 - x - 2y$ 10. $f(x, y) = 6 - x^2$ 11. $f(x, y) = \sqrt{16 - x^2 - y^2}$ 12. $f(x, y) = \sqrt{16 - 4x^2 - y^2}$ 13. $f(x, y) = 3 - x^2 - y^2$ 14. $f(x, y) = 2 - x - y^2$
2025/4/21
1. Problem Description
The problem asks us to describe the graphs of the following functions of two variables:
7. $f(x, y) = 6$
8. $f(x, y) = 6 - x$
9. $f(x, y) = 6 - x - 2y$
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0. $f(x, y) = 6 - x^2$
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1. $f(x, y) = \sqrt{16 - x^2 - y^2}$
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2. $f(x, y) = \sqrt{16 - 4x^2 - y^2}$
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3. $f(x, y) = 3 - x^2 - y^2$
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4. $f(x, y) = 2 - x - y^2$
2. Solution Steps
7. $f(x, y) = 6$. This is the plane $z = 6$.
8. $f(x, y) = 6 - x$. This is the plane $z = 6 - x$. This plane intersects the $xz$-plane ($y = 0$) at $z = 6 - x$ and is parallel to the $y$-axis.
9. $f(x, y) = 6 - x - 2y$. This is the plane $z = 6 - x - 2y$ or $x + 2y + z = 6$.
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0. $f(x, y) = 6 - x^2$. This is the parabolic cylinder $z = 6 - x^2$. It is parallel to the $y$-axis.
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1. $f(x, y) = \sqrt{16 - x^2 - y^2}$. So $z = \sqrt{16 - x^2 - y^2}$, which means $z^2 = 16 - x^2 - y^2$ or $x^2 + y^2 + z^2 = 16$. Since $z \ge 0$, this is the upper hemisphere of radius 4 centered at the origin.
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2. $f(x, y) = \sqrt{16 - 4x^2 - y^2}$. So $z = \sqrt{16 - 4x^2 - y^2}$, which means $z^2 = 16 - 4x^2 - y^2$ or $4x^2 + y^2 + z^2 = 16$. Dividing by 16, we have $x^2/4 + y^2/16 + z^2/16 = 1$. Since $z \ge 0$, this is the upper half of an ellipsoid with semi-axes 2, 4, and
4.
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3. $f(x, y) = 3 - x^2 - y^2$. So $z = 3 - x^2 - y^2$ or $z - 3 = - (x^2 + y^2)$. This is a paraboloid opening downwards with vertex at $(0, 0, 3)$.
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4. $f(x, y) = 2 - x - y^2$. So $z = 2 - x - y^2$ or $x = 2 - y^2 - z$. This is a parabolic cylinder opening along the $x$-axis.
3. Final Answer
7. Plane $z = 6$
8. Plane $z = 6 - x$
9. Plane $z = 6 - x - 2y$
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0. Parabolic cylinder $z = 6 - x^2$
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1. Upper hemisphere of radius 4: $z = \sqrt{16 - x^2 - y^2}$
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2. Upper half of the ellipsoid: $z = \sqrt{16 - 4x^2 - y^2}$
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3. Paraboloid opening downwards: $z = 3 - x^2 - y^2$
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