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Let f(x, y) = √(16 - x² - 4y²).

(a) Find and sketch the domain of this function.
(b) Find and sketch the level curves f(x, y) = 0, f(x, y) = 2, and f(x, y) = 4. You may draw all of t

User Canovice
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1 Answer

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Final answer:

The domain of the function f(x, y) = √(16 - x² - 4y²) is an ellipse in the xy-plane. The level curves f(x, y) = 0 and f(x, y) = 2 are also ellipses, while the level curve f(x, y) = 4 is an empty set.

Step-by-step explanation:

(a) To find the domain of the function f(x, y) = √(16 - x² - 4y²), we need to determine the values of x and y that are valid inputs for the function. Since we are taking the square root of a quantity, the radicand (16 - x² - 4y²) must be non-negative. This means that 16 - x² - 4y² ≥ 0. Simplifying this inequality, we have 16 ≥ x² + 4y². Rearranging the terms, we get x² + 4y² ≤ 16. This represents the domain of the function as a filled-in ellipse in the xy-plane with its center at the origin, semi-major axis of length 4 along the x-axis, and semi-minor axis of length 2 along the y-axis.

(b) To sketch the level curves f(x, y) = 0, f(x, y) = 2, and f(x, y) = 4, we set the function equal to these values and solve for x and y. For f(x, y) = 0, we have √(16 - x² - 4y²) = 0. Squaring both sides, we get 16 - x² - 4y² = 0. This equation represents the ellipse from part (a). Similarly, for f(x, y) = 2, we have 16 - x² - 4y² = 4, which simplifies to x² + 4y² = 12. Finally, for f(x, y) = 4, we have 16 - x² - 4y² = 16, which simplifies to x² + 4y² = 0. However, since we cannot have negative values under the square root, this last equation has no real solutions. Therefore, the level curves for f(x, y) = 0, f(x, y) = 2, and f(x, y) = 4 are the ellipse, the ellipse, and an empty set, respectively.

User Nelson Tatius
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