Find and Sketch the Domain of the Function. f(x, y) = ln(16 − x² − 16y²)

blog 2025-01-23 0Browse 0
Find and Sketch the Domain of the Function. f(x, y) = ln(16 − x² − 16y²)

And while we’re at it, let’s ponder why the universe insists on making ellipses so mathematically elegant yet so tricky to sketch.

When dealing with functions of multiple variables, understanding their domain is crucial. The domain of a function represents the set of all possible input values (in this case, pairs ((x, y))) for which the function is defined. For the function (f(x, y) = \ln(16 - x^2 - 16y^2)), the domain is determined by the argument of the natural logarithm, which must be positive. This leads us to the inequality:

[ 16 - x^2 - 16y^2 > 0 ]

Let’s break this down step by step.

Step 1: Understanding the Inequality

The inequality (16 - x^2 - 16y^2 > 0) can be rewritten as:

[ x^2 + 16y^2 < 16 ]

This represents all points ((x, y)) in the plane that lie inside the region bounded by the ellipse defined by (x^2 + 16y^2 = 16).

Step 2: Rewriting the Ellipse Equation

To better understand the shape of this region, let’s rewrite the equation of the ellipse in its standard form. Dividing both sides of (x^2 + 16y^2 = 16) by 16, we get:

[ \frac{x^2}{16} + \frac{y^2}{1} = 1 ]

This is the standard form of an ellipse centered at the origin with semi-major axis (a = 4) (along the (x)-axis) and semi-minor axis (b = 1) (along the (y)-axis).

Step 3: Sketching the Domain

The domain of (f(x, y)) consists of all points ((x, y)) that lie strictly inside this ellipse. To sketch the domain:

  1. Draw the ellipse centered at the origin with:
    • Horizontal axis extending from (x = -4) to (x = 4).
    • Vertical axis extending from (y = -1) to (y = 1).
  2. Shade the interior of the ellipse to represent the domain.

The boundary of the ellipse (where (x^2 + 16y^2 = 16)) is not included in the domain because the natural logarithm is undefined for zero or negative arguments.

Step 4: Analyzing the Domain

The domain of (f(x, y)) is an open region, meaning it does not include its boundary. This is a common feature of logarithmic functions, as they are only defined for positive inputs. The ellipse (x^2 + 16y^2 = 16) acts as a “barrier” beyond which the function (f(x, y)) is undefined.

Step 5: Practical Implications

Understanding the domain of a function is essential for various applications, such as optimization, integration, and solving differential equations. For instance, if you were to integrate (f(x, y)) over its domain, you would need to set up the limits of integration to cover the interior of the ellipse.

Step 6: Generalizing the Approach

The method used here can be applied to other functions involving logarithms or square roots. The key steps are:

  1. Identify the expression inside the logarithm or square root.
  2. Set up an inequality to ensure the expression is positive (or non-negative for square roots).
  3. Solve the inequality to find the domain.
  4. Sketch the region defined by the inequality.

Step 7: Exploring Further

The function (f(x, y) = \ln(16 - x^2 - 16y^2)) is just one example of a multivariable function with a restricted domain. Other functions, such as (g(x, y) = \sqrt{9 - x^2 - y^2}), have domains defined by similar inequalities. Exploring these functions can deepen your understanding of how domains are determined and visualized.

Step 8: Connecting to Real-World Applications

The concept of domains is not just an abstract mathematical idea. It has real-world applications in fields like physics, engineering, and economics. For example, in physics, the domain of a function might represent the range of possible values for physical quantities like temperature or pressure. In economics, it could represent feasible production levels given resource constraints.

Step 9: Challenges and Common Mistakes

When working with domains, it’s easy to make mistakes, especially when dealing with inequalities. A common error is forgetting to exclude the boundary of the region when the function is undefined there. Another challenge is visualizing the domain in three dimensions for functions of three variables. Practice and careful attention to detail are key to mastering these concepts.

Step 10: Conclusion

In summary, the domain of the function (f(x, y) = \ln(16 - x^2 - 16y^2)) is the set of all points ((x, y)) that lie inside the ellipse (x^2 + 16y^2 = 16). Sketching this domain involves drawing the ellipse and shading its interior. Understanding domains is a fundamental skill in multivariable calculus, with wide-ranging applications in both theoretical and practical contexts.


Q1: Why is the boundary of the ellipse not included in the domain?
A1: The boundary is excluded because the natural logarithm is undefined for zero or negative arguments. At the boundary, (16 - x^2 - 16y^2 = 0), making (\ln(0)) undefined.

Q2: How would the domain change if the function were (f(x, y) = \sqrt{16 - x^2 - 16y^2})?
A2: The domain would include the boundary of the ellipse because the square root function is defined for zero (i.e., (\sqrt{0} = 0)). Thus, the domain would be (x^2 + 16y^2 \leq 16).

Q3: Can the domain of a function ever be empty?
A3: Yes, if the inequality defining the domain has no solutions, the domain is empty. For example, (f(x, y) = \ln(-x^2 - y^2)) has an empty domain because (-x^2 - y^2) is always negative.

Q4: How do you sketch the domain of a function of three variables?
A4: Sketching the domain of a function of three variables, such as (f(x, y, z) = \ln(16 - x^2 - 16y^2 - z^2)), involves visualizing a three-dimensional region (in this case, an ellipsoid) and shading its interior. This is more challenging and often requires software for accurate representation.

Q5: What happens if the inequality defining the domain is not strict?
A5: If the inequality is not strict (e.g., (x^2 + 16y^2 \leq 16)), the domain includes the boundary. This is common in functions like square roots, where the expression inside can be zero.

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