Problem 11 - Calculus 3

Question

Find the range of the functions. $$ g(x, y)=\sqrt{16-4 x^{2}-y^{2}} $$

Step-by-Step Solution

Answer

The range of the function is [0, 4].

Step 1: Understand the Function

The given function is a function of two variables, represented by \(g(x, y) = \sqrt{16 - 4x^2 - y^2}\). For \(g(x, y)\) to be defined and have real values, the expression under the square root must be non-negative.

Step 2: Determine the Domain

We require the expression under the square root to be non-negative for \(g(x, y)\) to be real, so:\[16 - 4x^2 - y^2 \geq 0\]Rearranging gives:\[4x^2 + y^2 \leq 16\]This inequality is satisfied inside or on the boundary of an ellipse centered at (0, 0) with a semi-major axis of 4 along the x-direction and a semi-minor axis of 4 along the y-direction.

Step 3: Determine the Range

Since \(g(x, y) = \sqrt{16 - 4x^2 - y^2}\) takes non-negative values and decreases as \(4x^2 + y^2\) increases towards 16, the maximum value occurs when \(4x^2 + y^2 = 0\). At this point, \(g(0, 0) = \sqrt{16} = 4\).The minimum value occurs on the boundary where the inequality becomes an equality (\(4x^2 + y^2 = 16\)), giving \(g(x, y) = \sqrt{0} = 0\).Thus, as \(4x^2 + y^2\) varies from 0 to 16, \(g(x, y)\) decreases from 4 to 0.

Step 4: Conclude the Range

Based on the analysis, as \(g(x, y)\) takes values from 0 up to 4, we can deduce that the range of the function \(g(x, y)\) is the interval from 0 to 4, inclusive of both endpoints: \[ \text{Range of } g(x, y) = [0, 4] \]

Key Concepts

Ellipses

Ellipses are fundamental shapes in geometry, characterized by their unique oval form. An ellipse is essentially a stretched circle, and can be defined by two main axes: the major axis and the minor axis.

• The major axis is the longest diameter of the ellipse, while the minor axis is the shortest. These axes intersect at the center of the ellipse, creating a perfect balance between the two halves.
• For the function given by \(g(x, y) = \sqrt{16 - 4x^2 - y^2}\), the equation \(4x^2 + y^2 = 16\) forms the boundary of an ellipse.
• This ellipse is centered at the origin (0,0) and is aligned along the axes. The semi-major axis, stretching along the x-direction, is 2 units because it is linked with \(4x^2\), and the semi-minor axis, along the y-direction, is 4 units.

In this context, understanding the ellipse helps us describe the area where the original function \(g(x, y)\) is defined.

Domains and Ranges

In the realm of functions, the domain and range play crucial roles. The domain of a function refers to the complete set of possible input values (or x, y pairs in the case of multivariable functions), while the range is the set of possible output values.

• For \(g(x, y) = \sqrt{16 - 4x^2 - y^2}\), to ensure real numbers, the part under the square root must be non-negative. This stipulation determines the domain.
• We solve the inequality \(16 - 4x^2 - y^2 \geq 0\), which simplifies to \(4x^2 + y^2 \leq 16\). This indicates that all (x, y) points lying inside or on the ellipse satisfy the condition.

When analyzing the range, we consider the possible values that \(g(x, y)\) can output.

• Given that the expression under the square root decreases from 16 to 0, the range of \(g(x, y)\) is from 4 to 0. The function peaks at 4 when \(4x^2 + y^2 = 0\) and declines to 0 when \(4x^2 + y^2 = 16\).
• Thus, the function's range can be expressed with the interval [0, 4].

Square Roots

Square roots are a fundamental concept in mathematics that represent a value which, when multiplied by itself, gives the original number. The square root function is often seen in various forms and is crucial in defining the behavior of a function.

• For any expression under a square root to be valid and produce real numbers, it must be non-negative. This principle applies to the function \(g(x, y) = \sqrt{16 - 4x^2 - y^2}\).
• This requirement means \(16 - 4x^2 - y^2\) must be greater than or equal to zero to ensure the square root is defined.

Square roots inherently provide non-negative outputs.

• In the context of our function, \(g(x, y)\), the square root begins at 4 (when x and y are closest to the origin) and shrinks to 0 (as x and y reach the ellipse boundary).
• This behavior showcases the square root's property of transforming a positive input area into a range of real, non-negative outputs.