Problem 42 - Precalculus with Calculus Previews

Question

Find a function \(f\) that defines the indicated half-ellipse. Give the domain of each function. The equations are from Problems 1,3,9, and 12 $$ x^{2}+\frac{y^{2}}{16}=1 ; \text { upper half-ellipse } $$

Step-by-Step Solution

Answer

The function is \( f(x) = \sqrt{16(1-x^2)} \) with domain \([-1, 1]\).

Step 1: Identify the Original Equation

The given equation for the ellipse is \( x^2 + \frac{y^2}{16} = 1 \). This is in the standard form of an ellipse centered at the origin.

Step 2: Solve for y

To express \( y \) in terms of \( x \), rearrange the equation to solve for \( y^2 \): \( \frac{y^2}{16} = 1 - x^2 \). Multiply both sides by 16 to get \( y^2 = 16(1 - x^2) \).

Step 3: Determine the Expression for y

Take the square root of both sides to solve for \( y \): \( y = \pm \sqrt{16(1-x^2)} \). To represent only the upper half-ellipse, select the positive square root, so \( y = \sqrt{16(1-x^2)} \).

Step 4: Define the Function

The function \( f(x) \) representing the upper half-ellipse is \( f(x) = \sqrt{16(1-x^2)} \).

Step 5: Determine the Domain of the Function

The domain of \( f(x) = \sqrt{16(1-x^2)} \) is determined by the condition that the expression inside the square root must be non-negative: \( 16(1-x^2) \geq 0 \). Solve for \( x \) to find \( -1 \leq x \leq 1 \). Therefore, the domain is \([-1, 1]\).

Key Concepts

Domain of a Function

The domain of a function tells us all possible values of the variable, usually represented as "x", for which the function makes sense. In other words, it's the set of all allowable inputs. For a function to be valid, it must meet certain conditions. If a function involves a square root, for example, the expression inside the square root must be non-negative.For the exercise with the half-ellipse, the function is represented as \( f(x) = \sqrt{16(1-x^2)} \). Here, the term \( 16(1-x^2) \) is under a square root, so it must be \( \geq 0 \) to be valid:

• The expression \( 16(1-x^2) = 16 - 16x^2 \) should be greater than or equal to zero.
• This simplifies to \( 1-x^2 \geq 0 \), meaning \( -1 \leq x \leq 1 \).

Thus, the domain of this function is \([-1, 1]\). This range restricts the "x"-values to this interval, ensuring that the square root can be calculated without encountering a negative square root scenario.

Standard Form of an Ellipse

An ellipse is a stretched circle, often represented in its standard form. Understanding this form is key to identifying the properties of the ellipse, such as its center and its axes. For an ellipse centered at the origin, the standard form is usually written as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).Here's what each term represents:

• \( a^2 \) is associated with the semi-major axis if \( a > b \) or with the semi-minor axis if \( b > a \).
• \( b^2 \) represents the length squared of the other axis depending on its orientation.

In our exercise, the equation given is \( x^2 + \frac{y^2}{16} = 1 \), indicating:

• The center of the ellipse is at the origin \((0, 0)\).
• The semi-major axis along the y-direction is \( \sqrt{16} = 4 \), and the semi-minor axis along the x-direction is \( 1 \).

This structure helps solve for the equation in terms of either axis, allowing for further manipulations such as isolating the top or bottom half of the ellipse.

Solving Equations

To find functions derived from an ellipse, understanding how to manipulate and solve equations is crucial. When given a relationship like \( x^2 + \frac{y^2}{16} = 1 \), the goal often is to express one variable in terms of the other, usually "y" in terms of "x".Let's break down the process:

• Start by isolating \( y^2 \) on one side: \( \frac{y^2}{16} = 1 - x^2 \).
• Multiply both sides by 16 to clear the fraction: \( y^2 = 16(1-x^2) \).
• To solve for "y", take the square root of both sides, leading to \( y = \pm\sqrt{16(1-x^2)} \).

Choosing between the positive or negative square root tailors the solution to specific parts of the ellipse:

• For the upper half-ellipse, select the positive root: \( y = \sqrt{16(1-x^2)} \).

This method simplifies complex relation into understandable functions, enabling us to define equations for specific sections of the ellipse, like the upper or lower halves.