Problem 41 - Precalculus with Calculus Previews

Question

In Problems 41-44, 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 $$ \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 \text { ; lower half-ellipse } $$

Step-by-Step Solution

Answer

The function is \( f(x) = -\sqrt{9(1 - \frac{x^2}{25})} \) with domain \([-5, 5]\).

Step 1: Understand the Full Ellipse Equation

The given equation \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \) is the equation of an ellipse centered at the origin with a semi-major axis of 5 along the x-axis and a semi-minor axis of 3 along the y-axis. We need to isolate \( y \) to find the function.

Step 2: Isolate y from the Ellipse Equation

To focus on the lower half of the ellipse, solve for \( y \) by isolating it in the equation. Begin by subtracting \( \frac{x^2}{25} \) from both sides: \( \frac{y^2}{9} = 1 - \frac{x^2}{25} \). Next, multiply through by 9 to clear the fraction: \( y^2 = 9\left(1 - \frac{x^2}{25}\right) \).

Step 3: Solve for y Considering the Half-Ellipse

Take the square root of both sides to solve for \( y \). Since we want the lower half, we select the negative square root: \( y = -\sqrt{9\left(1 - \frac{x^2}{25}\right)} \). This form represents the lower half of the ellipse.

Step 4: Establish the Domain

For the expression under the square root to be valid, it must be non-negative: \( 1 - \frac{x^2}{25} \ge 0 \). Solve for \( x \) by the inequality \( \frac{x^2}{25} \le 1 \), which gives \( -5 \le x \le 5 \). Therefore, the domain for \( f(x) \) is \([-5, 5]\).

Key Concepts

Half-Ellipse

An ellipse is a shape that looks like a stretched circle, and it can be divided into two symmetrical parts called half-ellipses. Understanding half-ellipses is crucial in mathematics as sometimes we are interested only in one part of this shape. A half-ellipse essentially means taking one half of an ellipse; it could be the top or bottom half.

Imagine cutting an ellipse horizontally at its widest part. This would split it into an upper and lower half. In our specific exercise, we focus on the lower half of the ellipse. We track this by solving the equation of the ellipse, making sure to capture only the values of the lower part for our solution. This involves using the negative square root when solving for the y-values, to indicate values below the x-axis.

Domain of Functions

The domain of a function is all the input values (or x-values) for which the function is defined. When dealing with ellipses, finding the domain means identifying the range of x-values that satisfy the ellipse equation.

In the case with our half-ellipse, the function only exists where the values under the square root are non-negative. This ensures the results remain real numbers. Consequently, we solve for this by ensuring that the portion under the square root is greater than or equal to zero, leading to the inequality \(1 - \frac{x^2}{25} \ge 0\). Solving this inequality gives a domain for the function: \([-5, 5]\). This means the function is only valid for x-values ranging from -5 to 5.

Lower Half-Ellipse

When identifying the lower half-ellipse, we need to focus on the part beneath the x-axis. It's signified by solving the ellipse equation to isolate y, selecting the negative root which points downward.

With the given ellipse equation \(\frac{x^2}{25} + \frac{y^2}{9} = 1\), isolating y involves solving to get \(y = -\sqrt{9\left(1 - \frac{x^2}{25}\right)}\). Here, the negative sign before the square root indicates that we are working with the lower half of the ellipse. It's crucial because it specializes our function to the y-values below the x-axis, thereby representing just the lower curve of the ellipse. By focusing on this specific half, this allows us to analyze and utilize only the required segments of the ellipse in real-world applications.