Problem 47 - Algebra and Trigonometry with Analytic Geometry

Question

A sound receiving dish used at outdoor sporting events is constructed in the shape of a paraboloid, with its focus 5 inches from the vertex. Determine the width of the dish if the depth is to be 2 feet.

Step-by-Step Solution

Answer

The width of the dish is approximately 43.82 inches.

Step 1: Convert Units

First, convert the depth from feet to inches since the distance to the focus is given in inches. There are 12 inches in a foot, so 2 feet equals 24 inches.

Step 2: Use Parabola Equation

The equation of a parabola in vertex form is \( x^2 = 4py \), where \( p \) is the focal distance. Since the focus is 5 inches from the vertex, \( p = 5 \). Thus, the equation becomes \( x^2 = 20y \).

Step 3: Determine Edge of Dish

The depth of the dish is 24 inches. Substitute \( y = 24 \) into the parabola equation: \( x^2 = 20 \times 24 \). Thus, \( x^2 = 480 \).

Step 4: Solve for x

Solve the equation \( x^2 = 480 \) to find \( x \), which represents half the width of the dish. Thus, \( x = \sqrt{480} \). Calculating this gives \( x \approx 21.91 \) inches.

Step 5: Calculate Total Width

Since \( x \) is half the width of the dish, the total width is \( 2x \). Therefore, the width is \( 2 \times 21.91 \approx 43.82 \) inches.

Key Concepts

Parabola Equation

A parabola is a U-shaped curve that can open either upward, downward, to the left, or to the right. It is defined by its unique property: each point on the parabola is equidistant from a fixed point known as the focus, and a line known as the directrix. The general equation for a parabola is given in the form of \[ x^2 = 4py\]where \(x\) and \(y\) are the coordinates of any point on the parabola, and \(p\) is the focal distance, or the distance from the vertex of the parabola to its focus.

• The vertex is the point where the parabola changes direction.
• In our context, we use this equation to model a physical shape, like a paraboloid dish.

Understanding this equation is key to solving problems involving parabolic shapes, as it helps relate dimensions like depth and width to the position of the focus.

Vertex Form

The vertex form of a parabola's equation offers a handy way to express the parabola if you know the vertex coordinates \((h, k)\) and the focal distance. Sometimes, you may see this represented as \[ (x-h)^2 = 4p(y-k)\]This expression highlights how the shape changes with shifts vertically or horizontally, depending on \(h\) and \(k\). However, in simpler problems where the vertex is at the origin, \((0, 0)\), it simplifies to \[ x^2 = 4py\]

• Knowing this form allows you to quickly determine the coordinates of the vertex.
• All measurements are typically in relation to this vertex point.

This is particularly useful in exercises like the one provided, where simplifying the math helps in understanding and solving for specific dimensions.

Focal Distance

The focal distance \(p\) is an integral part of understanding parabolas. It represents the distance between the vertex of the parabola and its focus. For the equation \[x^2 = 4py\] the value of \(p\) directly scales the width and orientation of the parabola.

• A smaller value of \(p\) results in a steeper parabola.
• A larger \(p\) indicates a wider and more open curve.

In our exercise, the focal distance is given as 5 inches. This measurement helps refine the calculation for dimensions and angles, ensuring precision when constructing physical models like dishes or antennas.

Measurement Conversion

When working with parabolas or any geometrical problems, measurement conversion is often necessary to ensure all dimensions are in the same units. This straightforward step avoids calculation errors that could arise from using mixed units.

• As seen in the problem, depth was initially given in feet. Converting it to inches (12 inches per foot) makes calculations consistent.
• Converting the depth of 2 feet into 24 inches ensures that all lengths relate properly to the focal distance given in inches.

By keeping consistent units throughout, you maintain accuracy and simplify computations, making complex problems much more manageable to solve. Remembering to convert measurements before starting any equations can save a lot of confusion and potential errors later.