Problem 21
Question
MANUFACTURING The reflective surface in a flashlight has a parabolic shape with a cross section that can be modeled by \(y=\frac{1}{3} x^{2},\) where \(x\) and \(y\) are in centimeters. How far from the vertex should the filament of the light bulb be located?
Step-by-Step Solution
Verified Answer
The filament should be located \( \frac{3}{4} \) cm from the vertex.
1Step 1: Understanding the Vertex and Focus
In a parabola modeled by the equation \( y = ax^2 \), the vertex is at the origin, (0, 0). The distance from the vertex to the focus, \( p \), is related to \( a \) by the equation \( a = \frac{1}{4p} \). Here, \( a = \frac{1}{3} \).
2Step 2: Calculate the Value of p
Using the formula \( a = \frac{1}{4p} \) and substituting \( a = \frac{1}{3} \), we solve for \( p \):\[ \frac{1}{3} = \frac{1}{4p} \]Cross multiply to get \( 4p = 3 \), and then solve for \( p \) by dividing both sides by 4:\[ p = \frac{3}{4} \].
3Step 3: Interpret the Result
Since \( p = \frac{3}{4} \), the filament of the light bulb should be placed \( \frac{3}{4} \) centimeters from the vertex of the parabolic surface for optimal focus.
Key Concepts
ParabolaVertexFocusReflective Surfaces
Parabola
A parabola is a special type of curve, similar to a U-shape, that can be found in different mathematical and real-world contexts. In simple terms, it is a conic section formed by the intersection of a right circular cone and a plane that is parallel to the side of the cone. Parabolas have reflective properties which make them useful in devices like flashlights and satellite dishes.
They are defined mathematically by quadratic functions, commonly expressed as \(y = ax^2 + bx + c\) for vertical parabolas. Noticeably, the parabola always has a symmetry axis, and the curve is always equidistant from a single point called the focus and a straight line called the directrix.
It's essential to understand the parabola for applications involving focus and reflection due to its geometric structure.
They are defined mathematically by quadratic functions, commonly expressed as \(y = ax^2 + bx + c\) for vertical parabolas. Noticeably, the parabola always has a symmetry axis, and the curve is always equidistant from a single point called the focus and a straight line called the directrix.
It's essential to understand the parabola for applications involving focus and reflection due to its geometric structure.
Vertex
The vertex of a parabola is the point where the parabola changes direction. In other words, it is the tip or the highest or lowest point of the curve, depending on how the parabola is oriented. For the standard parabola \(y = ax^2\), the vertex is located at the origin point \((0, 0)\).
In mathematical terms, the vertex lies at \((h, k)\) for a parabola described by \(y = a(x-h)^2 + k\). For our specific case, \(h = 0\) and \(k = 0\), indicating that the parabola opens upward from the origin.
Understanding the vertex is crucial, as it's often the reference point for measuring distances in problems involving parabolas, like locating the focus.
In mathematical terms, the vertex lies at \((h, k)\) for a parabola described by \(y = a(x-h)^2 + k\). For our specific case, \(h = 0\) and \(k = 0\), indicating that the parabola opens upward from the origin.
Understanding the vertex is crucial, as it's often the reference point for measuring distances in problems involving parabolas, like locating the focus.
Focus
The focus of a parabola is a key concept, especially in applications involving reflection and concentration of waves or signals. It is a fixed point within the curve, and every point on the parabola is equidistant from the focus as it is from the directrix.
Knowing exactly where the focus lies is important in setups like flashlights, as light waves originating from this point are reflected off the parabolic surface in parallel lines, ensuring a concentrated beam. The relationship between the focus and the parabola's equation is determined by the parameter \(a\) and the formula \(a = \frac{1}{4p}\), where \(p\) is the distance between the vertex and the focus.
In our exercise, using the formula \(\frac{1}{3} = \frac{1}{4p}\), we've calculated \(p = \frac{3}{4}\), meaning the focus is \(\frac{3}{4}\) cm from the vertex.
Knowing exactly where the focus lies is important in setups like flashlights, as light waves originating from this point are reflected off the parabolic surface in parallel lines, ensuring a concentrated beam. The relationship between the focus and the parabola's equation is determined by the parameter \(a\) and the formula \(a = \frac{1}{4p}\), where \(p\) is the distance between the vertex and the focus.
In our exercise, using the formula \(\frac{1}{3} = \frac{1}{4p}\), we've calculated \(p = \frac{3}{4}\), meaning the focus is \(\frac{3}{4}\) cm from the vertex.
Reflective Surfaces
Reflective surfaces that take advantage of parabolic shapes have remarkable properties. Parabolas can direct light or sound waves precisely because of their geometric characteristics. Light rays emanating from the focus are reflected off the surface and travel parallel to the axis of symmetry, creating a concentrated beam.
This property is ingeniously utilized in several applications, such as car headlights, satellite dishes, and of course, flashlights. The efficiency of these devices largely depends on positioning the light source at the correct distance, further highlighting the importance of understanding distances like those between the vertex and the focus.
This property is ingeniously utilized in several applications, such as car headlights, satellite dishes, and of course, flashlights. The efficiency of these devices largely depends on positioning the light source at the correct distance, further highlighting the importance of understanding distances like those between the vertex and the focus.
- Efficient design: Light is focused and projected effectively, offering brighter illumination.
- Precision: Assists in focused transmission and reception, crucial for communication technologies.
- Simplicity: Requires fewer adjustments and components compared to other reflective designs.
Other exercises in this chapter
Problem 21
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