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Chapter 8: Problem 11

An electric heater is designed as a parabolic reflector that is \(5^{\prime \prime}\) deep. To prevent accidental burns, the centerline of the heating element, placed at the focus, must be set in \(1.5^{\prime \prime}\) from the base of the reflector. a. What equation could you use to design the reflector? (Hint: Tip the parabolic reflector so it opens upward.) b. How wide would the reflector be at its rim?

Short Answer

Expert verified
The equation of the parabola is \(y = \frac{1}{6}x^2\). The width of the reflector is approximately 10.96 inches.

Step by step solution

01

Identify the given parameters

The depth of the parabolic reflector is given as 5 inches, and the distance from the base to the focus is 1.5 inches.
02

Relate depth and focus with the vertex form of a parabola

Recall that the standard form of a parabola opening upward with vertex at \(h, k\) is \(y = a(x - h)^n + k\). For our problem, we can set the vertex at the origin \(0, 0\) so the equation simply becomes \(y = ax^2\). Since the focus is at \(0, 1.5\) and the vertex is at the origin, the relationship between \(a\) and the focal distance \(d\) is given by \(d = \frac{1}{4a}\). Plugging \(d = 1.5\), we get \(1.5 = \frac{1}{4a}\).
03

Solve for the value a

From the equation \(1.5 = \frac{1}{4a}\), we solve for \(a\) by isolating \(a\). We get \(4a = \frac{1}{1.5} \rightarrow a = \frac{1}{6}\). Thus, the equation of the parabola becomes \(y = \frac{1}{6}x^2\).
04

Determine the width of the reflector

The reflector is 5 inches deep, which means the corresponding \(y\) value at the rim is 5. Substitute \(y = 5\) into \(y = \frac{1}{6}x^2\) to find \(x\). Solving for \(x\), we get \(5 = \frac{1}{6}x^2 \rightarrow x^2 = 30 \rightarrow x = \sqrt{30} \rightarrow x \approx 5.48\). Since the parabolic reflector is symmetric about the y-axis, the total width is \(2 \times 5.48 = 10.96 \text{ inches}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

parabola equation
Parabolas have a unique property: symmetry. This is helpful when you need to determine dimensions and make precise calculations:
    • A parabola is symmetric about its axis of symmetry.
    • The axis of symmetry is a vertical line that passes through the vertex.

      For our parabolic reflector, this means that if \( x \approx 5.48 \) inches to one side, it will also be \( x \approx 5.48 \).

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Most popular questions from this chapter

(Graphing program optional.) The following function represents the relationship between time \(t\) (in seconds) and height \(h\) (in feet) for objects thrown upward on Pluto. For an initial velocity of \(20 \mathrm{ft} / \mathrm{sec}\) and an initial height above the ground of 25 feet, we get $$ h=-t^{2}+20 t+25 $$ a. Find the coordinates of the point where the graph intersects the \(h\) -axis. b. Find the coordinates of the vertex of the parabola. c. Sketch the graph. Label the axes. d. Interpret the vertex in terms of time and height. e. For what values of \(t\) does the mathematical model make sense?

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