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Chapter 14: Problem 20

A parabolic arch has a height of \(20 \mathrm{ft}\) and a width of \(36 \mathrm{ft}\) at the base. If the vertex of the parabola is at the top of the arch, at what height above the base is it \(18 \mathrm{ft}\) wide?

Short Answer

Expert verified
The height of the arch at 18 ft wide is 15 ft.

Step by step solution

01

Identify the vertex and equation form

The vertex of the parabola is at the top of the arch. Since the arch is symmetrical and highest at the center, the vertex is at (0, 20). The general form of a parabola with vertex (h, k) is given by y = a(x-h)^2 + k. Here, (h,k) = (0, 20), so the equation simplifies to y = ax^2 + 20.
02

Determine the value of a

At the base, the arch is 36 ft wide, meaning it stretches 18 ft on either side of the center (vertex). At these points, (x, y) = (-18, 0) and (x, y) = (18, 0). Substitute one set of these coordinates (e.g., (18, 0)) into the equation to find a: 0 = a(18)^2 + 20. Simplifies to 0 = 324a + 20. After rearranging, -20 = 324a, so a = -20/324 = -5/81.
03

Substitute a back into the equation

Substitute a = -5/81 into the parabolic equation: y = (-5/81)x^2 + 20.
04

Solve for the height at 18 ft width

The width of 18 ft corresponds to x = 卤9 ft (half the total width). Substitute x = 9 into the equation: y = (-5/81)(9)^2 + 20. This simplifies to y = (-5/81)(81) + 20 = -5 + 20 = 15 ft. Therefore, the height of the arch at a width of 18 ft is 15 ft.

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Key Concepts

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

Vertex Form of a Parabola
In problems involving parabolic arches, we often use a specific form of the parabolic equation known as the vertex form. This form is particularly useful because it directly gives us the vertex of the parabola, which is the highest or lowest point on the graph.
The vertex form of a parabola can be written as: \[ y = a(x-h)^2 + k \ \ \] Here: \[ (h, k) \ \ \] represents the vertex of the parabola
a is a coefficient that determines how 'wide' or 'narrow' the parabola is and whether it opens upwards or downwards
For our specific exercise, the vertex is at (0, 20). This makes our equation \[ y = ax^2 + 20 \] since the term (x-h) becomes x when h is 0.
Determining the Coefficient 'a'
To find the exact shape of our parabola, we need to determine the coefficient 'a'. This requires additional information about the parabola, such as a point on its curve.
In the given exercise, we know the arch is 36 feet wide at the base. This means at the ground level (y=0), the x-coordinates are -18 feet and 18 feet.
We substitute (18, 0) into our vertex form equation to find 'a': \[ 0 = a(18)^2 + 20 \ \ \] Simplifying this, we get: \[ 0 = 324a + 20 \ \ \] Solving for 'a', it follows that: \[ -20 = 324a \ \ \] \[ a = -\frac{20}{324} = -\frac{5}{81} \] Thus, the value of 'a' is -5/81. This negative value indicates that our parabola opens downwards.
Solving Quadratic Equations
Once we have the equation of our parabola: \[ y = -\frac{5}{81}x^2 + 20 \ \ \] we can solve for specific values to find different heights below the vertex. For instance, the exercise asks for the height at a width of 18 feet.
Given the symmetry of the parabola, 18 feet wide corresponds to 9 feet from the centre on each side (since 18/2 = 9).
We substitute x = 9 into the parabolic equation: \[ y = -\frac{5}{81}(9)^2 + 20 \ \ \] This simplifies to: \[ y = -\frac{5}{81}(81) + 20 \ \ \] \[ y = -5 + 20 = 15 \ \ \] Therefore, the height of the arch at a width of 18 feet is 15 feet.
Geometry of Parabolas
Understanding the geometry of parabolas helps us to make sense of their appearance and properties. Parabolas are defined as the set of all points equidistant from a fixed point called the 'focus' and a line called the 'directrix'.
In our case, the arch exemplifies the classic parabolic shape, with its vertex at the highest point.
Important properties include:
  • Symmetry: Parabolas are symmetric about their vertical axis passing through the vertex.
  • Vertex: The vertex is the maximum (or minimum) point.
  • Width: The coefficient 'a' impacts the width and direction of the parabola. A smaller absolute value of 'a' means a wider parabola, while a larger absolute value indicates a narrower shape.
These properties allow us to make quick calculations and visual approximations about the shape and dimensions of parabolas in various applications.

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Most popular questions from this chapter

The ceiling in a hallway \(20 \mathrm{ft}\) wide is in the shape of a semiellipse and is \(18 \mathrm{ft}\) high in the center and \(12 \mathrm{ft}\) high at the side walls. Find the height of the ceiling \(4 \mathrm{ft}\) from either wall.

The cost of production of a commodity is $$\$ 12$$ less per unit at a point \(A\) than it is at a point \(B\) and the distance between \(A\) and \(B\) is 100 miles. Assuming that the route of delivery of the commodity is along a straight line, and that the delivery cost is 20 cents per unit per mile, find the curve at any point of which the commodity can be supplied from either \(A\) or \(B\) at the same total cost. (HINT: Take points \(A\) and \(B\) at \((-50,0)\) and \((50,0)\), respectively.)

The cable of a suspension bridge hangs in the form of a parabola when the load is uniformly distributed horizontally. The distance between two towers is \(1500 \mathrm{ft}\), the points of support of the cable on the towers are \(220 \mathrm{ft}\) above the roadway, and the lowest point on the cable is \(70 \mathrm{ft}\) above the roadway. Find the vertical distance to the cable from a point in the roadway \(150 \mathrm{ft}\) from the foot of a tower.

Remove the \(x y\) term from the given equation by a rotation of axes. Draw a sketch of the graph and show both sets of axes. $$ 24 x y-7 y^{2}+36=0 $$

Remove the \(x y\) term from the given equation by a rotation of axes. Draw a sketch of the graph and show both sets of axes. $$ 6 x^{2}+20 \sqrt{3} x y+26 y^{2}=324 $$
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