I'm confused..

1. The cables of a suspension bridge are in the shape of a parabola. The towers supporting the cable are 600 feet apart and 80 feet high. If the cables touch the road surface midway between the towers, what is the height of the cable at a point 150 feet from the middle of the bridge?

2. A searchlight is shaped like a paraboloid of revolution. If the light source is located 2 feet from the base along the axis of symmetry and the opening is 5 feet across, how deep should the search light be?

What is be best to set up a graph for these? And what is a paraboloid?

Ah thanks!

The main cables of a suspension bridge uniformly distribute the weight of the bridge when in the form of a parabola. The main cables of a particular bridge are attached to towers that are 600 fi apart. The cables are attached to the towers at a height of 110 ft above the roadway and are 10 ft above the roadway at their lowest points. If vertical support cables are at 50-ft intervals along the level roadway, what are the lengths of these vertical cables?

1. Well, you're certainly "suspended" in confusion, aren't you? Let me shed some light on this parabolic puzzle!

To set up a graph for this problem, let's imagine the x-axis representing the distance from the middle of the bridge, and the y-axis representing the height of the cable above the road surface.
Since the cable touches the road surface midway between the towers, we can set up the equation of a parabola in the form of y = ax^2 + c, where a determines the shape of the curve.
To find the value of a, we can use the given information that the towers are 600 feet apart and 80 feet high. So, one of the towers would be located at (-300, 80) and the other at (300, 80).
Substituting these points into the equation will help us solve for a.

2. Now onto the paraboloid problem with the searchlight! It's time to unveil the depth of this illuminating enigma.
To plot a graph, we can use a three-dimensional coordinate system. Let's imagine the x-axis and y-axis forming a plane, and the z-axis representing the depth.
The light source is located 2 feet from the base along the axis of symmetry, so the vertex of the paraboloid would be at (0, 0, 2).
The opening is 5 feet across, which means it spans from x = -2.5 to x = 2.5, and we can determine the equation of the paraboloid using these coordinates and the vertex point.

Now, what is a paraboloid, you ask? Think of it as a funky-shaped object that resembles a hollow ball, with a parabola rotated around its axis. It's like a reflective surface shaped like a smiley-face bowl that focuses light or any other form of funny business.

I hope this clears up your confusion, or at least adds a touch of amusement to it!

To set up a graph for these problems, we need to understand the shape of a parabola and a paraboloid. Let's start by discussing these concepts:

1. Parabola: A parabola is a U-shaped curve that is symmetric about a line called the axis of symmetry. The standard equation of a parabola is y = ax^2 + bx + c, where "a", "b", and "c" are constants. In this case, the cables of the suspension bridge are in the shape of a parabola. We'll need to determine the equation of this parabola to solve the problem.

2. Paraboloid: A paraboloid is a three-dimensional shape that resembles a bowl or a hill. It is formed by rotating a parabola about its axis of symmetry. In the problem, the searchlight is shaped like a paraboloid of revolution. We'll need to find the equation of the paraboloid to determine its depth.

Now, let's move on to solving the problems:

1. To find the height of the cable at a point 150 feet from the middle of the bridge, we can set up a coordinate system with the origin at the middle of the bridge. Let's say the x-axis represents the distance from the middle of the bridge, and the y-axis represents the height of the cable.

Since we know the towers supporting the cable are 600 feet apart and 80 feet high, we can assume that the vertex of the parabola lies at the origin. This means the equation of the parabola will be in the form of y = ax^2.

To find the value of "a," we can use the given information that the cables touch the road surface midway between the towers. At x = 300 (midway point), the y-coordinate should be zero. So, we can substitute these values into the equation: 0 = a(300)^2.

Simplifying the equation, we get 0 = 90000a. Solving for "a," we find that a = 0.

This means that the equation of the parabola representing the cables is y = 0, indicating that the cables are essentially flat at the midpoint between the towers. Therefore, the height of the cable at a point 150 feet from the middle of the bridge will also be zero.

2. To determine the depth of the searchlight, we can again set up a coordinate system. Let's assume the axis of symmetry (the y-axis) is the vertical axis and the x-axis represents the distance from the base of the searchlight.

Given that the light source is located 2 feet from the base along the axis of symmetry and the opening is 5 feet across, we can consider the paraboloid to be symmetric about the y-axis.

The standard equation of a paraboloid in this context is given by: y = ax^2.

To find the value of "a," we can substitute the coordinates of a point on the paraboloid. Let's use the coordinates (2, 2.5) since the light source is located 2 feet from the base, and the opening is 5 feet across (which means the height at the opening should be half of the width).

Substituting these values into the equation, we get 2.5 = a(2)^2.

Simplifying the equation, we find that a = 0.625.

This means that the equation of the paraboloid representing the searchlight is y = 0.625x^2. To find the depth of the searchlight, we need to determine the maximum value of y, which occurs at the vertex of the paraboloid. The x-coordinate of the vertex can be found using the formula: x = -b/(2a).

Since there is no "b" term in this case, the x-coordinate of the vertex is 0. Plugging this value into the equation, we get y = 0.625(0)^2 = 0.

Hence, the depth of the searchlight is 0 feet.

In summary, for the suspension bridge problem, the height of the cable at a point 150 feet from the middle is 0 feet. For the searchlight problem, the depth of the searchlight is 0 feet as well.

Please note that the graph of a parabola or a paraboloid can be sketched based on the equations obtained for each problem.

1. The parabola shape is of the form y = a x^2, where y is the height and x is the horiziontal distance from the lowest point.

80 = a (300)^2
a = 80/(300)^2
When x = 150, y = 80*150^2/300^2 = 80*(1/4) = 20 feet

2. A paraboloid is the solid shape that you get when you rotate a parabola about its axis of symmetry. The light source will be located at the focus of the parabola so that the light that imerges is a parallel beam. (They should have told you that). Figure out the equation of the parabola from those facts and then calculate the depth of the parabola when the width is 5 feet.