Problem 69 - University Calculus: Early Transcendentals

Question

Find an equation for the set of all points equidistant from the planes \(y=3\) and \(y=-1\)

Step-by-Step Solution

Answer

Equations of the set are \( y = 3 \) and \( y = -1 \), covering all points between.

Step 1: Understanding the Problem

We need to find all points that are equidistant from the planes defined by the equations \( y = 3 \) and \( y = -1 \). These two planes are parallel and separated along the y-axis.

Step 2: Finding Midpoint

The midpoint between two planes equidistant on the y-axis is simply the average of the y-coordinates of these planes. Calculate: \[\text{Midpoint} = \frac{3+(-1)}{2} = \frac{2}{2} = 1\] Thus, the midpoint is \( y = 1 \).

Step 3: Finding Distance to Midpoint

Since points must be equidistant from both planes, the distance from any point to \( y = 1 \) is calculated such that it is equidistant to both the planes \( y = 3 \) and \( y = -1 \). This condition leads us to an expression for distance from \( y = 1 \):\[|y - 1| = 2\] Since both planes are parallel and equally spaced from \( y = 1 \), their equal distance is 2 on either side.

Step 4: Writing the Final Equation

The set of all points equidistant from the planes \( y = 3 \) and \( y = -1 \) implies using the distance from the midpoint, leading us to two possible solutions for y:\[|y - 1| = 2\] This implies:\[y - 1 = 2 \quad \text{or} \quad y - 1 = -2\]Solving these, we get:\[y = 3 \quad \text{or} \quad y = -1\] Thus, the required equation represents the entire region between \( y = 3 \) and \( y = -1 \).

Key Concepts

Parallel Planes

In geometry, parallel planes are two planes that never meet, no matter how far they extend. They run along the same direction at a consistent distance apart. For example, the planes given by the equations \( y = 3 \) and \( y = -1 \) are parallel because both are oriented sideways with respect to the y-axis and maintain a constant gap across the entire coordinate space. This gap creates a region of space between the planes that we can explore for specific points, such as those that are equidistant from both planes. Understanding parallel planes is critical in this problem because it sets the stage for identifying the midpoint on the y-axis between \( y = 3 \) and \( y = -1 \).

Midpoint Calculation

The concept of finding a midpoint comes into play when you need to determine a central point between two other points or planes. A midpoint is simply the average or mean of these two quantities. In our current problem, we have two y-coordinates: \( y = 3 \) and \( y = -1 \). Calculating the midpoint involves:

• Adding the two y-values: \( 3 + (-1) = 2 \)
• Dividing by two: \( \frac{2}{2} = 1 \)

After these steps, you find that the midpoint is \( y = 1 \). This midpoint is crucial as it represents the halfway position along the y-axis, equidistant from both planes. Understanding midpoint calculation helps you locate this central axis of symmetry between the parallel planes.

Distance Formula

The distance formula is used to calculate the length of the shortest path between two points. In one-dimensional cases, such as the y-axis in this exercise, it simplifies to the absolute difference between two y-values. Here, we use it to ensure that our selected points are equidistant from \( y = 3 \) and \( y = -1 \). To achieve this, consider the midpoint \( y = 1 \). Each point must satisfy:

• The distance from \( y = 1 \) to one of the target planes is \(|y - 1| = 2\).
• This represents a point that properly balances itself between the two original planes.

The absolute value notation \(|y - 1|\) implies equal spacing from the y-value \( y = 1 \), maintaining the condition of being equidistant to both planes.

Geometric Properties

Understanding geometric properties allows us to approach complex exercises methodically. This problem revolved around concepts such as symmetry, equidistance, and balance between geometric objects (planes). The property that points are equidistant between two parallel planes evokes an importance of balance in geometry. In this case, symmetry around \( y = 1 \) was pivotal.

• Equidistant points here lie in a middle zone extending both ways symmetrically from \( y = 1 \).
• The characteristics of these equidistant points tell us the region extends up to \( y = 3 \) and down to \( y = -1 \).

By understanding and applying these geometric properties, problems like this can be solved with ease and insight. Geometric understanding provides a foundation for visually conceptualizing and solving such spatial problems efficiently.