Problem 31 - Calculus with Concepts in Calculus

Question

Describe the boundaries of the following regions. a. The disk with center \((-3,2)\) and radius 6 b. The rectangular region with vertices \((0,0),(2,0)\), \((0,-3)\), and \((2,-3)\) c. The triangular region with vertices \((-1,1),(1,1)\), and \((0,-5)\) d. The upper half of the \(x y\) plane, consisting of all \((x, y)\) such that \(y \geq 0\) e. The graph of the parabola \(y=4 x^{2}\) f. The entire plane except the origin

Step-by-Step Solution

Answer

a. Circle: \( (x+3)^2 + (y-2)^2 = 36 \). b. Rectangular lines: \(x=0, x=2, y=0, y=-3\). c. Triangle lines: \(y=1\), \(y=-6x-5\), \(y=6x-5\). d. Upper plane: \(y \geq 0\). e. Parabola: \(y=4x^2\). f. All plane except origin.

Step 1: Identify the Center and Radius of the Disk

The disk is defined by its center and its radius. For part (a), the center is given as \((-3, 2)\) and the radius is 6. The boundary of the disk is the circle described by these parameters.

Step 2: Write Equation for the Disk Boundary

The equation of a circle in the coordinate plane is given by \( (x - h)^2 + (y - k)^2 = r^2 \),where \((h, k)\) is the center and \(r\) is the radius. Substituting \((-3, 2)\) for the center and 6 for the radius,\( (x + 3)^2 + (y - 2)^2 = 36 \).

Step 3: Determine the Boundaries of the Rectangle

The vertices of the rectangle in part (b) are given as \((0,0), (2,0), (0,-3),\) and \((2,-3)\). These points form a rectangle with horizontal sides from \((0,0)\) to \((2,0)\) and vertical sides from \((0,0)\) to \((0,-3)\). The boundaries are thus lines along \(x = 0, x = 2, y = 0,\) and \(y = -3\).

Step 4: Describe the Boundaries of the Triangle

For part (c), the triangular region is formed by vertices \((-1,1), (1,1), (0,-5)\). The boundary consists of line segments connecting these points. The equations of these lines must be determined using the slope-intercept form.

Step 5: Calculate the Line Equations of the Triangle

1. Line from \((-1,1)\) to \((1,1)\) is horizontal at \(y = 1\).2. Line from \((-1,1)\) to \((0,-5)\): Slope = \( \frac{-5 - 1}{0 + 1} = -6 \), Equation: \( y = -6x - 5 \).3. Line from \((0,-5)\) to \((1,1)\): Slope = \( \frac{1 + 5}{1 - 0} = 6 \), Equation: \( y = 6x - 5 \).

Step 6: Describe the Upper Half of the XY Plane

The upper half of the \(xy\) plane in part (d) is defined by \(y \geq 0\), meaning all area above, and including, the x-axis. Thus, the boundary line is \(y = 0\).

Step 7: Define the Boundary of the Parabola

Part (e) describes a parabola given by the equation \(y = 4x^2\). The boundary of this region is the graph of the parabola itself, which opens upwards.

Step 8: Define the Entire Plane Excluding the Origin

Part (f) includes all points in the plane except the origin \((0,0)\). The boundary is not a physical line or curve, it denotes the exclusion of this single point.

Key Concepts

Circle Equation

A circle in coordinate geometry is defined by its center and radius. The equation that represents a circle with a given center \(h, k\) and radius \r\ is given by the formula:\[(x - h)^2 + (y - k)^2 = r^2\]
In this formula:

• \(x\) and \(y\) are the coordinates of any point on the circle.
• \(h\) and \(k\) are the coordinates of the circle's center.
• \(r\) is the radius of the circle.

For example, the circle with a center at \((-3, 2)\) and a radius of 6 is described by substituting these values into our circle equation, giving us:\[(x + 3)^2 + (y - 2)^2 = 36\]This equation tells us that every point (\(x, y\)) that satisfies this equation lies on the boundary of the circle. A circle is a set of all points that are equidistant from a single point, which is its center.

Rectangle Boundaries

Rectangular boundaries in coordinate geometry are formed by connecting four vertices. These vertices define the sides of the rectangle, which are straight lines parallel to either the x-axis or y-axis.
For example, consider a rectangle with vertices at \(0,0\), \(2,0\), \(0,-3\), and \(2,-3\). The boundaries of this rectangle can be determined by:

• Horizontal sides running from \(x = 0\) to \(x = 2\) at \(y = 0\) and \(y = -3\).
• Vertical sides running from \(y = 0\) to \(y = -3\) at \(x = 0\) and \(x = 2\).

The lines that form these boundaries are given by the equations \x = 0\, \x = 2\, \y = 0\, and \y = -3\. Thus, the rectangular boundary is defined entirely by these linear equations.

Triangle Boundary Equations

A triangle in the coordinate plane can be described by the line segments connecting its vertices. These lines can be found using the slope-intercept form, which is expressed as \y = mx + b\, where \(m\) is the slope and \(b\) is the y-intercept.
Consider a triangle with vertices at \((-1,1)\), \(1,1\), and \(0,-5\):1. The line from \((-1,1)\) to \(1,1\) is horizontal, so the equation is simply \y = 1\.2. To find the line from \((-1,1)\) to \(0,-5\), calculate the slope: \m = \frac{-5 - 1}{0 + 1} = -6\. Thus, the equation is \y = -6x - 5\.3. For the line from \(0,-5\) to \(1,1\), the slope is \m = \frac{1 + 5}{1 - 0} = 6\, leading to the equation \y = 6x - 5\.Each of these equations represents a boundary of the triangle.

Parabolic Graph

A parabola is a symmetric curve shaped like an "U" or an "N" and is described by a quadratic equation of the form \y = ax^2 + bx + c\. It opens upwards if \(a > 0\) and downwards if \(a < 0\).
In part (e), we are given a parabola \y = 4x^2\ which describes a curve that opens upwards, since \(a = 4 > 0\) is positive. The vertex of this parabola lies at the origin \(0,0\), and because there are no linear or constant terms, this parabola is symmetric about the y-axis.
The graph of this parabola illustrates a continuous boundary of various regions above the x-axis, where every point \(x, y\) on the curve satisfies the equation \y = 4x^2\. Understanding parabolas is essential in coordinate geometry, as they help in modeling situations where distances, times, or other quantities have squared dependencies.