Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

An object's position \(P\) changes so that its distance from \((1,2,-3)\) is always twice its distance from \((1,2,3)\). Show that \(P\) is on a sphere and find its center and radius.

Step-by-Step Solution

Answer

The center of the sphere is (1,2,5) and its radius is 3.

Step 1: Understanding the Problem

We need to find the geometric place for the point P whose distance from the point (1,2,-3) is twice its distance from the point (1,2,3). To demonstrate that it forms a sphere, we will derive an equation based on these conditions.

Step 2: Set Up the Distances

Let the point P have coordinates (x, y, z). The distance from P to (1,2,-3) is given by \(d_1 = \sqrt{(x-1)^2 + (y-2)^2 + (z+3)^2}\) and the distance from P to (1,2,3) is given by \(d_2 = \sqrt{(x-1)^2 + (y-2)^2 + (z-3)^2}\). We are given that \(d_1 = 2d_2\).

Step 3: Square the Distances

To eliminate the square roots, square both sides of the equation \(d_1 = 2d_2\). This gives us: \((x-1)^2 + (y-2)^2 + (z+3)^2 = 4((x-1)^2 + (y-2)^2 + (z-3)^2)\).

Step 4: Simplify the Equation

Expand both sides: Left-hand side is \((x-1)^2 + (y-2)^2 + (z+3)^2\) and right-hand side is \(4((x-1)^2 + (y-2)^2 + (z-3)^2)\). Simplify to get: \((z+3)^2 = 4(z-3)^2\).

Step 5: Solve the Equation

Now, solve \((z+3)^2 = 4(z-3)^2\). This simplifies to \(z^2 + 6z + 9 = 4(z^2 - 6z + 9)\). Simplifying further, we get \(z^2 + 6z + 9 = 4z^2 - 24z + 36\). Rearrange this equation to get \(3z^2 - 30z + 27 = 0\).

Step 6: Factor the Quadratic Equation

Factoring \(3z^2 - 30z + 27 = 0\) gives \(3(z^2 - 10z + 9) = 0\). This further factors to \(3(z - 1)(z - 9) = 0\). Thus, the solutions are \(z = 1\) or \(z = 9\).

Step 7: Conclude Position of P

For each z value (\(z=1\) and \(z=9\)), the coordinates \((x, y)\) satisfy \((x-1)^2 + (y-2)^2 = 9\). This corresponds to a circle centered at (1,2) with radius 3 in each plane z=1 and z=9.

Step 8: Determine the Sphere

Both circles define the same sphere running parallel between planes z = 1 and z = 9. The sphere is centered at (1,2,5) (the midpoint of z=1 and z=9) with radius 3. Thus, the points on the sphere satisfy the equation \((x-1)^2 + (y-2)^2 + (z-5)^2 = 9\).

Key Concepts

Sphere

A sphere is a three-dimensional geometric shape that is perfectly symmetrical around its center. It can be described as the set of all points in space that are equidistant from a central point, known as the center of the sphere. In mathematics, the equation of a sphere centered at the point \( (h, k, l) \) with radius \( r \) is given by the formula: \[ (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \]
Spheres have several unique properties, one of which is their consistency in symmetry regardless of rotation. This makes spheres a common topic in calculus and geometry due to their simple yet profound geometric properties.
In the context of the problem provided, the point \( P \) moves such that its distances define a sphere. The center and radius of this sphere can be found by analyzing the geometric conditions given for the distances.

Distance Formula

The distance formula is a crucial concept in both two-dimensional and three-dimensional coordinate geometry. It allows us to find the distance between two points in space by using their coordinates. In three-dimensional space, the distance between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is calculated using:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
This formula derives from the Pythagorean theorem and is integral in deriving equations of geometric shapes, such as spheres.
In the given problem, the distance formula is used to establish the relationships between the point \( P \) and two other points. By equating and comparing distances, we deduce that \( P \) lies on a sphere by satisfying these specific geometric relationships.

Quadratic Equation

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. These equations can be solved using several methods, such as factoring, completing the square, or using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Quadratic equations often arise in problems involving symmetry and optimization, including defining the boundaries of geometric shapes.
In the problem, after setting up the condition of the sphere, the equation becomes quadratic. Solving this equation allows us to find specific values for \( z \) that satisfy the original distance condition, further leading us to the full equation of the sphere.

Geometric Place

The concept of a geometric place refers to the set of all points that satisfy a given condition or set of conditions. In geometry, understanding this concept is key to identifying and describing various shapes and forms.
In our problem, the geometric place is initially unknown but needs to be expressed in terms of a condition involving distances. Specifically, that one distance is twice another, leading to a series of solutions defining the points in space \( P \) could occupy.
Ultimately, the geometric place is shown to be a sphere, based on the distance constraints. Finding this involves interpreting the algebra into a more familiar geometric context, identifying the center, and radius of the sphere as derived from these conditions.