Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equations of the tangent spheres of equal radii whose centers are \((-3,1,2)\) and \((5,-3,6)\).

Step-by-Step Solution

Answer

The spheres' equations are \((x+3)^2 + (y-1)^2 + (z-2)^2 = 24\) and \((x-5)^2 + (y+3)^2 + (z-6)^2 = 24\).

Step 1: Determine the distance between the centers

Calculate the distance between the centers of the spheres located at \((-3, 1, 2)\) and \((5, -3, 6)\) using the distance formula: \[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\] Substitute the given values: \[d = \sqrt{(5+3)^2 + (-3-1)^2 + (6-2)^2} = \sqrt{8^2 + (-4)^2 + 4^2}\] \[d = \sqrt{64 + 16 + 16} = \sqrt{96}\] \[d = 4\sqrt{6}\] The distance between the centers is \(4\sqrt{6}\).

Step 2: Find the radius of the spheres

Since the spheres are tangent and have equal radii (\(r\)), their centers are separated exactly by two radii. Therefore, each radius must be half of the distance between the centers: \[r = \frac{4\sqrt{6}}{2} = 2\sqrt{6}\].Thus, the radius of each sphere is \(2\sqrt{6}\).

Step 3: Write the equations for the spheres

The general equation of a sphere centered at \((x_0, y_0, z_0)\) with radius \(r\) is: \[(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = r^2\]. For the first sphere with center \((-3, 1, 2)\) and radius \(2\sqrt{6}\), substitute into the equation:\[(x+3)^2 + (y-1)^2 + (z-2)^2 = (2\sqrt{6})^2 = 24\]. For the second sphere with center \((5, -3, 6)\) and the same radius, substitute into the equation: \[(x-5)^2 + (y+3)^2 + (z-6)^2 = 24\].

Key Concepts

Understanding the Distance Formula

To find the equation of tangent spheres, we first need to calculate the distance between their centers. The distance formula is crucial here. It's a way to determine the length between two points in space, especially useful in 3D geometry. The formula is: \[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\] where \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) are the coordinates of the two points. To use it:

• Subtract the x-coordinates to find the difference in the x-direction,
• Do the same for the y and z coordinates,
• Square each of these differences, add them together, then take the square root of the sum.

In our example, the centers are (-3, 1, 2) and (5, -3, 6), giving us a distance of \(4\sqrt{6}\) after calculations.

Deciphering the Sphere Equation

The equation of a sphere helps us describe its size and position in space. If a sphere is centered at a point \((x_0, y_0, z_0)\) with a radius \(r\), its equation is: \[(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = r^2\] This general formula represents all points \((x, y, z)\) that are exactly \(r\) units away from the center. For example, if the center is at (-3, 1, 2), then the equation becomes: \((x+3)^2 + (y-1)^2 + (z-2)^2 = r^2\). Spheres are symmetric, so in this form, it showcases that symmetry, pointing out a perfect roundness. For example:

• Subtract center coordinates from \(x, y, z\),
• Square these expressions, add them, and compare with the square of the radius.

This establishes the relationship between center, boundary, and points in 3D space.

Steps for Radius Calculation

To determine the radius of tangent spheres, understanding their spatial relationship is vital. Since tangent spheres just touch each other, the distance between their centers equals twice their radius. Therefore, if two spheres have equal radii and are tangent, their centers are separated by \(2r\). From our exercise, the distance between centers was \(4\sqrt{6}\). Therefore: \[r = \frac{4\sqrt{6}}{2} = 2\sqrt{6}\] This calculation gives each sphere a radius of \(2\sqrt{6}\). Here's how it connects:

• Take total distance between centers, since spheres are tangent, split it equally.
• Twice the radius matches this distance, simplifying the sphere construction around each center.

It's a crucial piece, blending geometry with algebra to build a precise 3D model.