Problem 20 - Algebra and Trigonometry

Question

Describe the trace of the sphere $$ x^{2}+(y-4)^{2}+(z-3)^{2}=144 $$ in (a) the \(x\) -plane and in (b) the plane \(z=-2\)

Step-by-Step Solution

Answer

In the \(x\)-plane, the trace is a circle of radius \(3\sqrt{15}\); in the plane \(z=-2\), it's a circle of radius \(\sqrt{119}\).

Step 1: Understanding the Sphere Equation

The given equation of the sphere is \(x^{2}+(y-4)^{2}+(z-3)^{2}=144\). This represents a sphere centered at \((0, 4, 3)\) with a radius of \(12\) (since \(144 = 12^2\)).

Step 2: Finding the Trace in the x-plane (z=0)

Substitute \(z = 0\) into the sphere equation: \[x^{2}+(y-4)^{2}+(0-3)^{2}=144\] Simplifying, we get: \(x^{2} + (y-4)^{2} + 9 = 144\). This reduces to \(x^{2} + (y-4)^{2} = 135\), representing a circle in the \(xy\)-plane centered at \((0, 4)\) with a radius of \(\sqrt{135}\).

Step 3: Calculating Radius in the x-plane

The radius of this circle is \(\sqrt{135}\). Simplify \(\sqrt{135}\): \(\sqrt{135} = \sqrt{9 \times 15} = 3\sqrt{15}\). Thus, the radius of the circle in the \(x\)-plane is \(3\sqrt{15}\).

Step 4: Finding the Trace in the Plane z=-2

Substitute \(z = -2\) into the sphere equation: \[x^{2}+(y-4)^{2}+(-2-3)^{2}=144\] Simplifying, we find \(x^{2} + (y-4)^{2} + 25 = 144\). This reduces to \(x^{2} + (y-4)^{2} = 119\), representing a circle in the \(xy\)-plane centered at \((0, 4)\) with a radius of \(\sqrt{119}\).

Step 5: Calculating Radius in the Plane z=-2

The radius of this circle is \(\sqrt{119}\). Therefore, the radius of the circle in the plane \(z=-2\) is \(\sqrt{119}\).

Key Concepts

Sphere Equation

The sphere equation is a fundamental concept in 3D geometry, capturing the essence of how a three-dimensional shape is defined mathematically. For a sphere centered at the point \(h, k, l\), with a radius \(r\), the equation takes the form:

• \( (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \)

This equation describes a spherical surface perfectly symmetrical around its center. In the given problem, \(x^2 + (y-4)^2 + (z-3)^2 = 144\), it shows a sphere centered at \( (0, 4, 3) \) with a radius of 12, because \(144 = 12^2\). The coordinates of the center and the radius are crucial in understanding and solving geometry problems involving spheres.

Circle in xy-plane

When analyzing the sphere's trace in the xy-plane (where the z-coordinate is held constant), it results in a circle. To find this trace, we substitute \(z = 0\) into the sphere equation:

• \( x^2 + (y-4)^2 + (0-3)^2 = 144 \)

After simplification, we obtain:

• \( x^2 + (y-4)^2 = 135 \)

This represents a circle in the xy-plane. This circle is centered at \( (0, 4) \) and has a radius of \( \sqrt{135} \), which simplifies to \( 3\sqrt{15} \). Understanding the circle's center and radius helps in visualizing the intersection of the sphere with the xy-plane.

Radius Calculation

Calculating the radius of circles resulting from the intersection of planes with spheres is key in spatial geometry problems. For any cross-section or trace of a sphere, once simplified, we arrive at an equation of the form:

• \( x^2 + y^2 = r^2 \)

where \(r\) is the radius. In our scenario, after substituting various z-values, such as 0 or -2, and simplifying, we need to calculate the radius of the resulting circles:

• For \(z = 0\), the radius is determined from \( x^2 + (y-4)^2 = 135 \), giving a radius of \( 3\sqrt{15} \).
• For \(z = -2\), simplify \( x^2 + (y-4)^2 = 119 \), resulting in a radius of \( \sqrt{119} \).

These calculations illustrate how alterations in plane positions affect sphere traces and their resulting circle sizes.

3D Geometry

Understanding the fundamentals of 3D geometry is essential when dealing with spheres and their traces. In three dimensions, objects have height, width, and depth, allowing for complex shapes and interactions with different planes.

• **Traces and Cross-Sections:** Analyzing traces involves observing how a sphere interacts with various planes, leading to visible cross-sections. These cross-sections can be circles, as seen in the problem.
• **Visualizing in 3D:** Imagining a sphere in a three-dimensional space aids in predicting how it will appear when intersected by different planes, such as the xy-plane or any parallel plane.
• **Practical Importance:** This 3D perspective is instrumental for fields like architecture, engineering, and even video game design, where spatial awareness impacts structure and functionality.

Through understanding the complexities of 3D geometry, one can better appreciate how mathematical equations represent real-world structures and their intersecting planes.