Problem 33 - Thomas Calculus

Question

In Exercises \(25-34\) , describe the given set with a single equation or with a pair of equations. The circle in which the plane through the point \((1,1,3)\) perpendicular to the \(z\) -axis meets the sphere of radius 5 centered at the origin

Step-by-Step Solution

Answer

The intersection is described by the equation \( x^2 + y^2 = 16 \).

Step 1: Understand the Given Problem

The problem describes a plane and a sphere. The sphere has a radius of 5 and is centered at the origin (0,0,0). You'll need to find the intersection of this sphere with the plane perpendicular to the z-axis and passing through the given point (1,1,3). This intersection will form a circle.

Step 2: Equation of the Sphere

The equation for a sphere centered at the origin with radius 5 is given by: \( x^2 + y^2 + z^2 = 25 \). This comes from the general formula for a sphere \((x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\) with center \((h,k,l)\) and radius \(r\).

Step 3: Equation of the Plane

The plane is perpendicular to the z-axis, which means it is parallel to the xy-plane. The equation of such a plane passing through a point (1,1,3) is simply \( z = 3 \).

Step 4: Find the Intersection

Replace \( z \) with 3 in the equation of the sphere \(x^2 + y^2 + z^2 = 25\). This gives the equation \( x^2 + y^2 + 3^2 = 25 \), which simplifies to: \( x^2 + y^2 = 16 \).

Step 5: Conclusion

The intersection of the sphere with the plane is described by the equation \( x^2 + y^2 = 16 \). This equation represents a circle in the xy-plane with a radius of 4 (since \(r^2 = 16\), so \(r = 4\)).

Key Concepts

Sphere Equation

A sphere in three-dimensional space is a perfectly symmetrical shape where every point on its surface is an equal distance from its center. This equal distance is called the radius. To describe a sphere mathematically, we use its general equation:
\[(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2\]

• \((h, k, l)\) are the coordinates of the center of the sphere.
• \(r\) is the radius of the sphere.

For the sphere centered at the origin \((0, 0, 0)\) with a radius of 5, the equation simplifies to:
\[x^2 + y^2 + z^2 = 25\]Here, the radius squared \(r^2\) is 25, meaning that the sphere extends 5 units from its center in all directions. This basic formula helps us visualize how a 3-dimensional object behaves when interacting with other geometric forms.

Plane Equation

A plane in three-dimensional space can be imagined as a flat, two-dimensional surface that stretches out infinitely. When a plane is perpendicular to one of the coordinate axes, its equation is much simpler. For a plane perpendicular to the \(z\)-axis, the equation is:
\[z = c\]where \(c\) is a constant. This means that the plane is parallel to the \(xy\)-plane and all points have the same \(z\)-coordinate.

• The equation \(z = 3\) implies that the plane passes through all points having a \(z\)-coordinate of 3.

This indicates that no matter what values \(x\) and \(y\) take, as long as \(z\) is 3, the point lies on this plane. This concept is pivotal because it shows how planes can intersect with other geometrical shapes like spheres.

Intersection of Surfaces

The intersection of surfaces in analytic geometry is when two or more geometric figures meet. In the context of this exercise, a plane intersects a sphere, creating a circle.

• To find this intersection, substitute the plane's equation into the sphere's equation.

For our sphere and plane:

• Start with the sphere: \(x^2 + y^2 + z^2 = 25\).
• Use the plane: \(z = 3\).
• Substituting \(z = 3\) gives \(x^2 + y^2 + 3^2 = 25\).

Simplifying this leads to:
\[x^2 + y^2 = 16\]This equation represents a circle with a radius of 4 in the \(xy\)-plane. This intersection proves how complex 3D figures can be analyzed using simple algebraic combinations to identify new shapes formed by their overlap.