Problem 51 - Calculus Early Transcendentals

Question

Consider the following equations of quadric surfaces. a. Find the intercepts with the three coordinate axes, when they exist. b. Find the equations of the \(x y-, x z^{-}\), and \(y z\) -traces, when they exist. c. Sketch a graph of the surface. $$x=y^{2}+z^{2}$$

Step-by-Step Solution

Answer

Based on the given quadric surface equation \(x = y^2 + z^2\), we found that the surface only has an intercept at the origin (0,0,0), the \(xy\)-trace has the equation \(x=y^2\), and the \(xz\)-trace has the equation \(x=z^2\). There is no \(yz\)-trace. The sketch of the surface would be a paraboloid opening toward the positive \(x\)-axis, with its vertex at the origin and expanding outward along the positive \(x\)-axis.

Step 1: Find the intercepts with the coordinate axes

To find the intercepts with the coordinate axes, we need to set two of the coordinates equal to zero and solve for the third one. To find the \(x\), \(y\), and \(z\) intercepts, set \((y,z)\), \((x,z)\), and \((x,y)\) equal to zero, respectively, and solve for the remaining variable. \(x\)-intercept: $$x=y^{2}+z^{2}$$ If \(y=0\) and \(z=0\), then \(x=0\). Thus, the \(x\)-intercept is \((0,0,0)\). \(y\)-intercepts: \(x=y^{2}+z^{2}\) If \(x=0\) and \(z=0\), then \(y^2=0\) and \(y=0\). Thus, there is only one \(y\)-intercept, and it is \((0,0,0)\). \(z\)-intercepts: \(x=y^{2}+z^{2}\) If \(x=0\) and \(y=0\), then \(z^2=0\) and \(z=0\). Thus, there is only one \(z\)-intercept, and it is \((0,0,0)\).

Step 2: Find the equations of the \(xy\), \(xz\), and \(yz\) traces

To find the equations of the \(xy\), \(xz\), and \(yz\) traces, we need to set one of the coordinates equal to zero and solve for the remaining variables. \(xy\)-trace: \(x=y^{2}+z^{2}\) If \(z=0\), then \(x=y^2\). Thus, the equation of the \(xy\)-trace is \(x=y^{2}\). \(xz\)-trace: \(x=y^{2}+z^{2}\) If \(y=0\), then \(x=z^2\). Thus, the equation of the \(xz\)-trace is \(x=z^{2}\). \(yz\)-trace: \(x=y^{2}+z^{2}\) Since neither \(y\) nor \(z\) can be set to zero in this case (because that would change the surface into a curve), there is no \(yz\)-trace.

Step 3: Sketch a graph of the surface

To sketch the graph of the surface, we will use the intercepts and traces found in steps 1 and 2. The surface has a single intercept at the origin \((0,0,0)\). The \(xy\)-trace is given by the equation \(x=y^2\), which describes a parabola in the \(xy\)-plane opening toward the \(x\)-axis. The \(xz\)-trace is given by the equation \(x=z^2\), which describes a parabola in the \(xz\)-plane opening toward the \(x\)-axis. There is no \(yz\)-trace. Considering these traces and the axis intercept, we can see that the surface is a paraboloid opening toward the positive \(x\)-axis. The graph will consist of parabolic curves forming a bowl shape, with its vertex at the origin and expanding outward along the positive \(x\)-axis.

Key Concepts

Coordinate Axes Intercepts

When dealing with quadric surfaces, understanding where they intersect with the coordinate axes is crucial for visualizing their geometry. To find these intercepts, you set two of the three variables to zero and solve for the third.

• The x-intercept is found by setting both \(y\) and \(z\) to zero. For the equation \(x = y^2 + z^2\), substituting \(y = 0\) and \(z = 0\) gives \(x = 0\). This means the x-intercept is at the origin, (0, 0, 0).
• Similarly, the y-intercept is found by setting \(x = 0\) and \(z = 0\). Solving \(x = y^2 + z^2\) under these conditions results in \(y = 0\), which also lands us at the origin, (0, 0, 0).
• For the z-intercept, set \(x = 0\) and \(y = 0\), which again yields \(z = 0\). Thus, the z-intercept is also at (0, 0, 0).

In this example, all intercepts coincide at the origin, indicating this is a central feature of the surface, providing a basis for further exploration.

Sketching Graphs

Sketching the graph of a quadric surface demands a clear understanding of its traces and intercepts. The surface equation \(x = y^2 + z^2\) provides clues for this. Starting with intercepts, as they all lie at the origin (0, 0, 0), the graph's center is easily identified. Traces, which are cross-sections of the surface, help form the 3D shape. In this context:

• The xy-trace is \(x = y^2\), making it a parabola in the xy-plane. It opens toward the positive x-axis.
• The xz-trace \(x = z^2\) is another parabola, seen in the xz-plane, also opening toward the positive x-axis.

To sketch the surface, visualize a paraboloid extending in the positive x-direction. Imagine a "bowl" shape expanding outward from the origin, defined by the parabolic traces. These features guide the sketching process, translating analytical findings into a 3D visual understood at a glance.

Paraboloids

A paraboloid is a 3D surface that has a unique curved "bowl-like" appearance. The general form of a paraboloid can be either elliptical or hyperbolic, and in this case, we are dealing with an elliptical paraboloid.For the surface given by \(x = y^2 + z^2\), the elliptical nature is clear:

• The cross-sections parallel to the xy and xz planes form parabolas (since they are traced by the conditions \(x=y^2\) and \(x=z^2\)).
• The x-axis acts as the "spine" of the paraboloid, with the surface curving around this axis.

This paraboloid opens along the x-axis. Understanding the characteristics of paraboloids can greatly assist in identifying such surfaces. Recognizing its symmetrical, rounded features not only complements the mathematical work but enriches the grasp of three-dimensional geometry.

Traces of Surfaces

Traces are intersections of a quadric surface with coordinate planes. They reveal crucial aspects of the surface's geometric behavior. For the equation \(x = y^2 + z^2\), examine each relevant trace:- The xy-trace occurs when \(z = 0\). This gives \(x = y^2\), a parabola opening along the x-axis. Its vertex is at (0, 0, 0).- The xz-trace happens when \(y = 0\). Here, \(x = z^2\) describes another parabola in the xz-plane.These two traces validate the elliptical paraboloid configuration of the surface. Their parabolic shapes indicate the outward curving nature as the x-values increase.There is no yz-trace because setting \(x = 0\) reduces the expression to a point at the origin, not a meaningful cross-section. Understanding and sketching traces solidify comprehension of how the surface behaves and changes within its spatial environment.