Problem 47 - 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. $$x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}=1$$

Step-by-Step Solution

Answer

#Conclusion# The given equation represents an ellipsoid centered at the origin with semi-principal axes of lengths 1, 2, and 3 along the x, y, and z axes, respectively. The intercepts are at (±1,0,0), (0,±2,0), and (0,0,±3). The equations of the traces in xy, xz, and yz planes are \(x^2 + \frac{y^2}{4} = 1\), \(x^2 + \frac{z^2}{9} = 1\), and \(\frac{y^2}{4} + \frac{z^2}{9} = 1\) respectively. A graph of the surface consists of an ellipsoid with the specified characteristics, including the plotted intercepts and the drawn ellipses in the plane traces.

Step 1: Part a: Finding the intercepts with the three coordinate axes

To find the intercepts with the x-axis, we will set y and z to 0 and solve for x: \(x^2 + \frac{0^2}{4} + \frac{0^2}{9} = 1\) \(x^2 = 1\) \(x = \pm1\) Thus, the x-intercepts are at \((1,0,0)\) and \((-1,0,0)\). Similarly, for the y-axis intercepts, we set x and z to 0 and solve for y: \(\frac{y^2}{4} = 1\) \(y^2 = 4\) \(y = \pm2\) The y-intercepts are at \((0,2,0)\) and \((0,-2,0)\). Finally, for the z-axis intercepts, we set x and y to 0 and solve for z: \(\frac{z^2}{9} = 1\) \(z^2 = 9\) \(z = \pm3\) The z-intercepts are at \((0,0,3)\) and \((0,0,-3)\).

Step 2: Part b: Finding the equations of the xy, xz, and yz traces

To find the equation of the xy-trace, we set z to 0 and obtain the following equation: \(x^2 + \frac{y^2}{4} = 1\) For the xz-trace, we set y to 0 and obtain the equation: \(x^2 + \frac{z^2}{9} = 1\) And for the yz-trace, we set x to 0 and obtain the equation: \(\frac{y^2}{4} + \frac{z^2}{9} = 1\)

Step 3: Part c: Sketch the graph of the surface

The given equation represents an ellipsoid centered at the origin with semi-principal axes along the x, y, z axes of lengths 1, 2, and 3, respectively. To sketch the graph: 1. Plot the x-, y-, and z-intercepts on a 3-dimensional coordinate system. 2. Draw ellipses in the xy, yz, and xz planes as per the equations obtained in part b. Each ellipse should pass through the corresponding axis intercepts. 3. Connect the ellipses to complete the ellipsoid. By following these steps, the graph of the surface will accurately represent the given equation of the ellipsoid.

Key Concepts

Intercepts

Intercepts help identify where a surface crosses the coordinate axes, giving us a foundational understanding of its shape. In the case of the ellipsoid represented by the equation:\[ x^2 + \frac{y^2}{4} + \frac{z^2}{9} = 1 \] we find the intercepts by setting two variables to zero and solving for the third.- **X-axis intercepts**: Set \( y = 0 \) and \( z = 0 \), leading to \( x^2 = 1 \). The solutions are \( x = \pm1 \), and the intercepts are at \((1,0,0)\) and \((-1,0,0)\).- **Y-axis intercepts**: Set \( x = 0 \) and \( z = 0 \), resulting in \( \frac{y^2}{4} = 1 \). Solving yields \( y = \pm2 \), placing intercepts at \((0,2,0)\) and \((0,-2,0)\).- **Z-axis intercepts**: Set \( x = 0 \) and \( y = 0 \), giving \( \frac{z^2}{9} = 1 \). The solutions are \( z = \pm3 \), with intercepts at \((0,0,3)\) and \((0,0,-3)\).These intercepts provide crucial points that assist in sketching and understanding the shape of the ellipsoid.

Traces

Traces are the cross-sections of a three-dimensional surface intersected by planes parallel to the coordinate planes. They aid in visualizing and understanding the form of a surface in each plane. For the equation:\[ x^2 + \frac{y^2}{4} + \frac{z^2}{9} = 1 \]we set one variable to zero to find each trace:- **XY-trace**: Setting \( z = 0 \), we get \( x^2 + \frac{y^2}{4} = 1 \). This equation represents an ellipse in the xy-plane. - **XZ-trace**: Setting \( y = 0 \), leads to \( x^2 + \frac{z^2}{9} = 1 \). Here, we have an ellipse in the xz-plane. - **YZ-trace**: When \( x = 0 \), the equation becomes \( \frac{y^2}{4} + \frac{z^2}{9} = 1 \), showing another elliptical shape in the yz-plane.These traces give insights into the surface's curvature and dimensions, helping to piece together the full 3D shape.

Ellipsoid

An ellipsoid is a three-dimensional surface that generalizes the shape of a sphere but elongated or compressed along different axes. The general form to describe an ellipsoid is:\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \] In this equation, \(a\), \(b\), and \(c\) denote the semi-principal axes' lengths along the x, y, and z axes respectively. The specific ellipsoid described in our equation:\[ x^2 + \frac{y^2}{4} + \frac{z^2}{9} = 1 \]features semi-principal axes 1, 2, and 3, indicating that it is stretched differently along each axis. This ellipsoid is centered at the origin of the coordinate system. By understanding the parameters \(a = 1\), \(b = 2\), and \(c = 3\), we capture the true essence of this shape's dimensions.

3D Graph Sketching

Sketching a 3D graph of a quadric surface like an ellipsoid can be challenging yet rewarding, as it brings abstract equations to life. To graph the ellipsoid\[ x^2 + \frac{y^2}{4} + \frac{z^2}{9} = 1 \],follow these steps:1. **Plot Intercepts**: Begin by marking the intercepts on the x, y, and z axes: - \((\pm1, 0, 0)\) for the x-axis - \((0, \pm2, 0)\) for the y-axis - \((0, 0, \pm3)\) for the z-axis2. **Draw Traces**: Sketch the ellipses corresponding to the xy, xz, and yz traces: - In the xy-plane, use the equation \( x^2 + \frac{y^2}{4} = 1 \) - In the xz-plane, use \( x^2 + \frac{z^2}{9} = 1 \) - In the yz-plane, follow \( \frac{y^2}{4} + \frac{z^2}{9} = 1 \)3. **Connect Ellipses**: Visually connect these ellipses, maintaining the relativity of their distances and dimensions, to form the complete ellipsoid.This process results in a 3D sketch that reflects the true form and size of the ellipsoid, enhancing understanding of the surface.