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Chapter 10: Problem 32

Sketch the appropriate traces, and then sketch and identify the surface. $$x^{2}-y^{2}+9 z^{2}=9$$

Short Answer

Expert verified
The surface represented by the equation \(x^{2}-y^{2}+9 z^{2}=9\) is a hyperboloid of two sheets with radii 3 along the x and y axes, and a distance of 1 between sheets along the z axis.

Step by step solution

01

Identify the values of a, b, c

Based on the standard formula of the hyperboloid of two sheets, we can see that each square term is divided by a constant. Here, the equation is \(x^{2}-y^{2}+9 z^{2}=9\). First, factor out the right side to one by dividing the entire equation by 9, we then get \((x^{2}/9)-(y^{2}/9)+(z^{2}/1) = 1\). From this, it can be observed that \(a = b = 3\) and \(c = 1\). This tells us the radius of the x and y traces, and the distance between sheets along the z axis.
02

Sketch the traces

The traces are the intersections of the surface with planes parallel to the coordinate planes. From the equation, we can see that the x- and y-traces are hyperbolas, because the z-term disappears when either x or y is equal to 0, and we get the formula for a hyperbola. The z-trace, however, seen when either x or y is 0, is an ellipse, as both x and y terms disappear and we get the formula for an ellipse. Draw these intersections on a graph to have a basic understanding of the shape of the surface.
03

Sketch the surface

The final step is to connect the traces together, essentially forming a three-dimensional figure. The 'sheets' of the hyperboloid span between the points of the z-trace ellipse, and the shape moves inwards towards the z-axis in regions where the x and y traces are. This forms the distinctive 'double cone' shape of a hyperboloid of two sheets, centered along the z-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

3D Surface Sketching
To understand 3D surface sketching, imagine visually plotting a given equation in three-dimensional space. It's like drawing a figure in a 3D coordinate system.
Each axis, namely x, y, and z, represents different dimensions of this space.
  • Think of sketching as connecting multiple two-dimensional graphs (or traces) to form a cohesive 3D model.
  • In the context of our hyperboloid of two sheets, you need to visualize how the different traces align along the z-axis to form the full surface.
Envision the hyperboloid as a surface expanding and contracting as per these 2D graphs' outlines. This not only helps in a successful visualization but is also crucial in solving 3D geometry problems, bringing equations to life.
Coordinate Geometry Traces
In coordinate geometry, traces are cross-sections of 3D surfaces along the coordinate planes. They are crucial to comprehending the full spatial structure of a surface.
With the equation \(x^{2}-y^{2}+9z^{2}=9\), traces aid in observing how the surface interacts with horizontal, vertical, or any other flat planes.
By individually setting x, y, or z to zero, these traces can be discerned graphically as follows:
  • Set \ x = 0 \: Observe the interaction of the surface with the y-z plane.
  • Set \ y = 0 \: Examine the x-z plane's cross-section.
  • Set \ z = 0 \: Identify how the surface intersects the xy-plane.
Using these traces provides a strategic method to decompose complex 3D surfaces into manageable two-dimensional pieces. This is a foundational skill in exploring more advanced geometric shapes.
Equation Transformation
Equation transformation involves modifying an equation to a more workable form, often revealing critical properties.
In our context, transforming the given equation \(x^{2}-y^{2}+9z^{2}=9\) by dividing the entire equation by 9 normalizes it, showing us the true nature of the surface.
Through the transformation: \[ \frac{x^{2}}{9} - \frac{y^{2}}{9} + \frac{z^{2}}{1} = 1 \] We achieve:
  • Unified Scaling: Reveals standardized coordinate relations for simplifying trace identification.
  • Immediate recognition: Instantly observes the hyperbolic properties explicitly.
Without such transformations, interpreting the fundamental structure or essential traits like centers and axes becomes cumbersome, making it hard to comprehend the geometry involved.
Hyperbola and Ellipse Traces
Given our equation \(x^{2}-y^{2}+9z^{2}=9\), we find a combination of hyperbolic and elliptic traces.
Understanding these traces builds upon prior knowledge of conic sections:
  • When either x or y variable is set to zero, the equations for hyperbolas, \(\frac{x^{2}}{9}-\frac{y^{2}}{9}=1\) and \(\frac{-y^{2}}{9}+\frac{x^{2}}{9}=1\) respectively, appear explicitly.
  • For the z-axis, setting x and y simultaneously zero yields ellipses due to \((z^{2}=1)\) reflecting elliptical behavior.
Hyperbolas and ellipses complement each other, guiding the surface depiction. This provides deeper insights into configuration and layout of the hyperboloid of two sheets, particularly emphasizing its double cone-like expansions around the z-axis. Leveraging these traces ensures accuracy in predicting the overall three-dimensional structure.

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Most popular questions from this chapter

If \(\quad a<0 \quad\) and \(\quad x=a \cosh s, y=b \sinh s \cos t \quad\) and \(z=c \sinh s \sin t,\) show that \((x, y, z)\) lies on the left half of the hyperboloid of two sheets \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1.\)

The racetrack at Bristol, Tennessee, is famous for being short with the steepest banked curves on the NASCAR circuit. The track is an oval of length 0.533 mile and the corners are banked at \(36^{\circ} .\) Circular motion at a constant speed \(v\) requires a centripetal force of \(F=\frac{m v^{2}}{r},\) where \(r\) is the radius of the circle and \(m\) is the mass of the car. For a track banked at angle \(A\) the weight of the car provides a centripetal force of \(m g\) sin \(A\) where \(g\) is the gravitational constant. Setting the two equal gives \(\frac{v^{2}}{r}=g \sin A .\) Assuming that the Bristol track is circular (it's not really) and using \(g=32 \mathrm{ft} / \mathrm{s}^{2},\) find the speed supported by the Bristol bank. Cars actually complete laps at over 120 mph. Discuss where the additional force for this higher speed might come from.

Suppose that (2,1,3) is a normal vector for a plane containing the point \((2,-3,4) .\) Show that an equation of the plane is \(2 x+y+3 z=13 .\) Explain why another normal vector for this plane is \(\langle-4,-2,-6\rangle .\) Use this normal vector to find an equation of the plane and show that the equation reduces to the same equation, \(2 x+y+3 z=13\)

Find an equation of the given plane. The plane containing the points (1,-2,1),(2,-1,0) and (3,-2,2)

Sketch the given plane. $$2 x-y+4 z=4$$
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