Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Describe with a sketch the sets of points \((x, y, z)\) satisfying the following equations. $$x^{2} y^{2} z^{2}>0$$

Short Answer

Expert verified
Question: Sketch the sets of points (x, y, z) that satisfy the equation x²y²z² > 0. Answer: The sets of points that satisfy the equation x²y²z² > 0 can be represented by the union of four non-adjacent octants: the first octant (positive x, y, z axes), the octant in the negative y and negative z plane but positive x-plane, the octant in the negative x and negative z plane but positive y-plane, and the octant in the negative x and negative y plane but positive z-plane.

Step by step solution

01

Analyze the equation

Recall the inequality \(x^{2} y^{2} z^{2}>0\). We can break it down by the fact that \(x^{2}\), \(y^{2}\), and \(z^{2}\) are all non-negative. - If either \(x\), \(y\), or \(z\) is equal to 0, then their square will be equal to 0 and the product \(x^{2} y^{2} z^{2}\) will also be 0. - If all three variables are positive then their squares will be positive and the product will be positive. - If all three variables are negative then their squares will be positive and the product will be positive. - We can also have two variables being negative, then their squares will be positive and the product of the three squares will still be positive. This inequality can be satisfied if all of the variables are positive or two of them are negative. In other words, we need to show the sets of points (\(x\),\(y\),\(z\)) where x*y*z > 0.
02

Define regions based on sign changes

From the analysis above we can now create a table with regions where the inequality \(x^{2} y^{2} z^{2}>0\) holds true: 1. x > 0, y > 0, z > 0 2. x > 0, y < 0, z < 0 3. x < 0, y > 0, z < 0 4. x < 0, y < 0, z > 0
03

Sketch the graph

To create the 3D graph, we want to illustrate the four regions in which the inequality holds: 1. x > 0, y > 0, z > 0: The first octant (positive x,y,z axes) 2. x > 0, y < 0, z < 0: The octant in the negative y and negative z plane but positive x-plane. 3. x < 0, y > 0, z < 0: The octant in the negative x and negative z plane but positive y-plane. 4. x < 0, y < 0, z > 0: The octant in the negative x and negative y plane but positive z-plane. Thus, the graph representing the sets of points where \(x^{2} y^{2} z^{2}>0\) is formed by combining these four non-adjacent octants.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Imagine three unit spheres (radius equal to 1 ) with centers at \(O(0,0,0), P(\sqrt{3},-1,0)\) and \(Q(\sqrt{3}, 1,0) .\) Now place another unit sphere symmetrically on top of these spheres with its center at \(R\) (see figure). a. Find the coordinates of \(R\). (Hint: The distance between the centers of any two spheres is 2.) b. Let \(\mathbf{r}_{i j}\) be the vector from the center of sphere \(i\) to the center of sphere \(j .\) Find \(\mathbf{r}_{O P}, \mathbf{r}_{O Q}, \mathbf{r}_{P Q}, \mathbf{r}_{O R},\) and \(\mathbf{r}_{P R}\).

Determine whether the following statements are true and give an explanation or counterexample. a. The line \(\mathbf{r}(t)=\langle 3,-1,4\rangle+t\langle 6,-2,8\rangle\) passes through the origin. b. Any two nonparallel lines in \(\mathbb{R}^{3}\) intersect. c. The curve \(\mathbf{r}(t)=\left\langle e^{-t}, \sin t,-\cos t\right\rangle\) approaches a circle as \(t \rightarrow \infty\). d. If \(\mathbf{r}(t)=e^{-t^{2}}\langle 1,1,1\rangle\) then \(\lim _{t \rightarrow \infty} \mathbf{r}(t)=\lim _{t \rightarrow-\infty} \mathbf{r}(t)\).

Parabolic trajectory In Example 7 it was shown that for the parabolic trajectory \(\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle, \mathbf{a}=\langle 0,2\rangle\) and \(\mathbf{a}=\frac{2}{\sqrt{1+4 t^{2}}}(\mathbf{N}+2 t \mathbf{T}) .\) Show that the second expression for \(\mathbf{a}\) reduces to the first expression.

For the given points \(P, Q,\) and \(R,\) find the approximate measurements of the angles of \(\triangle P Q R\). $$P(0,-1,3), Q(2,2,1), R(-2,2,4)$$

In contrast to the proof in Exercise \(81,\) we now use coordinates and position vectors to prove the same result. Without loss of generality, let \(P\left(x_{1}, y_{1}, 0\right)\) and \(Q\left(x_{2}, y_{2}, 0\right)\) be two points in the \(x y\) -plane and let \(R\left(x_{3}, y_{3}, z_{3}\right)\) be a third point, such that \(P, Q,\) and \(R\) do not lie on a line. Consider \(\triangle P Q R\). a. Let \(M_{1}\) be the midpoint of the side \(P Q\). Find the coordinates of \(M_{1}\) and the components of the vector \(\overrightarrow{R M}_{1}\) b. Find the vector \(\overrightarrow{O Z}_{1}\) from the origin to the point \(Z_{1}\) two-thirds of the way along \(\overrightarrow{R M}_{1}\). c. Repeat the calculation of part (b) with the midpoint \(M_{2}\) of \(R Q\) and the vector \(\overrightarrow{P M}_{2}\) to obtain the vector \(\overrightarrow{O Z}_{2}\) d. Repeat the calculation of part (b) with the midpoint \(M_{3}\) of \(P R\) and the vector \(\overline{Q M}_{3}\) to obtain the vector \(\overrightarrow{O Z}_{3}\) e. Conclude that the medians of \(\triangle P Q R\) intersect at a point. Give the coordinates of the point. f. With \(P(2,4,0), Q(4,1,0),\) and \(R(6,3,4),\) find the point at which the medians of \(\triangle P Q R\) intersect.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free