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Chapter 9: Problem 42

What does the equation \(z=x^{2}\) represent in the \(x z\) -plane? What does it represent in three-space?

Short Answer

Expert verified
The equation \(z=x^{2}\) represents an upwards-opening parabola in the \(xz\)-plane. In three-space, this equation represents a parabolic cylinder.

Step by step solution

01

Plot in the \(xz\)-plane

To understand what \(z=x^{2}\) represents in the \(xz\)-plane, we must first plot this equation. The \(x^{2}\) equation suggests a parabola shape. Indeed, in the \(xz\)-plane, this equation gives rise to a parabola that opens upwards.
02

Interpretation in the \(xz\)-plane

This upwards-opening parabola represents all the possible values of \(x\) and \(z\) satisfying the equation \(z=x^{2}\) in two dimensions.
03

Visualizing in Three-space

When it comes to visualizing this equation in three-space, there's an additional dimension compared to the previous scenario. Here, all the points \((x, y, z)\), where \(y\) can be any real number, and \(z = x^{2}\), will make up the graph of the equation in three dimensions.
04

Interpretation in three-space

The graph in three-space visually represents a surface that is parabolic in the \(xz\)-plane and constant along the \(y\)-axis. This surface is usually referred to as a parabolic cylinder because it's like a parabola extended along a third dimension

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Most popular questions from this chapter

Consider the vectors \(\mathbf{u}=\langle\cos \alpha, \sin \alpha, 0\rangle\) and \(\mathbf{v}=\langle\cos \beta, \sin \beta, 0\rangle\) where \(\alpha>\beta\) Find the dot product of the vectors and use the result to prove the identity \(\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\).

Use vectors to find the point that lies two-thirds of the way from \(P\) to \(Q\). \(P(1,2,5), \quad Q(6,8,2)\)

In Exercises 71 and \(72,\) determine the values of \(c\) that satisfy the equation. Let \(\mathbf{u}=\mathbf{i}+2 \mathbf{j}+\mathbf{3 k}\) and \(\mathbf{v}=\mathbf{2} \mathbf{i}+\mathbf{2} \mathbf{j}-\mathbf{k}\) \(\|c \mathbf{v}\|=5\)

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In Exercises 47 and \(48,\) the vector \(v\) and its initial point are given. Find the terminal point. \(\mathbf{v}=\langle 3,-5,6\rangle\) Initial point: (0,6,2)
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