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Chapter 11: Problem 11

Name and sketch the graph of each of the following equations in three-space. $$ y=e^{2 z} $$

Short Answer

Expert verified
The graph is a series of vertical planes, each with an exponential curve \(y = e^{2z}\), extending along the \(x\)-axis.

Step by step solution

01

Understand the Equation

The equation given is in three variables: \(x\), \(y\), and \(z\). It is \(y = e^{2z}\). Here \(y\) is expressed in terms of \(z\), and \(x\) does not appear in the equation, indicating the surface is independent of \(x\).
02

Analyze the Equation

For a fixed \(z\), \(y\) equals \(e^{2z}\). This suggests that in any cross-section parallel to the \(yz\)-plane (i.e., any plane where \(x\) is a constant), the curve will be an exponential growth curve, starting above the \(z\)-axis and increasing as \(z\) increases.
03

Sketch the Graph

Since \(x\) is not constrained by the equation, we can assume that the surface extends along the entire \(x\)-axis. As such, the surface consists of vertical planes stacked infinitely along the \(x\)-axis, each with an exponentially increasing curve \(y = e^{2z}\) in the \(yz\)-plane. The graph is a "sheaf" of exponential growth curves along the \(x\)-axis.

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Key Concepts

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

Exploring Exponential Functions
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In the exercise, we encounter the function \( y = e^{2z} \). Here, the base \( e \) represents Euler's number, approximately 2.718, and \( z \) is the variable exponent. This setup describes how \( y \) changes as \( z \) changes.

This type of function is characterized by rapid increases as the variable in the exponent increases, representing exponential growth. For our particular function, as \( z \) becomes larger, \( y \) grows extremely quickly, given by the factor of \( 2z \) in the exponent. Therefore, every increase in \( z \) results in a significant multiplication in \( y \), creating a steep upward curve in graphs.
  • Exponential functions display continuous growth.
  • They are used to model real-world scenarios such as population growth and radioactive decay.
  • The growth rate is proportional to the value of the function itself, causing the curve to get steeper rapidly.
Visualizing Three-Dimensional Surfaces
The equation \( y = e^{2z} \) helps us visualize a three-dimensional surface, even if it might seem initially challenging. Imagine the surface extending infinitely along one axis—in this case, the \( x \)-axis.

This means that for each value \( x \), the exponential curve in the \( yz \)-plane stays the same. Each cross-section parallel to the \( yz \)-plane reveals a standard exponential curve.

Visualizing a single two-dimensional curve growing steeply across the third dimension forms a surface that looks like multiple sheets or "sheaves" layered along the \( x \)-axis. This visualization can help us grasp more complicated three-dimensional shapes by considering simpler two-dimensional sections.
  • A three-dimensional surface can be thought of as a collection of similar two-dimensional curves.
  • The surface is represented in coordinate space, meaning it extends across different coordinate planes.
  • Surfaces have different appearances based on which cross-section we view.
Understanding Coordinate Planes
Coordinate planes form the foundation of three-dimensional graphing, serving as a reference for positions and structures in space. The exercise primarily involves the \( yz \)-plane, where the equation \( y = e^{2z} \) is considered while \( x \) remains constant.

Coordinate planes are imaginary flat surfaces created by two axes that intersect at a right angle. In three-dimensional space, we often refer to the \( xy \)-plane, \( yz \)-plane, and \( zx \)-plane.

Understanding these helps us to locate and describe points and shapes in space. As seen in the exercise, each coordinate plane offers a different perspective of the graph. The ability to set one variable at a constant (like \( x \) here) simplifies complex equations into a series of manageable two-dimensional slices, offering detailed insights into the behavior of mathematical functions.
  • Coordinate planes are helpful for visualizing components of three-dimensional surfaces.
  • They help break down complicated structures into simpler, easier-to-understand forms.
  • Each plane provides a unique view of the equation's behavior over different ranges.

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Most popular questions from this chapter

EXPL 48. In this exercise you will derive Kepler's First Law, that planets travel in elliptical orbits. We begin with the notation. Place the coordinate system so that the sun is at the origin and the planet's closest approach to the sun (the perihelion) is on the positive \(x\) -axis and occurs at time \(t=0\). Let \(\mathbf{r}(t)\) denote the position vector and let \(r(t)=\|\mathbf{r}(t)\|\) denote the distance from the sun at time \(t\). Also, let \(\theta(t)\) denote the angle that the vector \(\mathbf{r}(t)\) makes with the positive \(x\) -axis at time \(t\). Thus, \((r(t), \theta(t))\) is the polar coordinate representation of the planet's position. Let \(\mathbf{u}_{1}=\mathbf{r} / r=(\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j} \quad\) and \(\quad \mathbf{u}_{2}=(-\sin \theta) \mathbf{i}+(\cos \theta) \mathbf{j}\) Vectors \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are orthogonal unit vectors pointing in the directions of increasing \(r\) and increasing \(\theta\), respectively. Figure 12 summarizes this notation. We will often omit the argument \(t\), but keep in mind that \(\mathbf{r}, \theta, \mathbf{u}_{1}\), and \(\mathbf{u}_{2}\) are all functions of \(t .\) A prime in. dicates differentiation with respect to time \(t\). (a) Show that \(\mathbf{u}_{1}^{\prime}=\theta^{\prime} \mathbf{u}_{2}\) and \(\mathbf{u}_{2}^{\prime}=-\theta^{\prime} \mathbf{u}_{1}\). (b) Show that the velocity and acceleration vectors satisfy $$ \begin{array}{l} \mathbf{v}=r^{\prime} \mathbf{u}_{1}+r \theta^{\prime} \mathbf{u}_{2} \\ \mathbf{a}=\left(r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2}\right) \mathbf{u}_{1}+\left(2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime}\right) \mathbf{u}_{2} \end{array} $$ (c) Use the fact that the only force acting on the planet is the gravity of the sun to express a as a multiple of \(\mathbf{u}_{1}\), then explain how we can conclude that $$ \begin{aligned} r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2} &=\frac{-G M}{r^{2}} \\ 2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime} &=0 \end{aligned} $$ (d) Consider \(\mathbf{r} \times \mathbf{r}^{\prime}\), which we showed in Example 8 was a constant vector, say D. Use the result from (b) to show that \(\mathbf{D}=r^{2} \theta^{\prime} \mathbf{k} .\) (e) Substitute \(t=0\) to get \(\mathbf{D}=r_{0} v_{0} \mathbf{k}\), where \(r_{0}=r(0)\) and \(v_{0}=\|\mathbf{v}(0)\|\). Then argue that \(r^{2} \theta^{\prime}=r_{0} v_{0}\) for all \(t\). (f) Make the substitution \(q=r^{\prime}\) and use the result from (e) to obtain the first-order (nonlinear) differential equation in \(q\) : $$ q \frac{d q}{d r}=\frac{r_{0}^{2} v_{0}^{2}}{r^{3}}-\frac{G M}{r^{2}} $$ (g) Integrate with respect to \(r\) on both sides of the above equation and use an initial condition to obtain $$ q^{2}=2 G M\left(\frac{1}{r}-\frac{1}{r_{0}}\right)+v_{0}^{2}\left(1-\frac{r_{0}^{2}}{r^{2}}\right) $$ (h) Substitute \(p=1 / r\) into the above equation to obtain $$ \frac{r_{0}^{2} v_{0}^{2}}{\left(\theta^{\prime}\right)^{2}}\left(\frac{d p}{d t}\right)^{2}=2 G M\left(p-p_{0}\right)+v_{0}^{2}\left(1-\frac{p^{2}}{p_{0}^{2}}\right) $$

Show that for a plane curve \(\mathbf{N}\) points to the concave side of the curve. Hint: One method is to show that $$ \mathbf{N}=(-\sin \phi \mathbf{i}+\cos \phi \mathbf{j}) \frac{d \phi / d s}{|d \phi / d s|} $$ Then consider the cases \(d \phi / d s>0\) (curve bends to the left) and \(d \phi / d s<0\) (curve bends to the right).

. Find a curve given by a polynominal \(P_{5}(x)\) that provides a smooth transition between two horizontal lines. That is, assume a function of the form \(P_{5}(x)=a_{0}+a_{1} x+a_{2} x^{2}+\) \(a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}\), which provides a smooth transition between \(y=0\) for \(x \leq 0\) and \(y=1\) for \(x \geq 1\) in such a way that the function, its derivative, and curvature are all continuous for all values of \(x\). $$ y=\left\\{\begin{array}{ll} 0 & \text { if } \quad x \leq 0 \\ P_{5}(x) & \text { if } \quad 0

Make the required change in the given equation. \(\rho \sin \phi=1\) to Cartesian coordinates

, find the point of the curve at which the curvature is a maximum. $$ y=\sinh x $$
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