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A family of curves called hyperbolas (discussed in Section 10.4 ) has the parametric equations \(x=a\) tan \(t\) \(y=b \sec t,\) for \(-\pi

Short Answer

Expert verified
For the given parametric equations \(x = \tan t\) and \(y = \sec t\), we plotted points for selected values of \(t\) in the range \(-\pi < t < \pi\), avoiding \(t = \pm \pi/2\). We then connected the plotted points in the order they appear as \(t\) increases, showing the direction in which the hyperbola is generated. Finally, the connected points form the hyperbola curve with \(a = b = 1\).

Step by step solution

01

Write down the parametric equations

Given the parametric equations for hyperbola are \(x = a \tan t\) and \(y = b \sec t\), we need to plug in the values of \(a\) and \(b\) as 1. So we have: \(x = \tan t\) \(y = \sec t\)
02

Create a table of values containing \(t\), \(x\), and \(y\)

Let's generate some values of \(x\) and \(y\) for various values of \(t\) in the range \(-\pi < t < \pi\). Remember that \(|t| \neq \pi/2\), so we will exclude \(t = \pm \pi/2\) from the table. Choose, for example, \(t = -\pi + 0.1, -\pi/2 + 0.1, \pi/2 - 0.1, \pi - 0.1\): | \(t\) | \(x\) | \(y\) | |--------------|--------------|-------------| | \(-\pi + 0.1\) | \(\tan(\theta - \pi + 0.1)\) | \(\sec(\theta - \pi + 0.1)\) | | \(-\pi/2 + 0.1\) | \(\tan(\theta - \pi/2 + 0.1)\) | \(\sec(\theta - \pi/2 + 0.1)\) | | \(\pi/2 - 0.1\) | \(\tan(\theta + \pi/2 - 0.1)\) | \(\sec(\theta + \pi/2 - 0.1)\) | | \(\pi - 0.1\) | \(\tan(\theta + \pi - 0.1)\) | \(\sec(\theta + \pi - 0.1)\) | Calculate these values to obtain the points to plot.
03

Plot the points on the graph

Next, plot each point (\(x\), \(y\)) on a coordinate plane: 1. Point 1: \((\tan(\theta - \pi + 0.1), \sec(\theta - \pi + 0.1))\) 2. Point 2: \((\tan(\theta - \pi/2 + 0.1), \sec(\theta - \pi/2 + 0.1))\) 3. Point 3: \((\tan(\theta + \pi/2 - 0.1), \sec(\theta + \pi/2 - 0.1))\) 4. Point 4: \((\tan(\theta + \pi - 0.1), \sec(\theta + \pi - 0.1))\)
04

Indicate the direction of the curve

As \(t\) increases from \(t = -\pi\) to \(t = \pi\), draw arrows on the graph to visualize the direction in which the hyperbola is generated. You will see that the curve follows the order of the points: Point 1, Point 2, Point 3, and Point 4.
05

Connect the plotted points to form the curve

Now that we have plotted the points and indicated the direction in which the curve is generated, we can connect them to create the hyperbola curve. Note that the curve will not cross the asymptotes. The graph now shows the hyperbola with \(a = b = 1\).

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Key Concepts

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

Graphing Hyperbolas
Understanding hyperbolas begins with their distinctive shapes and the rules for graphing them. A hyperbola consists of two branches, which are mirror images, separated by a pair of lines called asymptotes.

When graphing hyperbolas with the equation given by the parametric equations \(x = a \tan t\) and \(y = b \sec t\), the first step is to identify the range for the parameter \(t\). Since the tangent and secant functions have undefined values (and hence the curve has asymptotes), we exclude \(t = \pm \frac{\pi}{2}\) from the range.

The next step involves plotting several points on the hyperbola by choosing values of \(t\) and calculating the corresponding \(x\) and \(y\) coordinates using the given equations. These points can then be plotted on the coordinate plane, and by connecting them, while respecting the asymptotes, the shape of the hyperbola emerges. It's also important to indicate the direction in which the curve is traced as the parameter \(t\) increases, often marked with arrows on the graph.
Trigonometric Identities
In the context of parametric equations, trigonometric identities are a set of equalities involving trigonometric functions that can simplify calculations and help in graphing curves. For instance, they can be used to transform the expressions of \(\tan t\) and \(\sec t\) or find relationships between different trigonometric functions based on the angle \(t\).

Some commonly used identities include \(\sec^2 t - \tan^2 t = 1\) and \(\tan(\frac{\pi}{2} \pm t) = \mp \cot t\). These identities are intrinsic to understanding the behavior of the hyperbola's parametric equations around the asymptotes. By applying these identities, one can avoid indeterminate forms that occur at the asymptotes and analyze the hyperbola's behavior at points close to \(t = \pm \frac{\pi}{2}\).
Parametric Equations
Parametric equations represent a set of equations where the system's variables are expressed as continuous functions of one or more independent parameters. In the case of a hyperbola, parametric equations allow us to express \(x\) and \(y\) coordinates in terms of a single parameter \(t\), which can be visually interpreted as time.

When graphing, we choose values for \(t\) and derive the corresponding \(x\) and \(y\) to generate points that outline the curve. Parametric equations are particularly useful for curves like hyperbolas, where the relationship between \(x\) and \(y\) is not a simple function and may include multiple values of \(y\) for a single \(x\) or vice versa. By using a parameter, we untangle this complexity and create a smooth representation of the curve.
Calculus
Calculus plays a central role when dealing with curves defined by parametric equations, especially when discussing properties such as tangents, areas, arc lengths, and curvature.

By using calculus, we can differentiate the parametric equations to find the slopes of the tangents at various points, thus understanding the steepness and direction of the curve. Additionally, we can integrate to find the area enclosed by a hyperbola or determine the length of a segment of the curve.

While the exercise provided doesn't delve into the full depths of calculus applications, understanding the fundamentals of differentiation and integration are vital for mastering the study of curves through parametric equations.

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Most popular questions from this chapter

Find an equation of the line tangent to the following curves at the given point. $$y^{2}-\frac{x^{2}}{64}=1 ;\left(6,-\frac{5}{4}\right)$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The hyperbola \(x^{2} / 4-y^{2} / 9=1\) has no \(y\) -intercepts. b. On every ellipse, there are exactly two points at which the curve has slope \(s,\) where \(s\) is any real number. c. Given the directrices and foci of a standard hyperbola, it is possible to find its vertices, eccentricity, and asymptotes. d. The point on a parabola closest to the focus is the vertex.

Consider the following Lissajous curves. Graph the curve and estimate the coordinates of the points on the curve at which there is (a) a horizontal tangent line and (b) a vertical tangent line. (See the Guided Project Parametric art for more on Lissajous curves.) $$\begin{aligned}&x=\sin 4 t, y=\sin 3 t\\\&0 \leq t \leq 2 \pi\end{aligned}$$

A plane traveling horizontally at \(100 \mathrm{m} / \mathrm{s}\) over flat ground at an elevation of \(4000 \mathrm{m}\) must drop an emergency packet on a target on the ground. The trajectory of the packet is given by $$x=100 t, \quad y=-4.9 t^{2}+4000, \quad \text { for } t \geq 0,$$ where the origin is the point on the ground directly beneath the plane at the moment of the release. How many horizontal meters before the target should the packet be released in order to hit the target?

The region bounded by the parabola \(y=a x^{2}\) and the horizontal line \(y=h\) is revolved about the \(y\) -axis to generate a solid bounded by a surface called a paraboloid (where \(a>0\) and \(h>0\) ). Show that the volume of the solid is \(\frac{3}{2}\) the volume of the cone with the same base and vertex.

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