/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Sketch the curve by eliminating ... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch the curve by eliminating the parameter, and indicate the direction of increasing \(t .\) $$ x=\sec t, y=\tan t \quad(\pi \leq t<3 \pi / 2) $$

Short Answer

Expert verified
The curve is the left branch of a hyperbola \( y^2 = x^2 - 1 \), moving left to right as \( t \) increases.

Step by step solution

01

Recall Trigonometric Identities

Recall the identity involving secant and tangent: \( \sec^2 t = 1 + \tan^2 t \). This is derived from the Pythagorean identity \( \cos^2 t + \sin^2 t = 1 \).
02

Express Secant in Terms of Tangent

Given \( x = \sec t \) and \( y = \tan t \), substitute \( x \) for \( \sec t \) in the identity, resulting in \( x^2 = 1 + y^2 \).
03

Solve for the Cartesian Equation

Rearrange the equation \( x^2 = 1 + y^2 \) to \( y^2 = x^2 - 1 \). This equation represents a hyperbola in the Cartesian plane.
04

Identify the Direction of Parameter t

Analyze the range of \( t \), which is \( \pi \leq t < 3\pi/2 \). In this range, both \( \sec t \) (x) and \( \tan t \) (y) move from negative infinity to inflection points where they switch signs as \( t \) increases.
05

Sketch the Curve

Sketch the curve \( y = \pm \sqrt{x^2 - 1} \), indicating the part of the hyperbola where \( \pi \leq t < 3\pi/2 \). Since \( \sec t \) is negative and \( \tan t \) transitions from negative to positive, indicate the direction by arrow signs along the curve from left to right.

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Key Concepts

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

Trigonometric Identities
Trigonometric identities are equations that relate the trigonometric functions such as sine, cosine, tangent, secant, etc., to one another. In this scenario, the identity \( \sec^2 t = 1 + \tan^2 t \) is particularly useful. This identity stems from the Pythagorean identity \( \cos^2 t + \sin^2 t = 1 \), which states that the sum of the squares of sine and cosine equals one. By dividing through by \( \cos^2 t \), we transform this into \( 1 + \tan^2 t = \sec^2 t \). This identity allows us to connect the functions in a way that eliminates the trigonometric parameter. Understanding these relationships helps in converting parametric equations into their Cartesian form, facilitating the sketching of graphs.
Elimination of Parameters
Eliminating a parameter means to find an equation for a curve that does not involve the parameter itself, giving a direct relationship between the variables. In our exercise, we have the parametric equations \( x = \sec t \) and \( y = \tan t \). By using the trigonometric identity \( \sec^2 t = 1 + \tan^2 t \), we can link \( x \) and \( y \) directly. Substitute \( x \) for \( \sec t \) into the identity to get \( x^2 = 1 + y^2 \). Simplifying this equation by solving for \( y^2 \), we find the Cartesian equation \( y^2 = x^2 - 1 \). This is now free of the parameter \( t \) and represents the same curve in the Cartesian coordinate system, allowing us to plot it without reference to \( t \).
Hyperbola
A hyperbola is a type of smooth curve lying in a plane, defined by its geometric properties or by equations that guarantee a particular shape. The standard form of a hyperbola's equation is either \( x^2 - y^2 = 1 \) or \( y^2 - x^2 = 1 \), depending on its orientation. From the elimination of parameters, we derived the relationship \( y^2 = x^2 - 1 \). This equation describes a hyperbola that opens horizontally. The hyperbola has two branches, but the specific range of \( t \) indicates which part of the hyperbola to sketch. The central characteristic of hyperbolas is their asymptotic lines, which they get infinitely close to but never intersect. Understanding this property is crucial for sketching and analyzing hyperbolic graphs.
Direction of Parameter
The direction of the parameter describes how the parametric equation traces the curve as the parameter changes. For \( t \) in the interval \( \pi \leq t < 3\pi/2 \), both \( x = \sec t \) and \( y = \tan t \) change as \( t \) varies. Within this range, \( \sec t \) passes from negative infinity towards zero, while \( \tan t \) moves from negative infinity to positive infinity. The effect is that the curve is traced in a specific direction from left to right. This is crucial for understanding motion or directionality on the graph, as it informs whether the curve is traced forwards or backwards as \( t \) increases. To sketch this properly, you should mark the direction with arrows indicating increasing \( t \). This aspect helps visualize dynamic processes represented by such curves, adding a layer of interpretation beyond merely plotting points.

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Most popular questions from this chapter

If \(f^{\prime}(t)\) and \(g^{\prime}(t)\) are continuous functions, and if no segment of the curve $$ x=f(t), \quad y=g(t) \quad(a \leq t \leq b) $$ is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the \(x\) -axis is $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ and the area of the surface generated by revolving the curve about the \(y\) -axis is $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5.] Use the formulas above in these exercises. The equations \(x=a \phi-a \sin \phi, \quad y=a-a \cos \phi \quad(0 \leq \phi \leq 2 \pi)\) represent one arch of a cycloid. Show that the surface area generated by revolving this curve about the \(x\) -axis is given by \(S=64 \pi a^{2} / 3\)

If \(f^{\prime}(t)\) and \(g^{\prime}(t)\) are continuous functions, and if no segment of the curve $$ x=f(t), \quad y=g(t) \quad(a \leq t \leq b) $$ is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the \(x\) -axis is $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ and the area of the surface generated by revolving the curve about the \(y\) -axis is $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5.] Use the formulas above in these exercises. $$ \begin{aligned} &\text { Find the area of the surface generated by revolving the curve }\\\ &x=\cos ^{2} t, y=\sin ^{2} t(0 \leq t \leq \pi / 2) \text { about the } y \text { -axis. } \end{aligned} $$

The region that is common to the circles \(r=2 \cos \theta\) and \(r=2 \sin \theta\)

(a) What is the area that is enclosed by one petal of the rose \(r=a \cos n \theta\) if \(n\) is an even integer? (b) What is the area that is enclosed by one petal of the rose \(r=a \cos n \theta\) if \(n\) is an odd integer? (c) Use a CAS to show that the total area enclosed by the rose \(r=a \cos n \theta\) is \(\pi a^{2} / 2\) if the number of petals is even. [Hint: See Exercise 78 of Section \(10.2 .]\) (d) Use a CAS to show that the total area enclosed by the rose \(r=a \cos n \theta\) is \(\pi a^{2} / 4\) if the number of petals is odd.

The region inside the rose \(r=2 a \cos 2 \theta\) and outside the circle \(r=a \sqrt{2}\)

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