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Chapter 9: Problem 6

In Problems, graph the curve traced by the given vector function. \(\mathbf{r}(t)=\cosh t \mathbf{i}+3 \sinh t \mathbf{j}\)

Short Answer

Expert verified
The curve is a hyperbola with equation \( x^2 - \frac{y^2}{9} = 1 \), opening along the x-axis.

Step by step solution

01

Understand the Vector Function

The vector function given is \( \mathbf{r}(t) = \cosh t \mathbf{i} + 3 \sinh t \mathbf{j} \). It describes a curve in the 2D plane with \( x = \cosh t \) and \( y = 3 \sinh t \). The components \( \cosh t \) and \( \sinh t \) are hyperbolic cosine and hyperbolic sine functions, respectively.
02

Relate Hyperbolic Functions to Cartesian Coordinates

Recall that for hyperbolic functions, \( \cosh^2 t - \sinh^2 t = 1 \). Using this, you can express \( x = \cosh t \) and \( y = 3 \sinh t \) in a relationship. Substitute \( \sinh t = \frac{y}{3} \) into the hyperbolic identity \( \cosh^2 t - \left( \frac{y}{3} \right)^2 = 1 \).
03

Express as Cartesian Equation

Substituting \( x = \cosh t \) and \( \sinh t = \frac{y}{3} \) into the identity, we get \( x^2 - \left( \frac{y}{3} \right)^2 = 1 \). Rearrange to the form of the equation of a hyperbola: \( x^2 - \frac{y^2}{9} = 1 \). This represents a hyperbola centered at the origin.
04

Sketch the Graph

The standard form \( x^2 - \frac{y^2}{9} = 1 \) shows it's a hyperbola opening along the x-axis. The vertices are at \( (1, 0) \) and \( (-1, 0) \), and the asymptotes are \( y = \pm 3x \). Sketch these key points and lines to visualize the curve.

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Key Concepts

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

Hyperbolic Functions
Hyperbolic functions are analogs to the trigonometric functions but for a hyperbola rather than a circle. They include the hyperbolic sine and cosine, denoted as \( \sinh t \) and \( \cosh t \) respectively. These functions are essential in various areas of mathematics, particularly in solving problems related to hyperbolic shapes and areas where exponential functions naturally appear.
A few key points to remember about hyperbolic functions:
  • \( \sinh t = \frac{e^t - e^{-t}}{2} \)
  • \( \cosh t = \frac{e^t + e^{-t}}{2} \)
  • They satisfy the identity \( \cosh^2 t - \sinh^2 t = 1 \).
Understanding these functions helps in rewriting expressions that involve exponential terms. In vector calculus, they can neatly describe curves like hyperbolas in a more natural and straightforward way.
Parametric Equations
Parametric equations express the coordinates of a set of points as functions of a variable, often denoted as \( t \), which is called a parameter. This form is useful for describing curves and surfaces that cannot be easily expressed using standard equations.
In this exercise involving the vector function \( \mathbf{r}(t) = \cosh t \mathbf{i} + 3 \sinh t \mathbf{j} \), the parametric equations are:
  • \( x(t) = \cosh t \)
  • \( y(t) = 3 \sinh t \)
These equations allow us to trace the path of a point in a plane as the parameter \( t \) changes. Parametric equations are particularly useful in describing complex shapes and motions in two dimensions.
Cartesian Coordinates
Cartesian coordinates use two perpendicular axes, commonly labeled \( x \) and \( y \), to uniquely determine the position of a point in a 2D plane. This system provides a straightforward way of representing geometric figures and is universally used in mathematics.
Converting parametric equations to Cartesian form means eliminating the parameter \( t \). In our example, using the identity \( \cosh^2 t - \sinh^2 t = 1 \) aids in this conversion.
From our parametric form:
  • \( x = \cosh t \)
  • \( y = 3 \sinh t \)
Eliminating \( t \) gives \( x^2 - \frac{y^2}{9} = 1 \), a classic Cartesian equation of a hyperbola. This conversion allows for easier graphing and analysis of the curve.
Hyperbola
A hyperbola is a type of conic section formed when a plane slice intersects both nappes of a double cone. Unlike ellipses and circles, hyperbolas are open curves with two separate branches.
In standard Cartesian form, a hyperbola can be expressed as:
\[ x^2 - \frac{y^2}{b^2} = 1 \] (for hyperbolas opening along the x-axis) or \[ \frac{y^2}{a^2} - x^2 = 1 \] (for hyperbolas opening along the y-axis).
Key characteristics include:
  • The center, vertices, and foci.
  • Symmetrical characteristics about the center.
  • Asymptotes, which are lines that the hyperbola approaches but never meets.
In our example, the equation \( x^2 - \frac{y^2}{9} = 1 \) identifies a hyperbola centered at the origin. This formula helps in sketching the curve and understanding its geometrical properties such as its asymptotes and directional orientation.

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Most popular questions from this chapter

In Problems \(75-78\), find the volume of the solid that is bounded by the graphs of the given equations. $$ z^{2}=3 x^{2}+3 y^{2}, x=0, \quad y=0, z=2, \text { first octant } $$

In Problems, evaluate the given iterated integral by changing to polar coordinates. $$ \int_{0}^{\sqrt{2} / 2} \int_{0}^{\sqrt{1-y^{2}}} \frac{y^{2}}{\sqrt{x^{2}+y^{2}}} d x d y $$

Evaluate the double integral \(\iint\left(\frac{x^{2}}{25}+\frac{y^{2}}{9}\right) d A\), where \(R\) is the elliptical region whose boundary is the graph of \(x^{2} / 25+y^{2} / 9=1\). Use the substitutions \(u=x / 5, v=y / 3\), and polar coordinates.

In Problems, find the indicated moment of inertia of the lamina that has the given shape and density. $$ \begin{aligned} &\text { Outside } r=1 \text { and inside } r=2 \sin 2 \theta, \text { first quadrant; }\\\ &\rho(r, \theta)=\sec ^{2} \theta ; I_{y} \end{aligned} $$

In Problems \(43-46\), convert the given equation to cylindrical coardinates. $$ x^{2}+y^{2}+z^{2}=25 $$
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