Problem 6
Question
In Problems, graph the curve traced by the given vector function. \(\mathbf{r}(t)=\cosh t \mathbf{i}+3 \sinh t \mathbf{j}\)
Step-by-Step Solution
Verified Answer
The curve is a hyperbola with equation \( x^2 - \frac{y^2}{9} = 1 \), opening along the x-axis.
1Step 1: 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.
2Step 2: 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 \( \cosh^2 t - \left( \frac{y}{3} \right)^2 = 1 \).
3Step 3: 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.
4Step 4: 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.
Key Concepts
Hyperbolic FunctionsParametric EquationsCartesian CoordinatesHyperbola
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:
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 \( \cosh^2 t - \sinh^2 t = 1 \).
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:
In this exercise involving the vector function \( \mathbf{r}(t) = \cosh t \mathbf{i} + 3 \sinh t \mathbf{j} \), the parametric equations are:
- \( x(t) = \cosh t \)
- \( y(t) = 3 \sinh t \)
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:
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:
- \( x = \cosh t \)
- \( y = 3 \sinh t \)
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:
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.
Other exercises in this chapter
Problem 6
Sketch some of the level curves associated with the given function. $$ f(x, y)=\tan ^{-1}(y-x) $$
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In Problems, find the gradient of the given function at the indicated point. $$ f(x, y)=\sqrt{x^{3} y-y^{4}} ;(3,2) $$
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\( \mathbf{r}(t)\) is the position vector of a moving particle. Graph the curve and the velocity and acceleration vectors at the indicated time. Find the speed
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Show that the given line integral is independent of the path. Evaluate in two ways: (a) Find a potential function \(\phi\) and then use Theorem 9.9.1, and (b) U
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