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Chapter 17: Problem 15

Draw a sketch of the graph of the given vector equation and find a cartesian equation of the graph. $$ \mathbf{R}(t)=3 \cosh t \mathbf{i}+5 \sinh t \mathbf{j} $$

Short Answer

Expert verified
The Cartesian equation is \[ \frac{x^2}{9} - \frac{y^2}{25} = 1. \ The graph is a hyperbola.

Step by step solution

01

- Understand the vector equation

The given vector equation is \[ \mathbf{R}(t)=3 \cosh t \mathbf{i}+5 \sinh t \mathbf{j} \. \] Here, \[ \cosh t \] and \[ \sinh t \] represent hyperbolic cosine and hyperbolic sine functions, respectively. These functions describe the x and y components of the vector as functions of the parameter t.
02

- Express the components separately

Write the components of the vector equation separately: \[ x = 3 \cosh t \] and \[ y = 5 \sinh t. \]
03

- Use hyperbolic function identities

Recall the identity for hyperbolic functions: \[ \cosh^2 t - \sinh^2 t = 1. \]
04

- Substitute hyperbolic functions' expressions

Express \[ \cosh t \] and \[ \sinh t \] in terms of x and y: \[ \cosh t = \frac{x}{3} \] and \[ \sinh t = \frac{y}{5}. \]
05

- Form the Cartesian equation

Substitute \[ \cosh t = \frac{x}{3} \] and \[ \sinh t = \frac{y}{5} \] into the hyperbolic identity: \[ \left( \frac{x}{3} \right)^2 - \left( \frac{y}{5} \right)^2 = 1. \]
06

- Simplify the Cartesian equation

Simplify the equation: \[ \frac{x^2}{9} - \frac{y^2}{25} = 1. \] This is the Cartesian equation of the given vector equation.
07

- Sketch the graph

The Cartesian equation \[ \frac{x^2}{9} - \frac{y^2}{25} = 1 \] represents a hyperbola. Sketch the graph by identifying the vertices and asymptotes of the hyperbola.

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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 of trigonometric functions but for a hyperbola rather than a circle. They are essential in various areas of calculus and complex analysis. The two primary hyperbolic functions are hyperbolic cosine, \(\textrm{cosh}(t)\), and hyperbolic sine, \(\textrm{sinh}(t)\).

Hyperbolic cosine and sine are defined as follows:
  • \(\textrm{cosh}(t) = \frac{e^t + e^{-t}}{2}\)
  • \(\textrm{sinh}(t) = \frac{e^t - e^{-t}}{2}\)


These functions have several useful identities, one of which is:
\[ \textrm{cosh}^2(t) - \textrm{sinh}^2(t) = 1. \]
This identity is crucial for converting vector equations with hyperbolic functions into Cartesian equations.
Cartesian Equation
A Cartesian equation is a way to describe curves using only the Cartesian coordinates \(x\text{ and }y\). It eliminates any parameter, like \(t\), by expressing the relationship between \(x\text{ and }y\) directly.

In the context of the given vector equation:
\[ \textbf{R}(t) = 3 \textrm{cosh}(t) \textbf{i} + 5 \textrm{sinh}(t) \textbf{j}, \]
the components can be isolated as:
  • \( x = 3 \textrm{cosh}(t) \)
  • \( y = 5 \textrm{sinh}(t) \)


Using the hyperbolic identity \( \textrm{cosh}^2(t) - \textrm{sinh}^2(t) = 1 \), we can express this as:
\[ \frac{x^2}{9} - \frac{y^2}{25} = 1. \]
This is a Cartesian equation that represents a hyperbola.
Hyperbola Graphing
Graphing a hyperbola requires understanding its standard Cartesian form. For the hyperbola given by the equation \[ \frac{x^2}{9} - \frac{y^2}{25} = 1, \] the key elements to identify are its vertices and asymptotes.

In this equation:
  • \textbf{Vertices}: The vertices are located at \( ( \text{卤}\frac{A} \text{,0} ) \), where \( A \) is the square root of the denominator under the \(x^2\) term, in this case, \( A = 3 \). So, the vertices are at \( (\text{卤}3 \text{,0}) \).
  • \textbf{Asymptotes}: The asymptotes are the lines along which the hyperbola approaches infinity. For hyperbolas of this form, the asymptotes are given by:
    \[ y = \text{卤} \frac{B}{A} x, \]
    where \( B \) is the square root of the denominator under the \(y^2\) term, here \( B = 5 \). Thus, the asymptotes are \( y = \text{卤} \frac{5}{3} x. \)


To sketch the hyperbola, first plot the vertices at \( (\text{卤}3 \text{,0}) \), draw the asymptote lines through the origin with slopes \( \text{卤} \frac{5}{3} \), and then sketch the branches of the hyperbola opening along these asymptotes. The hyperbola will extend infinitely along these branches while getting increasingly closer to the asymptotes.

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Most popular questions from this chapter

Find the indicated limit, if it exists. $$ \mathbf{R}(t)=\frac{t^{2}-2 t-3}{t-3} \mathbf{i}+\frac{t^{2}-5 t+6}{t-3} \mathbf{j} ; \lim _{t \rightarrow 3} \mathbf{R}(t) $$

Prove that two nonzero vectors are parallel if and only if the radian measure of the angle between them is 0 or \(\pi\).

A particle is moving along the curve having the given vector equation. In each problem, find the vectors \(\mathbf{V}(t), \mathbf{A}(t), \mathbf{T}(t)\), and \(\mathbf{N}(t)\), and the following scalars for an arbitrary value of \(t:|\mathbf{V}(t)|, A_{T}, A_{N}, K(t) .\) Also find the particular values when \(t=t_{1} .\) At \(t=t_{1}\), draw a sketch of a portion of the curve and representations of the vectors \(\mathbf{V}\left(t_{1}\right)\), \(\mathbf{A}\left(t_{1}\right), A_{T} \mathbf{T}\left(t_{1}\right)\), and \(A_{N} \mathbf{N}\left(t_{1}\right)\). $$ \mathbf{R}(t)=5 \cos 3 t \mathbf{i}+5 \sin 3 t \mathbf{j} ; t_{1}=\frac{1}{3} \pi $$

If \(\mathbf{A}=-8 \mathbf{i}+4 \mathbf{j}\) and \(\mathbf{B}=7 \mathbf{i}-6 \mathbf{j}\), find the vector projection of \(\mathbf{A}\) onto \(\mathbf{B}\).

Find an equation of the tangent line to the curve, \(y=5 \cos \theta, x=2 \sin \theta\), at the point where \(\theta=\frac{1}{3} \pi\).
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