Problem 6 - Advanced Engineering Mathematics

Question

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

Answer

The curve is a hyperbola with the equation \( 9x^2 - y^2 = 9 \).

Step 1: Interpret the Vector Function

The vector function given is \[ \mathbf{r}(t) = \cosh(t) \mathbf{i} + 3\sinh(t) \mathbf{j} \]where \( \cosh(t) \) is the hyperbolic cosine function and \( \sinh(t) \) is the hyperbolic sine function. The function describes a parametric curve in the plane.

Step 2: Rewrite in Component Form

Express the vector function in terms of its components for clarity. We have:\[x(t) = \cosh(t)\]\[y(t) = 3\sinh(t)\]These are the parameterizations of the curve in terms of \( t \).

Step 3: Express Using Hyperbolic Identities

Use the identity \( \cosh^2(t) - \sinh^2(t) = 1 \) to eliminate \( t \). Start by expressing \( \cosh(t) \) and \( \sinh(t) \) in terms of \( x \) and \( y \):\[ x = \cosh(t) \quad \text{and} \quad y = 3\sinh(t) \] Thus, \( \sinh(t) = \frac{y}{3} \).

Step 4: Apply Identity to Eliminate Parameter

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

Step 5: Simplify the Equation

Rearrange the equation derived to get:\[x^2 - \frac{y^2}{9} = 1\]Multiply through by 9 to clear the fraction:\[9x^2 - y^2 = 9\]This is the equation of a hyperbola.

Step 6: Graph the Equation

The equation \(9x^2 - y^2 = 9\) represents a hyperbola centered at the origin with the transverse axis along the x-axis. The hyperbola opens to the right and left. Use this information to sketch the curve.

Key Concepts

Hyperbolic Functions

Hyperbolic functions, similar to their trigonometric counterparts, are essential in various fields such as calculus, physics, and engineering. The functions

• \( \cosh(t) \), known as the hyperbolic cosine, is defined as \( \cosh(t) = \frac{e^t + e^{-t}}{2} \).
• The hyperbolic sine, \( \sinh(t) \), is \( \sinh(t) = \frac{e^t - e^{-t}}{2} \).

These definitions show that hyperbolic functions possess exponential properties which differentiate them from ordinary trigonometric functions.
Hyperbolic functions are important because they describe the shapes of hanging cables and the distribution of heat in a steady state. Another essential aspect of hyperbolic functions is their appearance in the formulas for curves, such as hyperbolas, particularly due to the identity \( \cosh^2(t) - \sinh^2(t) = 1 \). This identity plays a crucial role when transforming parametric equations into a more familiar Cartesian form.

Parametric Curves

Parametric curves provide a powerful way to represent a variety of shapes and paths through two or three-dimensional space. By defining both the \( x \) and \( y \) coordinates as functions of a third variable, typically \( t \), we gain greater control and flexibility over the curve's behavior.
In the vector function provided, \( x(t) = \cosh(t) \) and \( y(t) = 3\sinh(t) \), the curve is parameterized by \( t \), allowing us to describe an intricate geometric path rather than simply plotting points.
Parametric equations help in analyzing curves that do not have a single \( y \) value for every \( x \) value (or vice versa), which is the case in many complex shapes. A primary benefit of using parametric curves is the ability to handle cases such as loops, cusps, and vertical segments, making them indispensable in the study of motion and paths.

Graphing Hyperbolas

Graphing hyperbolas involves understanding their unique structure and properties. Hyperbolas are often related to functions involving subtraction, as shown in the identity \( x^2 - y^2 = 1 \), which characterizes their distinct form.
In our exercise, the equation \( 9x^2 - y^2 = 9 \), derived from applying hyperbolic function identities, is an equation of a hyperbola. To graph this, you follow these steps:

• Simplify the equation by dividing through by 9, leading to \( x^2 - \frac{y^2}{9} = 1 \).
• Recognize the hyperbola's standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where \( a^2 = 1 \) and \( b^2 = 9 \).
• Identify the center of the hyperbola, which in this case is the origin (0,0).
• Determine the orientation of the hyperbola's transverse axis, which is horizontal, meaning it opens left and right since \( x^2 \) appears first.
• Plot the vertices by calculating the points \((\pm a, 0)\), which here are \((\pm 1, 0)\), and sketch alongside the asymptotes, visualizing the overall shape of the curve.

By following these guidelines, you can effectively sketch and understand the hyperbola represented by the vector function's parametric form.