Problem 24 - Calculus 3

Question

Eliminate the parameter \(t\), write the equation in Cartesian coordinates, then sketch the graphs of the vector-valued functions. (Hint: Let \(x=2 t\) and \(y=t^{2} .\) Solve the first equation for \(x\) in terms of \(t\) and substitute this result into the second equation.) \(\mathbf{r}(t)=2(\sinh t) \mathbf{i}+2(\cosh t) \mathbf{j}, t>0\)

Step-by-Step Solution

Answer

The parametric functions describe a hyperbola.

Step 1: Setup and Variable Substitution

Given the vector-valued function \(\mathbf{r}(t)=2(\sinh t) \mathbf{i}+2(\cosh t) \mathbf{j}\), we need to express \(x\) and \(y\) in terms of \(t\). First, recognize that \(x = 2\sinh(t)\) and \(y = 2\cosh(t)\).

Step 2: Express \(t\) in terms of \(x\)

Solve for \(t\) in terms of \(x\). Given \(x = 2\sinh(t)\), we have \(\sinh(t) = \frac{x}{2}\). Using the identity \(\sinh(t) = \frac{e^{t} - e^{-t}}{2}\), solve \(\frac{x}{2} = \frac{e^{t} - e^{-t}}{2}\), leading to \(e^{2t} - 1 = x^{2}\).

Step 3: Express \(y\) in terms of \(t\)

Using the result from the hyperbolic identity, \(\cosh^2(t) - \sinh^2(t) = 1\), substitute \(\sinh(t) = \frac{x}{2}\) into \(y = 2\cosh(t)\). Since \(\cosh(t) = \sqrt{1 + \sinh^2(t)}\), substitute to find \(y = 2\sqrt{1 + \left(\frac{x}{2}\right)^2}\).

Step 4: Cartesian Equation Derivation

Substitute \(t\) using the relation \(\cosh^2(t) - \sinh^2(t) = 1\): \(\left(\frac{y}{2}\right)^2 - \left(\frac{x}{2}\right)^2 = 1\). Simplify to find \(\frac{y^2}{4} - \frac{x^2}{4} = 1\), which gives \(\frac{y^2}{4} = 1 + \frac{x^2}{4}\).

Step 5: Simplify and Sketch

Recognize this as a hyperbola. Rearrange to the standard form \(\frac{x^2}{4} - \frac{y^2}{4} = -1\) or \(\frac{y^2}{4} - \frac{x^2}{4} = 1\). Sketch this as a standard hyperbola opening along the y-axis.

Key Concepts

Hyperbolic functions

Hyperbolic functions are fascinating because they look similar to regular trigonometric functions but come with unique characteristics. They can be thought of as the analogs of sine and cosine for a hyperbola—hence their names. The main hyperbolic functions are the hyperbolic sine (\(\sinh\)) and hyperbolic cosine (\(\cosh\)). Here’s a brief overview:

• Hyperbolic sine, \(\sinh(t) = \frac{e^t - e^{-t}}{2}\).
• Hyperbolic cosine, \(\cosh(t) = \frac{e^t + e^{-t}}{2}\).

Both \(\sinh(t)\) and \(\cosh(t)\) have the beautiful property where their derivatives are still hyperbolic functions, much like their trigonometric counterparts. It’s crucial to also note the core identity these functions satisfy: \(\cosh^2(t) - \sinh^2(t) = 1\). This identity is what makes the parameter elimination and conversion to Cartesian coordinates possible.

Cartesian coordinates

Cartesian coordinates are the standard coordinate system used to graph points, lines, and curves in two-dimensional space. They rely on two perpendicular axes, the x-axis and y-axis, to define every point uniquely in a plane.

• Each point in this system is identified by a pair of numerical coordinates (x, y).
• This system allows us to convert complex vector equations into simpler equations, like those of a parabola or hyperbola, that can be easily graphed and understood.

In converting from a parameterized form like \(2 \sinh(t)\) and \(2 \cosh(t)\) to Cartesian coordinates, we eliminate the parameter (t) and express the relation solely in terms of x and y. This provides a clear, concise mathematical description of the entire curve in the coordinate plane.

Parameter elimination

Parameter elimination is a technique used to remove the parameter from equations describing curves. This process helps to switch from parameterized descriptions to coordinate-based equations. Here’s how it works in our exercise:

• We start with equations \(x = 2 \sinh(t)\) and \(y = 2 \cosh(t)\).
• The goal is to solve one equation—for instance, solve for \(t\) in terms of \(x\) using \(x = 2 \sinh(t)\).
• Once \(t\) is expressed in terms of \(x\), substitute it into the other equation \(y = 2 \cosh(t)\).

This operation transforms the parameterized set of equations into a single Cartesian equation that defines the relationship between x and y, like \(\frac{y^2}{4} - \frac{x^2}{4} = 1\), which is the general equation for a hyperbola.

Graphing hyperbolas

Graphing hyperbolas involves understanding their unique and symmetric features. They can be defined as a set of all points where the difference in distances to two fixed points (the foci) is constant. Here’s how you can sketch a hyperbola, based on our final equation:

• Start with the standard form of the hyperbola equation: \(\frac{y^2}{4} - \frac{x^2}{4} = 1\). This indicates a hyperbola that opens vertically because the term involving \(y\) is positive.
• The center of this hyperbola is at the origin, (0,0).
• Identify the asymptotes, which can be found by ignoring the constant on the right side: these are the lines \(y = \pm x\).

When sketching, always plot the center, foci, vertices, and asymptotes. A complete sketch will show the hyperbola branches opening along the y-axis, consistent with the equation form derived from eliminating the parameter.