Problem 30 - Calculus Volume 3

Question

Convert the parametric equations of a curve into rectangular form. No sketch is necessary. State the domain of the rectangular form. $$ x=2 \cosh t, \quad y=4 \sinh t $$

Step-by-Step Solution

Answer

Rectangular form: \(\frac{x^2}{4} - \frac{y^2}{16} = 1\). Domain: \(\frac{x^2}{4} \geq 1\).

Step 1: Write Down Parametric Equations

The given parametric equations are: \[ x = 2 \cosh t \quad \text{and} \quad y = 4 \sinh t \] Our goal is to convert these into a single equation involving only \(x\) and \(y\).

Step 2: Use Hyperbolic Identity

Recall the hyperbolic identity: \( \cosh^2 t - \sinh^2 t = 1 \).We will use this identity to eliminate the parameter \(t\).

Step 3: Express Cosh and Sinh

From the equation \(x = 2 \cosh t\), solve for \(\cosh t\):\[ \cosh t = \frac{x}{2}. \]From the equation \(y = 4 \sinh t\), solve for \(\sinh t\):\[ \sinh t = \frac{y}{4}. \]

Step 4: Substitute Expressions into Identity

Substitute \( \cosh t = \frac{x}{2} \) and \( \sinh t = \frac{y}{4} \) into the hyperbolic identity.\[ \left(\frac{x}{2}\right)^2 - \left(\frac{y}{4}\right)^2 = 1 \]

Step 5: Simplify the Equation

Simplify the equation:\[\frac{x^2}{4} - \frac{y^2}{16} = 1.\]This is the rectangular form of the given parametric equations.

Step 6: Determine the Domain

The expression \( \frac{x^2}{4} - \frac{y^2}{16} = 1 \) suggests that all values of \(x\) are valid as long as the equation holds true. There are no restrictions on \(y\), except for those inherent in the equation, which is valid for all real values that satisfy it.Thus, the domain is all real numbers for \(x\) such that \( \frac{x^2}{4} \geq 1 \).

Key Concepts

Rectangular Form

Parametric equations allow us to describe a curve by defining both the x- and y-coordinates in terms of a parameter, often denoted as "t". Converting these parametric equations into rectangular form means writing a single equation that relates x and y without the parameter. This can simplify analysis and provide a more direct perspective on the shape and properties of the curve.In the exercise, we started with:

• \( x = 2 \cosh t \)
• \( y = 4 \sinh t \)

To develop the rectangular form, we leveraged the identity related to hyperbolic functions: \( \cosh^2 t - \sinh^2 t = 1 \). By expressing \( \cosh t \) and \( \sinh t \) in terms of x and y and substituting them into the identity, we arrive at a rectangular equation: \[\frac{x^2}{4} - \frac{y^2}{16} = 1.\]This equation describes a hyperbola, demonstrating how hyperbolic functions directly relate to the hyperbolic curves in rectangular coordinates.

Hyperbolic Functions

Hyperbolic functions, similar to trigonometric functions, are crucial in describing certain types of curves and shapes, especially in mathematics and physics. The core hyperbolic functions include \(\sinh t\) (hyperbolic sine) and \(\cosh t\) (hyperbolic cosine). These are defined by exponential functions:

• \(\sinh t = \frac{e^t - e^{-t}}{2}\)
• \(\cosh t = \frac{e^t + e^{-t}}{2}\)

These functions provide the backbone for many calculations involving hyperbolas, just as sine and cosine functions do for circles.A key identity, \( \cosh^2 t - \sinh^2 t = 1 \), closely resembles the Pythagorean identity in trigonometry. It aids in eliminating the parameter "t" when converting parametric equations of hyperbolic nature into a rectangular form, as we did in the given exercise. Understanding the behavior of hyperbolic functions helps in dealing with not only curves but also in physics, especially relating to relativity and hyperbolic geometry.

Domain in Calculus

The domain of a function is a fundamental concept in calculus, describing all the possible values for which the function is defined. When converting from parametric equations to rectangular form, it becomes necessary to determine the domain of the new equation to fully understand where the graph can exist in the x-y plane.In the example provided, the rectangular form is:\[\frac{x^2}{4} - \frac{y^2}{16} = 1\]To find the domain of x, consider the restriction placed by the hyperbola. The domain consists of all x-values that satisfy \(\frac{x^2}{4} \geq 1\).

• Solving \(\frac{x^2}{4} \geq 1\) gives two ranges for x:
  • \(x \leq -2\)
  • \(x \geq 2\)

This criterion shapes the portions of the hyperbola that actually exist on a real graph. Understanding domains helps clarify not just what the function looks like but also where it is practically relevant in both mathematics and physical applications.