Problem 14 - A Graphical Approach to Precalculus with Limits

Question

Find a rectangular equation for each curve and describe the curve. $$x=\cot t, y=\csc t ; \text { for } t \text { in }(0, \pi)$$

Step-by-Step Solution

Answer

The curve is a hyperbola described by \(x = \pm \sqrt{y^2 - 1}\).

Step 1: Write down known equations

We are given the parametric equations: \[x = \cot t\] \[y = \csc t\] We need to eliminate the parameter \(t\) to find a relation between \(x\) and \(y\).

Step 2: Express in terms of sine and cosine

Recall that \(\cot t = \frac{\cos t}{\sin t}\) and \(\csc t = \frac{1}{\sin t}\). Substitute these into the known parametric equations:\[ x = \frac{\cos t}{\sin t} \] \[ y = \frac{1}{\sin t} \]

Step 3: Eliminate the parameter

We know \(y = \frac{1}{\sin t}\), so \( \sin t = \frac{1}{y} \). Substitute this into the equation for \(x\):\[ x = \frac{\cos t}{\sin t} = \cos t \cdot y \] Using \( \cos^2 t + \sin^2 t = 1 \), we find \( \cos t = \pm \sqrt{1 - \sin^2 t} = \pm \sqrt{1 - \left(\frac{1}{y}\right)^2} \).

Step 4: Substitute \(\cos t\)

From the last step, substitute \(\cos t\):\[ x = \pm y \sqrt{1 - \frac{1}{y^2}} \]\[ x = \pm \sqrt{y^2 - 1} \] The equation \(x = \pm \sqrt{y^2 - 1}\) is the rectangular form of the curve.

Step 5: Describe the curve

The rectangular equation \(x = \pm \sqrt{y^2 - 1}\) describes a hyperbola. Since \(t\) is in the interval \((0, \pi)\), both \(x\) and \(y\) are in the positive quadrants, limiting the hyperbola to intersect only positive-side regions.

Key Concepts

Parametric Equations

Parametric equations allow us to describe a curve by defining both of its components, usually marked as \(x\) and \(y\), in terms of a third variable, referred to as a parameter. In our given exercise, the parameter is \(t\), and the equations \(x = \cot t\) and \(y = \csc t\) represent a curve as it relates to this parameter. By using parametric equations, we can precisely define complex curves and shapes that may be difficult to express with a single equation involving \(x\) and \(y\) alone. To convert these equations to a rectangular form (i.e., eliminating the parameter \(t\)), we use relationships among trigonometric identities. This conversion helps us describe and analyze the curve in a Cartesian coordinate system, making it easier to understand its shape and behavior without having to rely on the parameter.

Hyperbola

A hyperbola is a type of conic section that appears as two separate curves, or "branches." In the rectangular equation \(x = \pm \sqrt{y^2 - 1}\), we see a depiction of the hyperbola. The equation suggests that the value of \(x\) is derived from the square root expression of \(y^2 - 1\), implying that the set of points do not form a closed curve. Instead, two symmetrical branches result, curving away from each other.This particular hyperbola is defined in such a way that \(y\) must be greater than 1 or less than -1 to keep the expression under the square root positive, ensuring the equation is valid and real numbers are used. Additionally, since \(t\) is constrained between \(0\) and \(\pi\), both \(x\) and \(y\) remain positive, restricting the hyperbola to the first quadrant of the Cartesian plane. This means only a small portion of the hyperbola is visible in this context.

Trigonometric Identities

Trigonometric identities are essential tools for simplifying and transforming trigonometric expressions. In the case of this exercise, the identities \(\cot t = \frac{\cos t}{\sin t}\) and \(\csc t = \frac{1}{\sin t}\) were employed to express both \(x\) and \(y\) in terms of sine and cosine functions of the parameter \(t\). Understanding these identities is crucial as they provide the basis for manipulating the expressions and ultimately eliminating the parameter to achieve a rectangular equation.Another critical trigonometric identity used is the Pythagorean identity \(\cos^2 t + \sin^2 t = 1\). This identity helped transform the equation \(x = \cos t \cdot y\) into a form that freed the expression from the parameter \(t\). By doing so, math enthusiasts can derive the rectangular equation \(x = \pm \sqrt{y^2 - 1}\) from the parametric form, giving a clearer picture of the curves represented by each parametric component."