Problem 13 - Calculus Early Transcendentals

Question

(a) Eliminate the parameter to find a Cartesian equation of the curve. (b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. \(x=\sin t, \quad y=\csc t, \quad 0< t<\pi / 2\)

Step-by-Step Solution

Answer

The Cartesian equation is \(y = \frac{1}{x}\). Right to left trace.

Step 1: Express csc(t) in terms of sin(t)

The problem gives us the equations \(x = \sin t\) and \(y = \csc t\). Recall that \(\csc t = \frac{1}{\sin t}\). So, substitute \(\sin t\) in terms of \(x\): \(y = \csc t = \frac{1}{x}\).

Step 2: Derive the Cartesian equation

In Step 1, we expressed \(y\) in terms of \(x\). The Cartesian equation is simply \(y = \frac{1}{x}\). This is the Cartesian equation of the curve, eliminating the parameter \(t\).

Step 3: Sketch the Curve

The Cartesian equation \(y = \frac{1}{x}\) resembles the normal function of a hyperbola. Between \(0 < x < 1\) because \(x = \sin t\) and for \(0 < t < \frac{\pi}{2}\), \(\sin t\) ranges from just above 0 to just below 1.

Step 4: Indicate the direction of tracing

As \(t\) increases from \(0\) to \(\frac{\pi}{2}\), \(\sin t\) increases from 0 to 1. Therefore, \(x\) ranges from 0 to 1 and the curve is traced from right to left (down toward the x-axis) as \(t\) increases.

Key Concepts

Cartesian equation

A Cartesian equation represents a curve using two variables, such as \(x\) and \(y\), without involving a parameter like \(t\). This allows us to relate \(x\) and \(y\) directly. In the given problem, we started with parametric equations: \(x = \sin t\) and \(y = \csc t\). The goal is to eliminate the parameter \(t\) to find a relationship between \(x\) and \(y\).
To do this, recall that \(\csc t = \frac{1}{\sin t}\). Given \(x = \sin t\), we replace \(\sin t\) with \(x\) in our equation for \(y\). This gives us \(y = \frac{1}{x}\), forming our Cartesian equation. This expression defines the same curve, but in terms of \(x\) and \(y\), making it easier to analyze and visualize.
Using a Cartesian equation simplifies the process of curve sketching and understanding its properties, as opposed to working with separate parametric equations.

trigonometric functions

Trigonometric functions play a crucial role in understanding and describing periodic phenomena. In our exercise, we deal with two trigonometric functions: sine \(\sin t\) and cosecant \(\csc t\).
- **Sine function** \(\sin t\): This function varies from 0 to 1 as \(t\) goes from 0 to \(\frac{\pi}{2}\). In our case, it's used to express the \(x\) coordinate, making \(x\) range between open interval (0, 1).
- **Cosecant function** \(\csc t\): This is the reciprocal of the sine function, hence \(\csc t = \frac{1}{\sin t}\). It describes the \(y\) coordinate in the problem and ranges from a large positive value to 1 as \(t\) approaches \(\frac{\pi}{2}\).
Understanding how these functions behave is essential for accurately sketching the curve and anticipating what happens as the parameter increases. As \(t\) increases, \(\sin t\) increases resulting in a decrease of \(\csc t\) (since \(y = \frac{1}{x}\)) making this mathematical journey particularly fascinating.

curve sketching

Curve sketching involves plotting a graph to visually represent an equation or function. Here, we sketch the curve of the Cartesian equation \(y = \frac{1}{x}\).
- **Understanding the curve**: This curve is a form of a hyperbola known for two branches. For our parameter range, it is critical to understand that due to \(x = \sin t\) where \(0 < x < 1\), we're looking at a specific section of this typical hyperbolic shape. The curve exists in the first quadrant.
- **Direction of tracing**: As \(t\) increases from 0 to \(\frac{\pi}{2}\), \(x\) moves from being almost 0 to almost 1. This means the curve is traced starting from left (high values of \(y\)) moving downwards to the right (smaller values of \(y\)), indicating a downward slope.
With these insights, you can create a mental image of how the curve looks and progresses as \(t\) changes, thus linking algebraic form to visual representation effectively.