Problem 21
Question
A pair of parametric equations is given. (a) Sketch the curve represented by the parametric equations. (b) Find a rectangular-coordinate equation for the curve by eliminating the parameter. $$x=\tan t, \quad y=\cot t, \quad 0 < t < \pi / 2$$
Step-by-Step Solution
Verified Answer
The rectangular equation is \( y = \frac{1}{x} \). Sketch starts at \((0, \infty)\) and approaches \((\infty, 0)\).
1Step 1: Understand the Parametric Equations
We are given two parametric equations: \( x = \tan t \) and \( y = \cot t \), where \( t \) is the parameter that ranges from \( 0 < t < \pi/2 \). This means as \( t \) changes within the given interval, \( x \) and \( y \) will change correspondingly.
2Step 2: Sketch the Curve
To sketch the curve, we need to analyze how \( x \) and \( y \) behave as \( t \) varies from 0 to \( \pi/2 \). \( \tan t \) increases from 0 to \( \infty \), while \( \cot t \) decreases from \( \infty \) to 0. Hence, the curve will start from the point \((0, \infty)\) and will approach \((\infty, 0)\). The sketch would show the path approaching a horizontal asymptote at \( y = 0 \) and a vertical asymptote at \( x = 0 \).
3Step 3: Eliminate the Parameter
To eliminate \( t \), use the identities of \( \tan t \) and \( \cot t \). Recall that \( y = \cot t = \frac{1}{\tan t} \). Thus, substitute the expression for \( \tan t \) in terms of \( x \): \( y = \frac{1}{x} \). This is the rectangular-coordinate equation.
4Step 4: Conclude the Rectangular Equation
With the above substitution, the rectangular-coordinate equation of the curve is \( y = \frac{1}{x} \). This tells us that the curve is a hyperbola in rectangular coordinates.
Key Concepts
Eliminating the ParameterSketching CurvesRectangular-Coordinate Equation
Eliminating the Parameter
Eliminating the parameter in parametric equations involves finding a way to relate the parametric variables directly without involving the parameter itself. In our original exercise, we have the equations \( x = \tan t \) and \( y = \cot t \), where \( t \) serves as the parameter.
First, note that \( \tan t \) and \( \cot t \) are reciprocal identities, meaning \( \cot t = \frac{1}{\tan t} \). By understanding this reciprocal relationship, we can express \( y \) directly in terms of \( x \) by substituting \( x \) for the \( \tan t \). This gives us:
First, note that \( \tan t \) and \( \cot t \) are reciprocal identities, meaning \( \cot t = \frac{1}{\tan t} \). By understanding this reciprocal relationship, we can express \( y \) directly in terms of \( x \) by substituting \( x \) for the \( \tan t \). This gives us:
- \( y = \cot t = \frac{1}{\tan t} = \frac{1}{x} \)
Sketching Curves
Sketching parametric curves is about visualizing how the point \((x, y)\) moves as the parameter changes. In this exercise, we are analyzing the curve generated by \( x = \tan t \) and \( y = \cot t \) as \( t \) varies from \( 0 \) to \( \pi/2 \).
When \( x = \tan t \), the values of \( x \) start from 0 and increase to infinity. Conversely, \( y = \cot t \) starts from infinity and decreases to 0. This specific behavior hints at the presence of asymptotic behavior:
When \( x = \tan t \), the values of \( x \) start from 0 and increase to infinity. Conversely, \( y = \cot t \) starts from infinity and decreases to 0. This specific behavior hints at the presence of asymptotic behavior:
- The curve starts at the point \((0, \infty)\).
- It approaches the horizontal asymptote as \( y = 0 \), since \( y \) gets closer to zero as \( t \) approaches \( \pi/2 \).
- There’s also a vertical asymptote at \( x = 0 \), as \( x \) tends towards zero when \( t \) is near the lower limit.
Rectangular-Coordinate Equation
A rectangular-coordinate equation expresses the relationship between two variables \( x \) and \( y \) in a Cartesian plane with no parameters involved. In our problem, starting with the parametric equations \( x = \tan t \) and \( y = \cot t \), we identified that a direct relationship can be derived.
Using the identity \( \cot t = \frac{1}{\tan t} \), we rearranged the equation to express \( y \) solely in terms of \( x \), resulting in \( y = \frac{1}{x} \). This conversion from a parametric form to a rectangular form simplifies understanding and handling of such equations.
Such a rectangular equation simplifies visualization. In particular, \( y = \frac{1}{x} \) is known as a hyperbola in typical Cartesian coordinates. Hyperbolas have unique properties that affect their graph:
Using the identity \( \cot t = \frac{1}{\tan t} \), we rearranged the equation to express \( y \) solely in terms of \( x \), resulting in \( y = \frac{1}{x} \). This conversion from a parametric form to a rectangular form simplifies understanding and handling of such equations.
Such a rectangular equation simplifies visualization. In particular, \( y = \frac{1}{x} \) is known as a hyperbola in typical Cartesian coordinates. Hyperbolas have unique properties that affect their graph:
- They display two asymptotes, which are lines that the curve approaches but never touches.
- The equation \( y = \frac{1}{x} \) specifically features a hyperbola with its asymptotes as the \( x \)- and \( y \)-axes.
Other exercises in this chapter
Problem 20
Sketch a graph of the polar equation, and express the equation in rectangular coordinates. $$\theta=5 \pi / 6$$
View solution Problem 21
Sketch the set in the complex plane. $$\\{z=a+b i | a \leq 0, b \geq 0\\}$$
View solution Problem 21
Sketch a graph of the polar equation, and express the equation in rectangular coordinates. $$r=6 \sin \theta$$
View solution Problem 22
Sketch the set in the complex plane. $$\\{z=a+b i | a>1, b>1\\}$$
View solution