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Chapter 12: Problem 24

Consider the following parametric equations. a. Eliminate the parameter to obtain an equation in \(x\) and \(y\). b. Describe the curve and indicate the positive orientation. $$x=\sin t, y=2 \sin t+1 ; 0 \leq t \leq \pi / 2$$

Short Answer

Expert verified
Question: Given the parametric equations \(x=\sin t\) and \(y=2\sin t+1\), find the Cartesian equation and describe the curve's shape and positive orientation. Answer: The Cartesian equation is \(y=2x+1\). The curve is a straight line with a slope of \(2\) and a \(y\)-intercept of \(1\). The positive orientation is in the positive \(x\)-direction and the positive \(y\)-direction.

Step by step solution

01

Find the relationship between \(x\) and \(y\)

First, let's find a relationship between \(x\) and \(y\) using the given parametric equations: $$x=\sin t, y=2\sin t+1.$$ We can solve the first equation for \(t\) by taking the arcsin of both sides: $$t=\arcsin x.$$ Now, we can substitute this expression for \(t\) into the second equation: $$y=2\sin(\arcsin x)+1.$$
02

Simplify the equation to obtain the Cartesian equation

We can simplify the equation we found in Step 1 by using the trigonometric identity: $$y=2\sin(\arcsin x)+1.$$ As we know that $$\sin(\arcsin x) = x,$$ so the equation becomes: $$y=2x+1.$$
03

Analyze the curve to understand its shape and orientation

The equation $$y=2x+1$$ represents a straight line with a slope of \(2\) and a \(y\)-intercept of \(1\). The positive orientation of the curve is determined by the direction in which the curve is traversed as the parameter \(t\) increases. As \(t\) increases from \(0\) to \(\pi/2\), \(\sin t\) will also increase from \(0\) to \(1\). Accordingly, \(x\) will take on values from \(0\) to \(1\). Since \(y=2x+1\), increasing \(x\) values will also result in increasing \(y\) values. Thus, as \(t\) increases, the curve moves in the positive \(x\)-direction and the positive \(y\)-direction. This gives us the positive orientation. In summary: a. The Cartesian equation is $$y=2x+1.$$ b. The curve is a straight line with a slope of \(2\) and a \(y\)-intercept of \(1\). The positive orientation is in the positive \(x\)-direction and the positive \(y\)-direction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cartesian equation
In mathematics, a Cartesian equation is one that expresses the relationship between the coordinates of points in a plane, usually in terms of the variables, typically noted as \(x\) and \(y\). When dealing with parametric equations, like the ones given in the exercise \(x=\sin t, y=2 \sin t+1\), the goal is often to eliminate the parameter, in this case, \(t\), so that you are left with an equation only involving \(x\) and \(y\).

This process is often called "eliminating the parameter." Here's how it works step by step:
  • Start by solving for the parameter in terms of one of the variables. In the exercise, you solve for \(t\) using \(x = \sin t\) to get \(t = \arcsin x\).
  • Substitute this expression for \(t\) into another given equation. Here, we plug \(t = \arcsin x\) into \(y = 2 \sin t + 1\).
  • Simplify the result using known mathematical identities. Use \( \sin(\arcsin x) = x \) to arrive at the final Cartesian equation \(y = 2x + 1\).
This final equation, \(y = 2x + 1\), can now be graphed directly in a 2-dimensional plane without involving the parameter \(t\). It represents a straight line, making it simpler to analyze.
Trigonometric identity
Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables within their domain. They are useful in simplifying or evaluating expressions, especially when dealing with parametric equations.

In the given problem, the identity \( \sin(\arcsin x) = x \) is used to simplify the expression for \(y\). Here's why it's crucial:
  • The arcsine function, \(\arcsin x\), is the inverse of \(\sin x\), meaning that it cancels out the effect of the sine function, bringing you back to the original \(x\).
  • This is why \(y = 2 \sin(\arcsin x) + 1\) simplifies neatly to \(y = 2x + 1\). All you're essentially doing here is recognizing that when you apply a function and its inverse, you arrive back at the input variable.
By using this trigonometric identity, the process of converting the parametric equations into a Cartesian equation becomes much simpler and more intuitive. This identity is part of the basic toolbox of trigonometry, enabling you to handle these types of mathematical problems effectively.
Positive orientation
Positive orientation refers to the direction in which a curve is traversed as the parameter counter (usually \(t\) in parametric equations), increases. Understanding positive orientation is essential for analyzing curves, especially when interpreting parametric equations.

In the given problem, the parameter \(t\) ranges from 0 to \(\pi/2\). As \(t\) increases:
  • \(\sin t\) transitions from 0 to 1, and therefore \(x = \sin t\) will increase from 0 to 1.
  • This also affects our Cartesian equation \(y = 2x + 1\), as it will result in \(y\) increasing smoothly, moving upward along the line.
  • The curve is described by a straight line starting at the point (0, 1) when \(t = 0\) and ending at the point (1, 3) when \(t = \pi/2\), showing a clear positive orientation towards both the positive \(x\) and \(y\) directions.
By understanding positive orientation, you can visualize not just where the curve lies, but how it's traced out in space over the course of the parameter's interval. This knowledge can assist in better interpreting and graphing parametric and Cartesian representations.

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Most popular questions from this chapter

Use a graphing utility to graph the hyperbolas \(r=\frac{e}{1+e \cos \theta}\) for \(e=1.1,1.3,1.5,1.7\) and 2 on the same set of axes. Explain how the shapes of the curves vary as \(e\) changes.

Find the arc length of the following curves on the given interval. $$x=t^{4}, y=\frac{t^{6}}{3} ; 0 \leq t \leq 1$$

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a. Show that an equation of the line \(y=m x+b\) in polar coordinates is \(r=\frac{b}{\sin \theta-m \cos \theta}\) b. Use the figure to find an alternative polar equation of a line. \(r \cos \left(\theta_{0}-\theta\right)=r_{0}\) Note that \(Q\left(r_{0}, \theta_{0}\right)\) is a fixed point on the line such that \(O Q\) is perpendicular to the line and \(r_{0} \geq 0\) \(P(r, \theta)\) is an arbitrary point on the line.

Determine whether the following statements are true and give an explanation or counterexample. a. The area of the region bounded by the polar graph of \(r=f(\theta)\) on the interval \([\alpha, \beta]\) is \(\int_{\alpha}^{\beta} f(\theta) d \theta\). b. The slope of the line tangent to the polar curve \(r=f(\theta)\) at a point \((r, \theta)\) is \(f^{\prime}(\theta)\). c. There may be more than one line that is tangent to a polar curve at some points on the curve.
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