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Give parametric equations and parameter intervals for the motion of a particle in the \(x y\) -plane. Identify the particle's path by finding a Cartesian equation for it. Graph the Cartesian equation. (The graphs will vary with the equation used.) Indicate the portion of the graph traced by the particle and the direction of motion. $$x=\sin t, \quad y=\cos 2 t, \quad-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$$

Short Answer

Expert verified
The path is a parabola: \(y = 1 - 2x^2\), from \(x = -1\) to \(x = 1\), moving left to right.

Step by step solution

01

Identify Given Parametric Equations

The problem gives us parametric equations for a particle in the xy-plane: \(x = \sin t\) and \(y = \cos 2t\), where \(-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}\). These describe the position of the particle at any time \(t\).
02

Eliminate the Parameter

To find the Cartesian equation, we need to eliminate the parameter \(t\). Start with the equation \(x = \sin t\). Solving for \(t\) gives us \(t = \arcsin x\). Substitute \(t\) in \(y = \cos 2t\): when \(y = \cos(2 \arcsin x)\), use the identity \(\cos 2\theta = 1 - 2\sin^2 \theta\) to express \(y\) in terms of \(x\). Thus, \(\sin t = x\) implies \(\sin^2 t = x^2\), leading to \(\cos 2t = 1 - 2x^2\).
03

Formulate the Cartesian Equation

Based on the steps above, the Cartesian equation that describes the particle's path is \(y = 1 - 2x^2\). This is a quadratic equation, specifically a downward-facing parabola.
04

Determine Path Interval and Direction

Given the interval \(-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}\), the values of \(x\) span from \(-1\) to \(1\) as \(t\) moves through this range. The direction of motion on the graph can be determined by evaluating \(x(t) = \sin t\) and \(y(t) = \cos 2t\) for small values of \(t\) to see how \(y\) changes with \(x\). As \(t\) increases from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), \(x\) increases from \(-1\) to \(1\).
05

Graph the Cartesian Equation

Graph the equation \(y = 1 - 2x^2\). The path traced by the particle includes \(-1 \leq x \leq 1\) on the parabola. Starting from \(x = -1\) and moving to \(x = 1\), the particle moves upwards to the vertex at \(x = 0\), and then descends, indicating the direction of motion from left to right.

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Key Concepts

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

Cartesian equation
When working with parametric equations, like those given in problems describing the motion of a particle, it's often useful to convert these into a Cartesian equation. This equation describes the path of the particle in the familiar Cartesian coordinate system.
The Cartesian equation allows you to express the relationship between the coordinates without directly involving the parameter, usually denoted by \(t\). In this context, the Cartesian equation we derived is \(y = 1 - 2x^2\). This equation tells us that for every \(x\), there is a corresponding \(y\), thereby eliminating \(t\) from the picture.
Converting parametric equations to a Cartesian equation usually involves:
  • Identifying the parametric equations, for instance, \(x = \sin t\) and \(y = \cos 2t\).
  • Finding an expression for \(t\) in terms of one of the parametric variables. In this example, solving \(x = \sin t\) gives \(t = \arcsin x\).
  • Substituting the expression back into the other parametric equation and using trigonometric identities to consolidate into one equation in terms of \(x\) and \(y\).
This process ensures that we understand the mathematical shape and position of the path in the \(xy\)-plane.
particle motion
Understanding particle motion through parametric equations is key in visualizing how particles travel along their paths. These equations tell you where the particle is at any particular time \(t\). The motion is defined in two dimensions, \(x\) and \(y\), in terms of the parameter \(t\).
In this problem, as the parameter \(t\) changes from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), the equations \(x = \sin t\) and \(y = \cos 2t\) describe the trajectory, specifying the horizontal and vertical coordinates at each moment.
The direction of motion is an important detail, which can be traced by checking how changes in \(t\) affect \(x\) and \(y\). We see that as \(t\) increases, \(x\) starts from \(-1\) and goes to \(1\). Observing \(y\), particularly its changes based on \(t\), helps to understand that the movement starts at one end of the parabola (\(-1\)), peaks, and then descends down the other side, confirming a left to right movement.
  • Extremely useful in physics and engineering for modeling trajectories.
  • Can easily account for changing velocities and directions with simple adjustments to parameters.
This understanding bridges theoretical equations and real-world path tracing.
parabola graph
The graph of a parabola is a visual representation of its equation. For the equation \(y = 1 - 2x^2\), the curve is quadratic and symmetric about the y-axis, opening downwards due to the negative coefficient of \(x^2\).
In the context of the parametric problem, graphing the parabola helps us visualize the path traced by the particle. The vertex of this parabola is at the point \((0, 1)\), which represents the highest point the particle reaches along its journey.
Graphing reveals several characteristics to keep in mind:
  • The range of \(x\) from \(-1\) to \(1\) outlines the domain over which the particle travels.
  • Because the parabola is symmetrical, half the graph shows movement from one side to the vertex, and the other half shows movement away from the vertex.
  • The direction can be indicated by an arrow along the path, starting from \(-1\) towards \(1\), reflecting left-to-right motion.
By visualizing this graph, one gains insight into the particle's dynamics—where it speeds up, slows down, or changes direction—making the abstract equations come to life.

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Most popular questions from this chapter

The asymptotes of \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) Show that the vertical distance between the line \(y=(b / a) x\) and the upper half of the right-hand branch \(y=(b / a) \sqrt{x^{2}-a^{2}}\) of the hyperbola \(\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\) approaches 0 by showing that $$ \lim _{x \rightarrow \infty}\left(\frac{b}{a} x-\frac{b}{a} \sqrt{x^{2}-a^{2}}\right)=\frac{b}{a} \lim _{x \rightarrow \infty}\left(x-\sqrt{x^{2}-a^{2}}\right)=0 $$ Similar results hold for the remaining portions of the hyperbola and the lines \(y=\pm(b / a) x .\)

Use a CAS to perform the following steps for the given curve over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure \(11.15 . )\) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for \(n=2,4,8\) with the actual length given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$ x=2 t^{3}-16 t^{2}+25 t+5, \quad y=t^{2}+t-3, \quad 0 \leq t \leq 6 $$

Replace the Cartesian equations in Exercises \(53-66\) with equivalent polar equations. $$x=7$$

Exercises \(29-36\) give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$e=1, \quad x=2$$

The witch of Maria Agnesi The bell-shaped witch of Maria Agnesi can be constructed in the following way. Start with a circle of radius \(1,\) centered at the point \((0,1),\) as shown in the accompanying figure. Choose a point \(A\) on the line \(y=2\) and connect it to the origin with a line segment. Call the point where the segment crosses the circle \(B\) . Let \(P\) be the point where the vertical line through \(A\) crosses the horizontal line through \(B\) . The witch is the curve traced by \(P\) as \(A\) moves along the line \(y=2\) . Find parametric equations and a parameter interval for the witch by expressing the coordinates of \(P\) in terms of \(t\) the radian measure of the angle that segment \(O A\) makes with the positive \(x\) -axis. The following equalities (which you may assume) will help. $$\begin{array}{ll}{\text { a. } x=A Q} & {\text { b. } y=2-A B \sin t} \\\ {\text { c. } A B \cdot O A=(A Q)^{2}}\end{array}$$

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