Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In which direction is the curve \(x=-2 \sin t, y=2 \cos t,\) for \(0

Short Answer

Expert verified
Answer: The curve is generated in the clockwise direction.

Step by step solution

01

Analyze the parametric equations

First, let's take a look at the parametric equations \(x = -2 \sin t\) and \(y = 2 \cos t\). These equations define the coordinates of a point \((x, y)\) on the curve as a function of the parameter \(t\). As \(t\) varies in the interval \(0 < t < 2 \pi\), the point moves along the curve.
02

Determine the relation between x and y coordinates

Now let's try to eliminate the parameter \(t\) and see if we can find a relation between \(x\) and \(y\). We can use the identity \(\sin^2 t + \cos^2 t = 1\) as follows: Square both equations and sum them: \((-2 \sin t)^2 + (2 \cos t)^2 = 4\sin^2 t + 4\cos^2 t\) Now, we can factor out 4 and use the identity: \(4 (\sin^2 t + \cos^2 t) = 4 (1) = 4\) This gives us \(x^2+y^2=4\), which is the equation for a circle with radius 2 centered at the origin.
03

Determine the behavior of x and y individually

Now let's analyze how \(x\) and \(y\) change as \(t\) goes from \(0\) to \(2 \pi\). \(x = -2 \sin t\) Since the sine function is positive in the first and second quadrants and negative in the third and fourth quadrants, \(-2\sin t\) is negative in the first and second quadrants and positive in the third and fourth quadrants. This means that as \(t\) increases, the \(x\)-coordinate will go from negative to positive, and then back to negative as \(t\) approaches \(2\pi\). \(y = 2 \cos t\) Cosine function is positive in the first and fourth quadrants and negative in the second and third quadrants. Thus, as \(t\) goes from \(0\) to \(2\pi\), the \(y\)-coordinate starts from positive, goes to negative, and then back to positive.
04

Determine the direction of the curve generation

Now, we can combine the analysis from steps 3 and make some conclusions about the direction of the curve. At the beginning of the interval, when \(t=0\), \(x = -2\sin(0) = 0\) and \(y=2\cos(0)=2\). This means that the curve starts at point (0,2). As the curve goes from positive to negative \(y\) values, the \(x\) goes from negative to positive, and then back to negative values. This indicates that the generation of the curve is in the clockwise direction. In conclusion, the curve defined by the parametric equations \(x = -2 \sin t, y = 2 \cos t\) for the interval \(0 < t < 2 \pi\) is generated in the clockwise direction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sector of a hyperbola Let \(H\) be the right branch of the hyperbola \(x^{2}-y^{2}=1\) and let \(\ell\) be the line \(y=m(x-2)\) that passes through the point (2,0) with slope \(m,\) where \(-\infty < m < \infty\). Let \(R\) be the region in the first quadrant bounded by \(H\) and \(\ell\) (see figure). Let \(A(m)\) be the area of \(R .\) Note that for some values of \(m\) \(A(m)\) is not defined. a. Find the \(x\) -coordinates of the intersection points between \(H\) and \(\ell\) as functions of \(m ;\) call them \(u(m)\) and \(v(m),\) where \(v(m) > u(m) > 1 .\) For what values of \(m\) are there two intersection points? b. Evaluate \(\lim _{m \rightarrow 1^{+}} u(m)\) and \(\lim _{m \rightarrow 1^{+}} v(m)\) c. Evaluate \(\lim _{m \rightarrow \infty} u(m)\) and \(\lim _{m \rightarrow \infty} v(m)\) d. Evaluate and interpret \(\lim _{m \rightarrow \infty} A(m)\)

Suppose that the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\) is revolved about the \(x\) -axis. What is the volume of the solid enclosed by the ellipsoid that is generated? Is the volume different if the same ellipse is revolved about the \(y\) -axis?

Consider the following Lissajous curves. Graph the curve and estimate the coordinates of the points on the curve at which there is (a) a horizontal tangent line and (b) a vertical tangent line. (See the Guided Project Parametric art for more on Lissajous curves.) $$\begin{aligned}&x=\sin 2 t, y=2 \sin t\\\&0 \leq t \leq 2 \pi\end{aligned}$$

Show that the graph of \(r=a \sin m \theta\) or \(r=a \cos m \theta\) is a rose with \(m\) leaves if \(m\) is an odd integer and a rose with \(2 m\) leaves if \(m\) is an even integer.

Equations of the form \(r=a \sin m \theta\) or \(r=a \cos m \theta,\) where \(a\) is a real number and \(m\) is a positive integer, have graphs known as roses (see Example 6 ). Graph the following roses. \(r=\sin 2 \theta\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free