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Chapter 10: Problem 24

Eliminate the parameter to find a description of the following circles or circular arcs in terms of \(x\) and \(y .\) Give the center and radius, and indicate the positive orientation. $$x=2 \sin t-3, y=2 \cos t+5 ; 0 \leq t \leq 2 \pi$$

Short Answer

Expert verified
Answer: The description of the circle in terms of \(x\) and \(y\) is \((x + 3)^2 + (y - 5)^2 = 4\). Its center is at the point \((-3, 5)\), and it has a radius of 2.

Step by step solution

01

Rewrite the parametric equations

Rewrite \(x\) and \(y\) in terms of \(\sin t\) and \(\cos t\) $$x = 2\sin t - 3 \Rightarrow \sin t = \frac{x + 3}{2}$$ $$y = 2\cos t + 5 \Rightarrow \cos t = \frac{y - 5}{2}$$
02

Use trigonometric identity to eliminate the parameter

Now we can use the trigonometric identity \(\sin^2 t + \cos^2 t = 1\) $$\left(\frac{x + 3}{2}\right)^2 + \left(\frac{y - 5}{2}\right)^2 = 1$$
03

Simplify the equation

Simplify the equation to get the circle formula in terms of \(x\) and \(y\) $$\left(x + 3\right)^2 + \left(y - 5\right)^2 = 2^2$$ $$\left(x + 3\right)^2 + \left(y - 5\right)^2 = 4$$
04

Find the center and radius of the circle

Now, we can find the center and radius of the circle by comparing the equation obtained in step 3 with the general equation of a circle. The general equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius. Comparing, we get the center as \((-3, 5)\) and radius as \(2\). So, the description of the circle in terms of \(x\) and \(y\) is: $$(x + 3)^2 + (y - 5)^2 = 4$$ with the center at \((-3, 5)\), radius of \(2\), and positive orientation.

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Key Concepts

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

Trigonometric Identities
Understanding trigonometric identities helps not only in solving parametric equations but also in simplifying complex algebraic expressions. A fundamental identity we encounter is the Pythagorean identity, which states that for any angle \(t\), \(\sin^2 t + \cos^2 t = 1\). This identity arises from the Pythagorean theorem in right-angled triangles and connects the trigonometric functions of sine and cosine.

In the context of the exercise, this identity allows us to amalgamate the separate parametric equations involving \(\sin t\) and \(\cos t\) into a single equation without the parameter \(t\). By expressing \(x\) and \(y\) in terms of \(\sin t\) and \(\cos t\), we set the stage for utilizing the Pythagorean identity to eliminate the parameter and reveal the underlying equation of a circle.
Circle Equations
The equation of a circle in the coordinate plane captures all the points that are a fixed distance (radius) from a central point (the center). The standard form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) represents the coordinates of the center and \( r \) is the radius.

By transforming the parametric equations into this standard form, we can straightforwardly identify the center and radius of the circle. In the exercise, once we expressed \(x\) and \(y\) in terms of sine and cosine and employed the Pythagorean identity, we were able to rearrange and compare our results to this standard form to find the center at \( (-3, 5) \) and a radius of 2 units. This critical step bridges the gap between parametric representation and the familiar Cartesian coordinate system, enabling a deeper comprehension of the geometry involved.
Coordinate Geometry
Coordinate geometry, or analytic geometry, is the study of geometry using a coordinate system. This approach combines algebra and geometry to describe the position of points, lines, and figures in the two-dimensional plane. The method allows us to solve geometrical problems algebraically and to deduce geometrical facts using algebraic computations.

In our case, translating parametric equations into a standard Cartesian circle equation is an example of utilizing coordinate geometry. After finding the center and radius, we can plot these elements on the Cartesian plane and visualize the circle. This visual representation not only helps us better understand the geometric entity in question but also strengthens our grasp of the relationship between algebraic expressions and their geometric interpretations.

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Most popular questions from this chapter

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.

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.) $$\begin{aligned}&x=\sin 2 t, y=2 \sin t\\\&0 \leq t \leq 2 \pi\end{aligned}$$

Points at which the graphs of \(r=f(\theta)\) and \(r=g(\theta)\) intersect must be determined carefully. Solving \(f(\theta)=g(\theta)\) identifies some-but perhaps not all-intersection points. The reason is that the curves may pass through the same point for different values of \(\theta .\) Use analytical methods and a graphing utility to find all the intersection points of the following curves. \(r=2 \cos \theta\) and \(r=1+\cos \theta\)

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 \(\lim _{m \rightarrow \infty} A(m)\)

Water flows in a shallow semicircular channel with inner and outer radii of \(1 \mathrm{m}\) and \(2 \mathrm{m}\) (see figure). At a point \(P(r, \theta)\) in the channel, the flow is in the tangential direction (counterclockwise along circles), and it depends only on \(r,\) the distance from the center of the semicircles. a. Express the region formed by the channel as a set in polar coordinates. b. Express the inflow and outflow regions of the channel as sets in polar coordinates. c. Suppose the tangential velocity of the water in \(\mathrm{m} / \mathrm{s}\) is given by \(v(r)=10 r,\) for \(1 \leq r \leq 2 .\) Is the velocity greater at \(\left(1.5, \frac{\pi}{4}\right)\) or \(\left(1.2, \frac{3 \pi}{4}\right) ?\) Explain. d. Suppose the tangential velocity of the water is given by \(v(r)=\frac{20}{r},\) for \(1 \leq r \leq 2 .\) Is the velocity greater at \(\left(1.8, \frac{\pi}{6}\right)\) or \(\left(1.3, \frac{2 \pi}{3}\right) ?\) Explain. e. The total amount of water that flows through the channel (across a cross section of the channel \(\theta=\theta_{0}\) ) is proportional to \(\int_{1}^{2} v(r) d r .\) Is the total flow through the channel greater for the flow in part (c) or (d)?
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