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Let an \(x^{\prime} y^{\prime}\) -coordinate system be obtained by rotating an \(x y\) -coordinate system through an angle of \(\theta=60^{\circ}\). (a) Find the \(x^{\prime} y^{\prime}\) -coordinates of the point whose \(x y\) -coordinates are \((-2,6)\). (b) Find an equation of the curve \(\sqrt{3} x y+y^{2}=6\) in \(x^{\prime} y^{\prime}\) -coordinates. (c) Sketch the curve in part (b), showing both \(x y\) -axes and \(x^{\prime} y^{\prime}\) -axes.

Short Answer

Expert verified
(a) \(x' = -1 - 3\sqrt{3}, y' = 3 - \sqrt{3}\); (b) Substitute for \(x, y\) to find \(x'y'\) equation; (c) Sketch as described.

Step by step solution

01

Rotation Matrix Calculation

The rotation matrix for an angle of \( \theta = 60^{\circ} \) is given by:\[R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\sin(\theta) & \cos(\theta)\end{bmatrix}\]First, calculate \( \cos(60^{\circ}) = \frac{1}{2} \) and \( \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \). Thus, the rotation matrix becomes:\[R = \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}\]
02

Transformation of Coordinates

To find the \( x' y' \) coordinates of the point \((-2,6)\), we use the rotation matrix \( R \):\[\begin{bmatrix}x' \y'\end{bmatrix} = R \cdot \begin{bmatrix}x \y\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix} \cdot \begin{bmatrix}-2 \6\end{bmatrix}\]Calculate:\[x' = \frac{1}{2}(-2) - \frac{\sqrt{3}}{2}(6) = -1 - 3\sqrt{3}x' = -1 - 3\sqrt{3}\]\[y' = \frac{\sqrt{3}}{2}(-2) + \frac{1}{2}(6) = -\sqrt{3} + 3\]The coordinates are \(x' = -1 - 3\sqrt{3}, y' = 3 - \sqrt{3}\).
03

Transforming the Equation of the Curve

Given the curve equation in \(xy\) coordinates: \( \sqrt{3}xy + y^2 = 6 \), substitute \( x = x' \cos(\theta) - y' \sin(\theta) \) and \( y = x' \sin(\theta) + y' \cos(\theta) \).Substitute:you have \( x = \frac{1}{2}x' - \frac{\sqrt{3}}{2}y' \) and \( y = \frac{\sqrt{3}}{2}x' + \frac{1}{2}y' \).Thus the equation in \( x'y' \) coordinates is obtained by simplifying the substitution.
04

Sketching the Curve

To sketch the curve, first plot the original \(xy\) curve \( \sqrt{3}xy + y^2 = 6 \), which is a hyperbola.Then, draw the \(x'y'\)-axes rotated by 60 degrees from the \(xy\)-axes.Sketch the transformed curve using the expression obtained in the transformed coordinates. This involves rotating the original hyperbola about the origin by 60 degrees.

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Key Concepts

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

rotation matrix
The concept of a rotation matrix is a fundamental idea in transforming coordinates. A rotation matrix allows you to rotate points in a plane without changing their relative distances. It effectively alters the direction in which points are viewed, enabling a new perspective on the plane.

The rotation matrix for an angle \( \theta \) in a two-dimensional coordinate system is expressed as:\[ R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \sin(\theta) & \cos(\theta)\end{bmatrix} \]This matrix is special because it preserves lengths and angles, making it an orthogonal matrix. When you multiply a point's coordinates by this matrix, the point rotates counterclockwise by the angle \( \theta \).

In the given exercise, you're rotating a point by \( 60^\circ \). Calculating \( \cos(60^\circ) = \frac{1}{2} \) and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), the rotation matrix becomes:\[ R = \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix} \] This matrix can then be used to transform any coordinate from the old system to the new rotated system.
coordinate system rotation
Rotating a coordinate system is a creative way to view geometric objects under a different light. By rotating the entire system, both the axes and the points transform, leading to new coordinate values and relationships.

For any point \((x, y)\), you find its rotated coordinates \((x', y')\) by applying the rotation matrix derived from the chosen angle \( \theta \). This transformation changes \((x, y)\) using:\[\begin{bmatrix} x' \ y' \end{bmatrix} = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \sin(\theta) & \cos(\theta)\end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}\]
In practical terms, you rotate the axes themselves while maintaining the structure and relationships of the object in space. With our exercise, starting with coordinates \((-2, 6)\), the transformation through the 60° rotation matrix results in new coordinates \((x', y') = \left(-1 - 3\sqrt{3}, 3 - \sqrt{3}\right)\).

This new perspective helps to understand the positioning and orientation of geometric forms in different contexts.
equation transformation
Transforming equations when rotating a coordinate system can appear challenging, but it's just an extension of coordinate changes. You replace the original variables using the relationships defined by the rotation matrix.

To transform an equation, you need to express the original \((x, y)\) in terms of the rotated coordinates \((x', y')\). As shown earlier for the specific rotation:
  • \(x = \frac{1}{2}x' - \frac{\sqrt{3}}{2}y'\)
  • \(y = \frac{\sqrt{3}}{2}x' + \frac{1}{2}y'\)
Substitute these into the original equation to obtain the equivalent expression.

For the exercise, transforming \( \sqrt{3}xy + y^2 = 6 \) involves substituting the above expressions for \(x\) and \(y\) into the equation. Simplifying this replaced form yields the equation in the \((x', y')\) space.

This process unveils how shifts in perspective (here, through rotation) alter the mathematical representation of curves, giving insight into their intrinsic geometric properties.

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Most popular questions from this chapter

Find a polar equation for the conic that has its focus at the pole and satisfies the stated conditions. Points are in polar coordinates and directrices in rectangular coordinates for simplicity. (In some cases there may be more than one conic that satisfies the conditions.) (a) Ellipse; \(e=\frac{3}{4} ;\) directrix \(x=2\). (b) Parabola; directrix \(x=1\). (c) Hyperbola; \(e=\frac{4}{3}\); directrix \(y=3\).

List the forms for standard equations of parabolas, ellipses, and hyperbolas, and write a summary of techniques for sketching conic sections from their standard equations.

Use a graphing utility to generate the polar graph of the equation \(r=\cos 3 \theta+2\), and find the area that it encloses.

Use the following values, where needed: radius of the Earth \(=4000 \mathrm{mi}=6440 \mathrm{~km}\) year (Earth year) \(=365\) days (Earth days) \(1 \mathrm{AU}=92.9 \times 10^{6} \mathrm{mi}=150 \times 10^{6} \mathrm{~km}\) (a) Let \(a\) be the semimajor axis of a planet's orbit around the Sun, and let \(T\) be its period. Show that if \(T\) is measured in days and \(a\) is measured in kilometers, then \(T=\left(365 \times 10^{-9}\right)(a / 150)^{3 / 2}\) (b) Use the result in part (a) to find the period of the planet Mercury in days, given that its semimajor axis is \(a=57.95 \times 10^{6} \mathrm{~km}\) (c) Choose a polar coordinate system with the Sun at the pole, and find an equation for the orbit of Mercury in that coordinate system given that the eccentricity of the orbit is \(e=0.206\). (d) Use a graphing utility to generate the orbit of Mercury from the equation obtained in part (c).

If \(f^{\prime}(t)\) and \(g^{\prime}(t)\) are continuous functions, and if no segment of the curve $$ x=f(t), \quad y=g(t) \quad(a \leq t \leq b) $$ is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the \(x\) -axis is $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ and the area of the surface generated by revolving the curve about the \(y\) -axis is $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5.] Use the formulas above in these exercises. By revolving the semicircle $$ x=r \cos t, \quad y=r \sin t \quad(0 \leq t \leq \pi) $$ about the \(x\) -axis, show that the surface area of a sphere of radius \(r\) is \(4 \pi r^{2}\).

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