Problem 38
Question
Eliminate the parameter t. Then use the rectangular equation to sketch the plane curve represented by the given parametric equations. Use arrows to show the orientation of the curve corresponding to increasing values of t. (If an interval for t is not specified, assume that \(-\infty< t <\infty.\)) \(x=\sqrt{t}+2, y=\sqrt{t}-2\)
Step-by-Step Solution
Verified Answer
The rectangular equation for the given parametric equations is \(x^2-y^2=64t\). The graph of this equation, when sketched, would be a hyperbola extending indefinitely to the right and to the left.
1Step 1 - Convert the parametric equations into rectangular form
Start by noting that we have \(x=\sqrt{t}+2\) and \(y=\sqrt{t}-2\). We can square both to get rid of the square roots, which results in \(x^2=(\sqrt{t}+2)^2\) and \(y^2=(\sqrt{t}-2)^2\). Simplify further to get \(x^2=t+4\sqrt{t}+4\) and \(y^2=t-4\sqrt{t}+4\). Subtracting y-squared from x-squared gives the final equation \(x^2-y^2=8\sqrt{t}\).
2Step 2 - Rearrange the equation to express it in rectangular form
Square both sides to eliminate the square root, yielding \(x^2-y^2 = 8*8t = 64t\). Thus, our final rectangular equation is \(x^2-y^2=64t\).
3Step 3 - Sketch the graph
The equation we have obtained is a hyperbola. It opens to the right and left, which means that the 't' increases in those directions. Therefore, the graph of the equation (\(x^2-y^2=64t\)) will be a hyperbola which extends indefinitely to the right and to the left.
Key Concepts
Rectangular Form ConversionSketching Plane CurvesHyperbola GraphingOrientation of a Curve
Rectangular Form Conversion
Understanding the translation from parametric to rectangular forms is a vital skill in mathematics, particularly when visualizing complex curves. Parametric equations, like the ones given \(x=\sqrt{t}+2, y=\sqrt{t}-2\), represent a set of points in a plane by expressions for both the x and the y coordinates, which are both dependent on a third variable, the parameter t. To convert these into rectangular form, we seek to eliminate the parameter and express y solely in terms of x or vice versa.
Our exercise starts with the realization that both x and y depend on \(\sqrt{t}\). By squaring both equations and performing algebraic manipulations, we eliminate the root and get closer to the desired rectangular form. Rectangular form is invaluable as it allows us to sketch the curve directly on the standard xy-coordinate plane without needing to calculate specific points for certain t values. This simplifies the graphical representation of the curve, rendering it accessible to traditional analysis.
Our exercise starts with the realization that both x and y depend on \(\sqrt{t}\). By squaring both equations and performing algebraic manipulations, we eliminate the root and get closer to the desired rectangular form. Rectangular form is invaluable as it allows us to sketch the curve directly on the standard xy-coordinate plane without needing to calculate specific points for certain t values. This simplifies the graphical representation of the curve, rendering it accessible to traditional analysis.
Sketching Plane Curves
Sketching the graph of a plane curve translates abstract equations into visual understandings. When converting parametric equations to their rectangular counterpart, as we do with the hyperbolic equations in the example, it is a stepping stone towards creating a tangible representation of the curve.
After obtaining the rectangular form \(x^2-y^2=64t\), we recognize it as a hyperbola, a conic section. Sketching such curves involves plotting a set of points that satisfy the equation on the Cartesian plane and connecting them smoothly. This process is often facilitated by recognizing familiar shapes within the equations, such as circles, ellipses, parabolas, and hyperbolas, or by plotting a few key points and using symmetry. For the given hyperbola, recognizing the orientation and direction—whether it opens up and down or right and left—is crucial for an accurate sketch.
After obtaining the rectangular form \(x^2-y^2=64t\), we recognize it as a hyperbola, a conic section. Sketching such curves involves plotting a set of points that satisfy the equation on the Cartesian plane and connecting them smoothly. This process is often facilitated by recognizing familiar shapes within the equations, such as circles, ellipses, parabolas, and hyperbolas, or by plotting a few key points and using symmetry. For the given hyperbola, recognizing the orientation and direction—whether it opens up and down or right and left—is crucial for an accurate sketch.
Hyperbola Graphing
Hyperbolas are unique in their two-part structure, opening in opposing directions. In our current example, the equation \(x^2-y^2=64t\) implies that as t increases, the curve moves indefinitely to the right and left. To graph such a hyperbola, one must remember that it consists of two symmetrical branches.
To accurately draw these branches, we should find the center, asymptotes, and vertices. Primarily, for the equation \(x^2 - y^2 = 64t\), a clear way to find these components is to set t to zero, which would help us locate the center and the asymptotes. Once these components are identified, the sketching becomes a matter of connecting the dots, making sure not to cross the asymptotes, as they represent directions in which the branches approach but never meet.
To accurately draw these branches, we should find the center, asymptotes, and vertices. Primarily, for the equation \(x^2 - y^2 = 64t\), a clear way to find these components is to set t to zero, which would help us locate the center and the asymptotes. Once these components are identified, the sketching becomes a matter of connecting the dots, making sure not to cross the asymptotes, as they represent directions in which the branches approach but never meet.
Orientation of a Curve
The orientation of a curve signifies the direction in which the curve progresses as the parameter t increases. This concept is a critical aspect of understanding the dynamic nature of parametric curves. For instance, in the given exercise, the orientation of the hyperbola will be indicated by arrows.
Since increasing values of t make both \(x\) and \(y\) larger in magnitude but preserve their signs, the hyperbola branches will be drawn with arrows pointing outwards from the origin, showing the curve's orientation with respect to an increasing parameter t. This aspect is important not only in sketching but also in fields like physics and engineering, where the direction of motion along a path is often significant.
Since increasing values of t make both \(x\) and \(y\) larger in magnitude but preserve their signs, the hyperbola branches will be drawn with arrows pointing outwards from the origin, showing the curve's orientation with respect to an increasing parameter t. This aspect is important not only in sketching but also in fields like physics and engineering, where the direction of motion along a path is often significant.
Other exercises in this chapter
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use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$ \frac{(y+2)^{2}}{4}-\frac{(x-1)^{
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In Exercises \(37-50,\) graph each ellipse and give the location of its foci. $$ \frac{(x-1)^{2}}{16}+\frac{(y+2)^{2}}{9}=1 $$
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In Exercises 35–42, find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$ (x+2)^{2}=-8(y+2) $$
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use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$ \frac{(y-2)^{2}}{36}-\frac{(x+1)^
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