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Chapter 10: Problem 80

Sketch the graph of the equation. $$r^{2}=1 / \theta$$

Short Answer

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Question: Sketch the graph of the polar equation $$r^{2}=\frac{1}{\theta}$$. Answer: To sketch the graph of the polar equation $$r^{2}=\frac{1}{\theta}$$, you need to: 1. Understand the relationship between $$r$$ and $$\theta$$ in the given equation. 2. Rewrite the equation in parametric form: $$x(\theta) = \sqrt{\frac{1}{\theta}}\cos\theta$$ and $$y(\theta) = \sqrt{\frac{1}{\theta}}\sin\theta$$. 3. Plot points for various values of $$\theta$$ (example: $$\frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \pi, \frac{3\pi}{2},$$ and $$2\pi$$). 4. Connect the points and sketch the graph. The graph is an outward spiral with $$r$$ increasing as $$\theta$$ decreases.

Step by step solution

01

Understand the Polar Equation

We are given the polar equation $$r^{2}=\frac{1}{\theta}$$. Here, $$r$$ is the distance from the origin to a point on the graph and $$\theta$$ is the angle between the positive $$x$$-axis and the line segment joining the origin to that point.
02

Rewrite the equation in parametric form

To sketch the graph, it will be helpful to convert the polar equation to parametric form. Recall that, in polar coordinates, the relationship between Cartesian ($$x$$, $$y$$) and polar ($$r$$, $$\theta$$) coordinates are given by: $$x = r\cos\theta$$ $$y = r\sin\theta$$ Rewrite the given polar equation as $$r = \sqrt{\frac{1}{\theta}}$$ Now, substitute the expressions of $$r$$ in terms of $$x$$ and $$y$$, we get $$x(\theta) = \sqrt{\frac{1}{\theta}}\cos\theta$$ $$y(\theta) = \sqrt{\frac{1}{\theta}}\sin\theta$$ Now we have the parametric equations of the graph.
03

Plot points for various values of $$\theta$$

Now, we will plot the points corresponding to various values of $$\theta$$. To cover all cases, we can choose some values for $$\theta$$ between $$0$$ and $$2\pi$$. For example, let's find the points for $$\theta = \frac{\pi}{4}$$, $$\frac{\pi}{3}$$, $$\frac{\pi}{2}, \pi, \frac{3\pi}{2},$$ and $$2\pi$$: For $$\theta = \frac{\pi}{4}$$, $$x = \sqrt{\frac{1}{\frac{\pi}{4}}}\cos\frac{\pi}{4} \approx 1.0517$$ $$y = \sqrt{\frac{1}{\frac{\pi}{4}}}\sin\frac{\pi}{4} \approx 1.0517$$ For $$\theta = \frac{\pi}{3}$$, $$x = \sqrt{\frac{1}{\frac{\pi}{3}}}\cos\frac{\pi}{3} \approx 1.0746$$ $$y = \sqrt{\frac{1}{\frac{\pi}{3}}}\sin\frac{\pi}{3}\approx 1.8603$$ And so on for the other values of $$\theta$$.
04

Sketch the graph

Plot the points from step 3 in the Cartesian plane, and connect them to create a smooth curve. This curve represents the graph of the polar equation $$r^{2}=\frac{1}{\theta}$$. Keep in mind that the curve will look like an outward spiral, as $$r$$ increases as $$\theta$$ decreases.

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Key Concepts

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

Polar Equation
Understanding a polar equation is like learning a new language in the world of mathematics. Instead of using the familiar Cartesian coordinates, a polar equation represents points on a plane using distance from a fixed point (known as the radius, r) and an angle (θ) from a fixed direction. The given polar equation r2 = 1/θ is intriguing because it relates the square of the radial distance directly to the reciprocal of the angle.

As θ gets bigger, the radial distance becomes smaller, which suggests that our graph will show a curve that winds around the origin by decreasing its distance from it as the angle increases. This characteristic will be important when we plot the curve. Understanding this decay in the radius is crucial for visualizing the shape of the graph as it gives us a hint about the nature of the spiral pattern it will follow.
Parametric Equations
To visualize a polar equation on a Cartesian plane, it's helpful to convert it to parametric equations. Parametric equations define a group of quantities as functions of one or more independent variables called parameters. In the polar-to-Cartesian context, the parameters are x(t) and y(t), which represent the coordinates of points on the graph as functions of the angle θ.

Converting our polar equation into parametric form involves expressing r as a function of θ, and then using the formulas x = r*cos(θ) and y = r*sin(θ) to describe the horizontal and vertical displacements. This transition bridges the gap between polar and Cartesian worldviews, making it easier to plot the shape described by the polar equation on familiar (x, y) axes.
Polar to Cartesian Conversion
The magic of polar to Cartesian conversion lies in its ability to translate the curvy language of polar coordinates into the strict grid of Cartesian coordinates. This process allows us to visualize complex shapes on a familiar coordinate plane and solve polar equations using the tools we already know.

To perform this translation, we use two key conversion formulas: x = r*cos(θ) and y = r*sin(θ). By applying these formulas to our polar equation, we have created a bridge to the Cartesian world. Now, every (θ, r) point in polar form can be mapped to an (x, y) point on the Cartesian plane. This straightforward conversion empowers students to unlock the visual puzzles posed by polar equations and to bring their geometric shapes to life with ease.
Plotting Polar Equations
The act of plotting polar equations is much akin to connecting the dots, but with a twist. As we venture into plotting the polar equation r2 = 1/θ, we must remember that each point we place corresponds to a specific angle θ and calculated radius r. By plotting points for various values of θ, as demonstrated in the step-by-step solution, we lay down the breadcrumbs that will soon form the path of our graph.

After plotting enough points, these individual markers begin to suggest the outline of a curve - our graph in this case resembles an outward spiral. Connecting these points with a smooth curve will unveil the intricate dance between the angular position and radial distance defined by our original equation. It's this disciplined approach to plotting 鈥� calculate, mark, and connect 鈥� that will consistently guide students through the graphing of even the most complex polar equations.

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Most popular questions from this chapter

If \(a>b>0,\) then the eccentricity of the ellipse $$\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1 \quad \text { or } \quad \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$$ is the number \(\frac{\sqrt{a^{2}-b^{2}}}{a} .\) Find the eccentricity of the ellipse whose equation is given. $$\frac{x^{2}}{18}+\frac{y^{2}}{25}=1$$

If \(a>b>0,\) then the eccentricity of the ellipse $$\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1 \quad \text { or } \quad \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$$ is the number \(\frac{\sqrt{a^{2}-b^{2}}}{a} .\) Find the eccentricity of the ellipse whose equation is given. $$\frac{(x-3)^{2}}{10}+\frac{(y-9)^{2}}{40}=1$$

Explain why the following symmetry tests for the graphs of polar equations are valid. (a) If replacing \(\theta\) by \(-\theta\) produces an equivalent equation, then the graph is symmetric with respect to the line \(\theta=0\) (the \(x\) -axis). (b) If replacing \(\theta\) by \(\pi-\theta\) produces an equivalent equation, then the graph is symmetric with respect to the line \(\theta=\pi / 2(\text { the } y\) -axis ) (c) If replacing \(r\) by \(-r\) produces an equivalent equation, then the graph is symmetric with respect to the origin.

Use a calculator in degree mode and assume that air resistance is negligible. A golf ball is hit off the ground at an angle of \(\theta\) degrees with an initial velocity of 100 feet per second. (a) Graph the path of the ball when \(\theta=30^{\circ}\) and when \(\theta=60^{\circ} .\) In which case does the ball land farthest away? (b) Do part (a) when \(\theta=25^{\circ}\) and \(\theta=65^{\circ}\) (c) Experiment further, and make a conjecture as to the results when the sum of the two angles is \(90^{\circ} .\) (d) Prove your conjecture algebraically. [Hint: Find the value of \(t\) at which a ball hit at angle \(\theta\) hits the ground (which occurs when \(y=0\) ); this value of \(t\) will be an expression involving \(\theta .\) Find the corresponding value of \(x\) (which is the distance of the ball from the starting point). Then do the same for an angle of \(90^{\circ}-\theta\) and use the cofunction identities (in degrees) to show that you get the same value of \(x .]\)

(a) Graph the curve given by \(x=3 \sin 2 t \quad\) and \(\quad y=2 \cos k t \quad(0 \leq t \leq 2 \pi)\) when \(k=1,2,3,4 .\) Use the window with \(-3.5 \leq x \leq\) 3.5 and \(-2.5 \leq y \leq 2.5\) and \(t\) -step \(=\pi / 30\) (b) Predict the shape of the graph when \(k=5,6,7,8 .\) Verify your predictions graphically.
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