Question

Graph the following equations. Then use arrows and labeled points to indicate how the curve is generated as \(\theta\) increases from 0 to \(2 \pi\). $$r=\frac{1}{1+2 \cos \theta}$$

Short answer

Answer: The curve of the given polar equation resembles a lemniscate. As \(\theta\) increases, the curve starts at the origin, moves rightwards & upwards, reaches its maximum y-coordinate, moves leftwards & downwards, reaches its maximum negative x-coordinate, continues leftwards & downwards, reaches its minimum y-coordinate, and finally moves back to the origin.

Step-by-step solution

1. Convert Polar to Cartesian Equation

First of all, we should convert the polar equation into a Cartesian equation. To do that, we will use the following conversion formulas: \( x = r\cos\theta\) and \(y = r\sin\theta\) Recall that our polar equation is: \(r = \frac{1}{1+2\cos\theta}\) Multiply both sides by \(r(1+2\cos\theta)\): \(r^2 (1+2\cos\theta) = 1\) Now, substitute the conversion formulas: \((x^2+y^2) (1+2\frac{x}{x^2+y^2}) = 1\) Now, we have a Cartesian equation: \((x^2+y^2)(x^2+y^2+2x) = x^2+y^2\)

2. Plot the Curve

To plot the curve and visualize how it's generated, we can use the Cartesian equation and plot it using a graphing software/tool or do it manually. It should give us a graph similar to that of a lemniscate.

3. Show Curve Generation with Arrows and Labeled Points

To show how the curve is generated, we can use arrows and labeled points on the graph. As \(\theta\) increases from \(0\) to \(2\pi\), the curve starts at the origin, and following this sequence, we can place our arrows: 1. Start at \((0,0)\) when \(\theta = 0\) 2. Move rightwards and then upwards for \(\theta\in(0, \pi)\) 3. Reach maximum y-coordinate at \(\theta = \pi/2\) 4. Move downwards and continue leftwards for \(\theta\in(\pi/2, \pi)\) 5. Reach maximum negative x-coordinate (y=0) at \(\theta = \pi\) 6. Move leftwards and then downwards for \(\theta\in(\pi, 3\pi/2)\) 7. Reach minimum y-coordinate at \(\theta = 3\pi/2\) 8. Move upwards and continue rightwards for \(\theta\in (3\pi/2, 2\pi)\) 9. We return to the origin at \(\theta = 2\pi\) Now that we have indicated the direction and points, the curve generation process should be visually clear on the plot.