Problem 32 - The Calculus with Analytic Geometry

Question

Draw a sketch of the graph of the given equation.\(r^{2}=-4 \sin 2 \theta\) (lemniscate)

Step-by-Step Solution

Answer

Plot key points and draw a smooth curve to form a figure-eight, which is the shape of the lemniscate.

Step 1: Identify the Equation Type

The given equation is in polar form: \[r^{2}=-4 \sin 2 \theta\]. This represents a type of lemniscate, which is a figure-eight or infinity-shaped curve.

Step 2: Analyze the Symmetry

Note that the sine function indicates symmetry about the origin. Since \(\sin 2\theta\) is an even function, the curve will be symmetrical about both the x-axis and the y-axis.

Step 3: Convert Equation to Cartesian Coordinates

The polar equation can be converted to Cartesian coordinates for better understanding. Using \(r\sin\theta = y\) and \(r\cos\theta = x\), express \(\theta\) in terms of \(x\) and \(y\). However, for sketching purposes, plotting points in polar coordinates may provide a quicker path.

Step 4: Find Key Points

Identify key points where \(\theta = 0, \frac{\pi}{4}, \frac{\pi}{2}, \text{and so on}\). Solve for \(r\) at these points: For \(\theta = 0\), \[r^{2}=-4 \sin 0 = 0 \rightarrow r = 0 \]For \(\theta = \frac{\pi}{4}\), \[r^{2}=-4 \sin \frac{\pi}{2} = -4 \rightarrow no \ real \ solutions \]*(since \(r^2\) must be non-negative). Similarly, solve for other key points.

Step 5: Plot the Points

Using the solutions from step 4, plot the points on polar graph paper. For instance, if \(\theta = \frac{\pi}{4}, \theta =\frac{3\pi}{4}\) leads to imaginary points, skip. You will notice non-trivial points fit a lemniscate shape.

Step 6: Draw the Lemniscate

Once key points are plotted, draw a smooth curve through these points to complete the lemniscate shape (Figure-eight).

Key Concepts

lemniscate

A lemniscate is a fascinating curve in polar coordinates that resembles a figure-eight or infinity sign (∞). It's important to identify such curves because they often have unique properties and symmetry.
The equation given, \[r^{2} = -4 \sin 2 \theta\], describes a lemniscate. This specific form suggests that the curve will have a symmetrical, looping shape.
Key points to remember about lemniscates:

• They emerge from equations like \[r^{2} = a^{2} \sin 2 \theta\] or \[r^{2} = a^{2} \cos 2 \theta\].
• They usually form around the origin (0,0) in the polar coordinate system.
• The loops intersect in the middle, creating a central point.

In our problem, step 4 involves finding key points by solving for \(r\) with various \(\theta\) values. However, make sure \(r^2\) remains non-negative for real solutions.

symmetry in polar coordinates

Analyzing symmetry in polar coordinates is crucial for understanding the overall shape and properties of the graph.
For the equation \[r^{2} = -4 \sin 2 \theta\], look for symmetries to predict the complete graph.
Symmetry types to consider in polar coordinates are:

• Symmetry about the x-axis: If replacing \((r, \theta)\) with \((r, -\theta)\) yields the same equation, the graph is symmetric about the x-axis.
• Symmetry about the y-axis: If replacing \((r, \theta)\) with \((r, \pi - \theta)\) gives an equivalent equation, the graph is symmetric about the y-axis.
• Symmetry about the origin: If \((r, \theta)\) replaced by \((-r, \theta + \pi)\) results in the same equation, the graph is symmetric about the origin.

In our given equation, the sine function and the symmetry of \sin 2 \theta ensure that the graph will be symmetric about both the x-axis and the y-axis, leading to the distinct lemniscate shape.

converting polar to Cartesian

Converting from polar coordinates to Cartesian coordinates can provide a different perspective and facilitate graphing.
Here’s how to convert polar to Cartesian coordinates:

• Use the formulas \(x = r \cos \theta\) and \(y = r \sin \theta\).
• Replace \(r\) and \(\theta\) in the polar equation with their Cartesian equivalents.

For the equation \[r^{2} = -4 \sin 2 \theta\]:
Start with the polar identity for sine: \sin 2 \theta = 2 \sin \theta \cos \theta\.
Substitute \(r \sin \theta = y\) and \(r \cos \theta = x\):

\[r^{2} = -4(2 \sin \theta \cos \theta)\]
\[r^{2} = -8 \sin \theta \cos \theta\]
\[r^{2} = -8 \frac{y}{r} \frac{x}{r} \rightarrow r^{4} = -8xy\]
Thus, after simplifying: \[x^{2} + y^{2} = -2xy\].
This step shows that recognizing and converting coordinates is useful for dual perspectives on the graph.