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Chapter 6: Problem 27

Test for symmetry and then graph each polar equation. $$r=4 \sin 3 \theta$$

Short Answer

Expert verified
The polar function \( r=4 \sin 3\theta \) is symmetric with respect to the origin. The graph starts from origin and has maxima at \( r = 4 \) and minima at \( r = -4 \). It completes three cycles over a period of \( \frac{2 \pi}{3} \).

Step by step solution

01

Test for Symmetry

There are three types of symmetry in polar equations: symmetry about the x-axis, symmetry about the y-axis, and symmetry about the origin. Test the given function for these three types. The polar equation \( r=4 \sin 3 \theta\) resembles the form \( r=a \sin n\theta \). This form of polar equation is symmetric with respect to the origin, because when we substitute \( -\theta \) for \( \theta \), the equation is unchanged. That is, \( r=4 \sin 3(-\theta) \) equals \( r= -4 \sin 3\theta = -r\). So, it is symmetric with respect to the origin.
02

Determine properties for graph

The amplitude of the function is 4 and the period of the function is \( \frac{2 \pi}{3} \). The function completes 3 cycles in an interval of \( 2\pi \). The maxima occur when \( \sin(3 \theta) = 1 \), giving \( r = 4 \), and the minima occur when \( \sin(3 \theta) = -1 \), giving \( r = -4 \). When theta equals to 0, r equals to 0, hence the graph starts from origin.
03

Graph the polar equation

Use the properties of the function and sketch the polar graph. Begin at \( \theta = 0 \) at the origin. Then find the points corresponding to the values of r for each value of theta through one period ranging from 0 to \( \frac{2 \pi}{3} \) and plot those points. Take note of the symmetry with respect to the origin and graph the function through three full cycles to complete the graph.

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Most popular questions from this chapter

Graph \(r_{1}\) and \(r_{2}\) in the same polar coordinate system. What is the relationship between the two graphs? $$r_{1}=4 \cos 2 \theta, r_{2}=4 \cos 2\left(\theta-\frac{\pi}{4}\right)$$

Explain how to find the dot product of two vectors.

Let $$\mathbf{u}=-\mathbf{i}+\mathbf{j}, \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j}, \quad \text { and } \quad \mathbf{w}=-5 \mathbf{j}$$ Find each specified scalar or vector. $$4 \mathbf{u} \cdot(5 \mathbf{v}-3 \mathbf{w})$$

Find \(\text {pro}_{\mathbf{w}} \mathbf{V}\) Then decompose v into two vectors, \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2},\) where \(\mathbf{v}_{1}\) is parallel to \(\mathbf{w}\) and \(\mathbf{v}_{2}\) is orthogonal to \(\mathbf{w}.\) $$\mathbf{v}=3 \mathbf{i}-2 \mathbf{j}, \quad \mathbf{w}=2 \mathbf{i}+\mathbf{j}$$

Determine whether v and w are parallel, orthogonal, or neither. $$\mathbf{v}=3 \mathbf{i}-5 \mathbf{j}, \quad \mathbf{w}=6 \mathbf{i}+10 \mathbf{j}$$
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