Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(1-32\), sketch the graph of the given polar equation and verify its symmetry (see Examples 1-3). \(r=3 \sin 5 \theta\) (five-leaved rose)

Step-by-Step Solution

Answer

The graph is a five-leaved rose symmetrically about the origin.

Step 1: Understand the Polar Equation

The given equation is a polar equation: \(r = 3 \sin 5\theta\). This type of equation often results in a graph with symmetrical petals or leaves. Here, \(r\) is the radial distance from the origin and \(\theta\) is the angle from the positive x-axis. The factor \(5\) in the function indicates that it will have \(5\) petals, since it is of the form \(r = a \sin (n\theta)\). The amplitude, \(3\), dictates the maximum length of each petal.

Step 2: Determine Symmetry

To verify the symmetry of the graph, we test for the three possible types of symmetry in polar graphs: symmetry about the polar axis (x-axis), symmetry about the line \(\theta = \frac{\pi}{2}\) (y-axis), and symmetry about the pole (origin). For a rose curve \(r = a\sin(n\theta)\), it will have symmetry about the origin when \(n\) is odd. Since \(5\) is odd, this graph is symmetrical with respect to the origin.

Step 3: Sketch the Graph

Start by plotting key points. At \(\theta = 0\), \(r = 0\) because \(\sin(0) = 0\). Compute values for \(\theta = \frac{\pi}{10}, \frac{2\pi}{10}, \frac{3\pi}{10}, \dots, \frac{10\pi}{10}\) and plot \(r = 3 \sin 5\theta\). Points such as \(r = 3\sin(\frac{5\pi}{10}) = 3\sin(\frac{\pi}{2}) = 3\) indicate where the petals extend from the origin. Continue plotting as \(\theta\) increases to trace out each petal.

Step 4: Characteristics of the Graph

Having plotted the points, the graph should display 5 equally spaced and sized petals of length \(3\). Since the value of \(r\) is negative when \(\theta = \frac{\pi}{10}\) through \(\frac{5\pi}{10}\) (producing a full cycle), the graph extends these petals symmetrically through the origin. Each petal forms as \(\theta\) ranges through \([0, \frac{2\pi}{5}], [\frac{2\pi}{5}, \frac{4\pi}{5}], \) etc. for a complete 360-degree rotation.

Key Concepts

Graph Symmetry

In polar equations, like the one given in the exercise \( r = 3 \sin(5\theta) \), understanding graph symmetry is crucial. Graph symmetry in polar coordinates entails checking if the graph looks the same under certain transformations.

There are three main types of symmetries to consider:

• Symmetry about the polar axis (x-axis)
• Symmetry about the line \( \theta = \frac{\pi}{2} \) (y-axis)
• Symmetry about the origin (pole)

To test for symmetry about the origin, substitute \( \theta \) with \( \theta + \pi \). In our equation, this reflects across the origin when \( n \) (in \( n\theta \)) is odd, as is the case with \( 5 \). This ensures the rose curve looks the same when you rearrange around the origin, making it beautifully symmetrical with 5 petals.

This concept is useful as it simplifies graphing, ensuring that once you understand the symmetry, you only need to sketch a portion of the graph, and the rest will mirror or extend from it.

Polar Coordinates

Polar coordinates provide a unique way to define points on a plane. Unlike Cartesian coordinates that use \(x\) and \(y\), a point in polar coordinates is defined by:

• \( r \) - the radial distance from the origin
• \( \theta \) - the angle from the positive x-axis

In the equation \( r = 3 \sin(5\theta) \), each point is determined by converting \( \theta \) to an angle and calculating the corresponding \( r \) value. For \( \theta = 0 \), \( r = 0 \), indicating the point is at the origin.

As \( \theta \) varies between \( 0 \) and \( 2\pi \), all possible points of a petal are plotted. High values of \( r \) show where the petals of our rose curve peak, giving the graph its distinctive shape.

Understanding how \( \theta \) and \( r \) interact allows for precise plotting of the full graph, showing its beauty and symmetry elegantly.

Rose Curves

Rose curves are fascinating sets of polar graphs that resemble flowers. The given equation \( r = 3 \sin(5\theta) \) is a great example. In rose curves:

• \( n \) in \( a\sin(n\theta) \) or \( a\cos(n\theta) \) affects how many petals the curve has.
• \( a \) affects the length of these petals.

If \( n \) is odd, as in this case, the rose curve has \( n \) petals. When \( n \) is even, it has \( 2n \) petals. Thus, for \( r = 3\sin(5\theta) \), you see exactly 5 petals, stretched out symmetrically.
Each petal is traced out as \( \theta \) sweeps through its range from 0 to \( 2\pi \). The complete graph features petals that start and end at the origin, which is intrinsic to rose curves' visually appealing nature.
Rose curves demonstrate how simple trigonometric modifications in polar equations can produce complex and beautiful graphs through basic properties.