Problem 85 - Algebra and Trigonometry

Question

Use a graphing utility to graph \(r=1+2 \sin n \theta\) for \(n=1,2,3,4,5,\) and \(6 .\) Use a separate viewing screen for each of the six graphs. What is the pattern for the number of large and small petals that occur corresponding to each value of \(n ?\) How are the large and small petals related when \(n\) is odd and when \(n\) is even?

Step-by-Step Solution

Answer

In the polar graphs of \(r=1+2 \sin n \theta\), for odd \(n\) there are \(2n\) equal-sized petals, whereas for even \(n\) there are \(n\) petals, alternating between large and small in size, starting with a large petal on the positive horizontal axis.

Step 1: Graphing the Equation for n=1, 2,...6

Firstly, use a graphing utility to plot the function \(r=1+2 \sin n \theta\) on polar coordinates using separate screens for each value of \(n\) from 1 to 6. Observe each graph closely to identify the shape and structure of each plot, also the number of petals.

Step 2: Identify the Pattern

Look for patterns in the number of petals, their sizes and any shift in orientation. Typically, when \(n\) is even, there will be \(n\) identical petals and when \(n\) is odd there will be \(2n\) petals. Petals tend to alternate between large and small in size, starting with a large petal on the positive horizontal axis.

Step 3: Find the Relation between Petals Size

Analyze the relationship between the size of the petals when \(n\) is odd and when \(n\) is even. Typically, when \(n\) is odd, there are \(2n\) petals that are equal in size. Meanwhile, for even \(n\), there are \(n\) petals, with each petal alternating between large and small, starting with a large petal on the positive horizontal axis.