Question

match each graph with its equation
a. $r = 1 + 2\sin(\theta)$
b. $r = 3\cos(3\theta)$
c. $r = 1 + 2\cos(\theta)$
d. $r = 3\sin(2\theta)$
e. $r = 3\cos(2\theta)$

Explanation:

• For polar equations of the form \(r = a\sin(n\theta)\) or \(r=a\cos(n\theta)\):
• If \(n\) is even, the number of petals is \(2n\). If \(n\) is odd, the number of petals is \(n\).
• For \(r = 3\sin(2\theta)\), \(n = 2\) (even), so there are \(2\times2=4\) petals. The graph of \(r = a\sin(n\theta)\) is symmetric about the \(y\) - axis (since \(\sin(-\theta)=-\sin(\theta)\)).
• For \(r = 3\cos(2\theta)\), \(n = 2\) (even), there are \(2\times 2 = 4\) petals. The graph of \(r=a\cos(n\theta)\) is symmetric about the \(x\) - axis (since \(\cos(-\theta)=\cos(\theta)\)).
• For limaçons of the form \(r=a + b\sin\theta\) or \(r=a + b\cos\theta\):
• If \(\frac{a}{b}<1\), the limaçons have an inner loop. \(r = 1+2\sin\theta\) is symmetric about the \(y\) - axis (\(\sin(-\theta)=-\sin(\theta)\)), and \(r = 1 + 2\cos\theta\) is symmetric about the \(x\) - axis (\(\cos(-\theta)=\cos(\theta)\)).

Answer:

• The first graph (with 4 petals symmetric about the \(y\) - axis) matches \(d. r = 3\sin(2\theta)\)
• The second graph (with 4 petals symmetric about the \(x\) - axis) matches \(e. r=3\cos(2\theta)\)
• The third graph (a limacon with an inner loop symmetric about the \(y\) - axis) matches \(a. r = 1+2\sin(\theta)\)