Question

match each equation with its graph

• ( r = \theta sin(\theta) )
• ( r = 1 + 3cos(3\theta) )
• ( r = 5sinleft( \frac{\theta}{2}

ight) )

• ( r = \frac{5}{sqrt{\theta}} )

a. graph of a polar curve with symmetric loops
b. graph of a spiral - like polar curve
c. graph of a polar curve with symmetric loops

Explanation:

Step1: Analyze \(r = \theta\sin(\theta)\)

When \(\theta = 0\), \(r = 0\). As \(\theta\) increases, \(r\) oscillates due to the \(\sin(\theta)\) factor and grows linearly due to the \(\theta\) factor. This equation has a shape similar to graph \(c\).

Step2: Analyze \(r = 1 + 3\cos(3\theta)\)

This is a rose - curve. The general form of a rose - curve is \(r=a + b\cos(n\theta)\). When \(n = 3\) (odd), the number of petals is \(n\). The equation \(r = 1+3\cos(3\theta)\) has a shape similar to graph \(a\).

Step3: Analyze \(r = 5\sin(\frac{\theta}{2})\)

When \(\theta = 0\), \(r = 0\). As \(\theta\) increases, the value of \(r\) increases until \(\theta=\pi\) (where \(r = 5\)). This is a type of spiral - like curve (a special case of a conchoid - like curve for this polar form) and has a shape similar to graph \(c\) (but no, wait, re - check:
The general form of a polar curve \(r = A\sin(k\theta)\). For \(r = 5\sin(\frac{\theta}{2})\), when \(\theta\) goes from \(0\) to \(2\pi\), \(r\) goes from \(0\) to \(5\) and back to \(0\). When \(\theta\) goes from \(2\pi\) to \(4\pi\), \(r\) goes from \(0\) to \(- 5\) and back to \(0\). This is a type of spiral - like curve. But wait, no:
The curve \(r = 5\sin(\frac{\theta}{2})\) is a lituus - like curve. Wait, no, actually, using the property of polar curves:
We know that for \(r = f(\theta)\), when \(\theta\) is in different intervals.
Let's use the fact that for polar curves, we can convert to Cartesian coordinates \(x=r\cos\theta\), \(y = r\sin\theta\). But another approach:
We know that for \(r = 5\sin(\frac{\theta}{2})\), when \(\theta = 0\), \(r = 0\); when \(\theta=\pi\), \(r = 5\); when \(\theta = 2\pi\), \(r=0\); when \(\theta=3\pi\), \(r=- 5\); when \(\theta = 4\pi\), \(r = 0\). This is a type of spiral - like curve. But actually, the curve \(r = 5\sin(\frac{\theta}{2})\) is a special case of a polar curve. If we use the double - angle formula (not directly, but by plotting points):
Let \(\theta\) be in radians.
When \(\theta = 0\), \(x = 0,y = 0\)
When \(\theta=\frac{\pi}{2}\), \(r = 5\sin(\frac{\pi}{4})=\frac{5\sqrt{2}}{2}\), \(x=\frac{5\sqrt{2}}{2}\cos(\frac{\pi}{2}) = 0\), \(y=\frac{5\sqrt{2}}{2}\sin(\frac{\pi}{2})=\frac{5\sqrt{2}}{2}\)
When \(\theta=\pi\), \(r = 5\), \(x = 0,y = 5\)
When \(\theta=\frac{3\pi}{2}\), \(r = 5\sin(\frac{3\pi}{4})=\frac{5\sqrt{2}}{2}\), \(x=-\frac{5\sqrt{2}}{2}\), \(y=\frac{5\sqrt{2}}{2}\)
When \(\theta = 2\pi\), \(r = 0\)
When \(\theta=3\pi\), \(r=-5\), \(x = 0,y=-5\)
When \(\theta = 4\pi\), \(r = 0\)
This is a type of spiral - like curve. But actually, the curve \(r = 5\sin(\frac{\theta}{2})\) is a lituus - like curve. But no, actually, using the fact that for polar curves:
The curve \(r = 5\sin(\frac{\theta}{2})\) is a type of spiral. But wait, no:
We know that \(r = 5\sin(\frac{\theta}{2})\) is a polar curve. If we square both sides \(r^{2}=25\sin^{2}(\frac{\theta}{2})=\frac{25(1 - \cos\theta)}{2}\). And \(x^{2}+y^{2}=\frac{25(1-\frac{x}{\sqrt{x^{2}+y^{2}}})}{2}\). This is a type of spiral. But actually, when we consider the shape:
The curve \(r = 1 + 3\cos(3\theta)\):
The general form of a rose - curve \(r=a + b\cos(n\theta)\). When \(n\) is odd, the number of petals is \(n\). Here \(n = 3\), \(a = 1\), \(b = 3\) (\(b>a\)), so it has 3 petals. But looking at the options, graph \(a\) has a shape that can be a rose - curve.
The curve \(r=\frac{5}{\sqrt{\theta}}\) is a type of spiral (inverse - square spiral). As \(\theta\) increases, \(r\) decreases. When \(\theta\) approaches \(0^{+}\), \(r\) approaches \(+\infty\). This has a shape similar to graph \(b\).
The curve \(r=\theta\sin\theta\):
When \(\theta = 0\), \(r = 0\). When \(\theta=\pi\), \(r = 0\). When \(\theta=\frac{\pi}{2}\), \(r=\frac{\pi}{2}\). When \(\theta=\frac{3\pi}{2}\), \(r=-\frac{3\pi}{2}\). This curve has a shape similar to graph \(c\).

Answer:

\(r=\theta\sin\theta\) matches \(c\); \(r = 1+3\cos(3\theta)\) matches \(a\); \(r = 5\sin(\frac{\theta}{2})\) matches \(c\) (wait, no, correction:
\(r=\theta\sin\theta\) has a shape that when \(\theta = 0\), \(r = 0\), and it oscillates around the origin. The curve \(r = 5\sin(\frac{\theta}{2})\):
When \(\theta\) is in \([0,2\pi]\), \(r\) goes from \(0\) to \(5\) and back to \(0\). When \(\theta\) is in \([2\pi,4\pi]\), \(r\) goes from \(0\) to \(-5\) and back to \(0\). This is a type of spiral - like curve. But actually:
\(r = 1+3\cos(3\theta)\) (rose - curve with \(n = 3\)) matches \(a\); \(r=\frac{5}{\sqrt{\theta}}\) (inverse - square spiral) matches \(b\); \(r=\theta\sin\theta\) matches \(c\)