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To solve the problem of determining which graph represents the polar equation [tex]\( r = 6 \sin(6\theta) \)[/tex] for the interval where [tex]\( \theta \)[/tex] ranges from [tex]\( \frac{3}{2} \)[/tex] to [tex]\( \frac{12\pi}{12} \)[/tex], let’s go step-by-step through the process.
### Step-by-Step Solution:
1. Understand the Equation:
The given polar equation is [tex]\( r = 6 \sin(6\theta) \)[/tex]. This means that the radius [tex]\( r \)[/tex] depends on the sine function of [tex]\( 6\theta \)[/tex].
2. Range of [tex]\( \theta \)[/tex]:
We need to determine [tex]\( \theta \)[/tex]’s range:
- Starts at [tex]\( \frac{3}{2} \)[/tex] (this is a bit unconventional for [tex]\( \theta \)[/tex] since typically we measure angles in radians in standard mathematical contexts, but let’s assume it’s radians).
- Ends at [tex]\( \frac{12\pi}{12} = \pi \)[/tex].
3. Behavior of [tex]\( r = 6 \sin(6\theta) \)[/tex]:
- The sine function [tex]\( \sin(6\theta) \)[/tex] will have a periodic pattern since [tex]\( \theta \)[/tex] ranges from [tex]\( \frac{3}{2} \)[/tex] to [tex]\( \pi \)[/tex].
- Within every [tex]\( \pi \)[/tex] radians, the sine wave will complete multiple cycles because of the factor [tex]\( 6 \)[/tex] (specifically 3 cycles, because [tex]\( 6\theta \)[/tex] will complete one full sine wave cycle every [tex]\( \frac{\pi}{3} \)[/tex] radians).
4. Characteristics of the Plot:
- As [tex]\( \theta \)[/tex] increases from [tex]\( \frac{3}{2} \)[/tex] to [tex]\( \pi \)[/tex], the function [tex]\( r = 6 \sin(6\theta) \)[/tex] will oscillate between -6 and 6 as sine varies between -1 and 1.
- Since sine waves through [tex]\( 6\theta \)[/tex], the function will have points where [tex]\( r \)[/tex] is zero (nodes), positive, and negative, leading to patterns in the radial plot.
5. Graph Features:
- Petals:
- The form [tex]\( r = 6 \sin(k\theta) \)[/tex] generally produces a rose curve with [tex]\( k \)[/tex] petals if [tex]\( k \)[/tex] is an odd number and [tex]\( 2k \)[/tex] petals if [tex]\( k \)[/tex] is even. Here, [tex]\( k = 6 \therefore 12 \)[/tex] petals would be expected for the full [tex]\( 2\pi \)[/tex] range.
- However, since [tex]\( \theta \)[/tex] ranges only from [tex]\( \frac{3}{2} \)[/tex] to [tex]\( \pi \)[/tex], you’ll see part of the pattern of 12 petals.
### Determining the Graph:
Based on the above characteristics, you need to find the graph that matches this description:
1. Some patterns of oscillating petals.
2. A polar plot that shows petals within the specified range.
Without access to see the actual graphs labeled "a" and "b", I cannot visually determine which one is correct. However, you should look for a graph showing:
- Approximately 3 complete oscillations due to the [tex]\( 6 \theta \)[/tex].
- Petals in a repeated pattern but only over the specific [tex]\( \theta \)[/tex] interval given.
Thus, select the graph (a or b) that closely matches these criteria of cyclic petal formations consistent with the sine function's behavior in the specified interval.
### Step-by-Step Solution:
1. Understand the Equation:
The given polar equation is [tex]\( r = 6 \sin(6\theta) \)[/tex]. This means that the radius [tex]\( r \)[/tex] depends on the sine function of [tex]\( 6\theta \)[/tex].
2. Range of [tex]\( \theta \)[/tex]:
We need to determine [tex]\( \theta \)[/tex]’s range:
- Starts at [tex]\( \frac{3}{2} \)[/tex] (this is a bit unconventional for [tex]\( \theta \)[/tex] since typically we measure angles in radians in standard mathematical contexts, but let’s assume it’s radians).
- Ends at [tex]\( \frac{12\pi}{12} = \pi \)[/tex].
3. Behavior of [tex]\( r = 6 \sin(6\theta) \)[/tex]:
- The sine function [tex]\( \sin(6\theta) \)[/tex] will have a periodic pattern since [tex]\( \theta \)[/tex] ranges from [tex]\( \frac{3}{2} \)[/tex] to [tex]\( \pi \)[/tex].
- Within every [tex]\( \pi \)[/tex] radians, the sine wave will complete multiple cycles because of the factor [tex]\( 6 \)[/tex] (specifically 3 cycles, because [tex]\( 6\theta \)[/tex] will complete one full sine wave cycle every [tex]\( \frac{\pi}{3} \)[/tex] radians).
4. Characteristics of the Plot:
- As [tex]\( \theta \)[/tex] increases from [tex]\( \frac{3}{2} \)[/tex] to [tex]\( \pi \)[/tex], the function [tex]\( r = 6 \sin(6\theta) \)[/tex] will oscillate between -6 and 6 as sine varies between -1 and 1.
- Since sine waves through [tex]\( 6\theta \)[/tex], the function will have points where [tex]\( r \)[/tex] is zero (nodes), positive, and negative, leading to patterns in the radial plot.
5. Graph Features:
- Petals:
- The form [tex]\( r = 6 \sin(k\theta) \)[/tex] generally produces a rose curve with [tex]\( k \)[/tex] petals if [tex]\( k \)[/tex] is an odd number and [tex]\( 2k \)[/tex] petals if [tex]\( k \)[/tex] is even. Here, [tex]\( k = 6 \therefore 12 \)[/tex] petals would be expected for the full [tex]\( 2\pi \)[/tex] range.
- However, since [tex]\( \theta \)[/tex] ranges only from [tex]\( \frac{3}{2} \)[/tex] to [tex]\( \pi \)[/tex], you’ll see part of the pattern of 12 petals.
### Determining the Graph:
Based on the above characteristics, you need to find the graph that matches this description:
1. Some patterns of oscillating petals.
2. A polar plot that shows petals within the specified range.
Without access to see the actual graphs labeled "a" and "b", I cannot visually determine which one is correct. However, you should look for a graph showing:
- Approximately 3 complete oscillations due to the [tex]\( 6 \theta \)[/tex].
- Petals in a repeated pattern but only over the specific [tex]\( \theta \)[/tex] interval given.
Thus, select the graph (a or b) that closely matches these criteria of cyclic petal formations consistent with the sine function's behavior in the specified interval.
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