Answer :
Let's solve the problem step by step.
Given:
- Diameter of the can: 4 inches
- Time taken to roll: 5 seconds
- Number of rotations: 4
To find the suitable function [tex]\( h(t) \)[/tex] representing the price tag's height relative to the table top, we'll need to determine several key factors:
### 1. Determine the radius and amplitude:
The diameter of the can is 4 inches, so the radius [tex]\( r \)[/tex] is:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{4}{2} = 2 \text{ inches} \][/tex]
The amplitude [tex]\( A \)[/tex] of the function will be equal to the radius:
[tex]\[ A = r = 2 \text{ inches} \][/tex]
### 2. Find the angular frequency ([tex]\(\omega\)[/tex]):
The angular frequency [tex]\(\omega\)[/tex] relates to how many radians per second the can rotates. This is given by:
[tex]\[ \omega = \frac{2 \pi \times (\text{number of rotations})}{\text{time}} \][/tex]
Here, the number of rotations is 4 and the time is 5 seconds:
[tex]\[ \omega = \frac{2 \pi \times 4}{5} = \frac{8 \pi}{5} \text{ radians per second} \][/tex]
### 3. Determine the vertical shift [tex]\( C \)[/tex]:
Given the can rolls on its side level to the table, the center of the can (which is 2 inches above the table) represents the midline about which the height function oscillates. Therefore, the vertical shift [tex]\( C \)[/tex] is equal to the radius:
[tex]\[ C = 2 \text{ inches} \][/tex]
### 4. Construct the possible functions:
Given the information, we want a trigonometric function involving cosine or sine with the above-calculated parameters. Specifically, the correct function should reflect these values:
[tex]\[ h(t) = A \cos(\omega t) + C \quad \text{ or } \quad h(t) = A \sin(\omega t) + C \][/tex]
Hence, combining the values we found:
[tex]\[ A = 2, \quad \omega = \frac{8 \pi}{5}, \quad C = 2 \][/tex]
Now, reviewing the options provided:
- Option A: [tex]\( h(t) = -2 \cos \left(\frac{8 \pi}{5} t\right) + 2 \)[/tex]
- Amplitude: 2
- [tex]\(\omega\)[/tex]: [tex]\( \frac{8 \pi}{5} \)[/tex]
- Vertical shift: 2
This matches our parameters perfectly (ignoring the negative sign on the cosine, which just indicates a phase shift).
- Option B: [tex]\( h(t) = 2 \cos \left(\frac{5 \pi}{2} t\right) + 2 \)[/tex]
- [tex]\(\omega\)[/tex] is not equal to [tex]\(\frac{8 \pi}{5}\)[/tex]
- Option C: [tex]\( h(t) = 2 \sin \left(\frac{8 \pi}{5} t\right) + 2 \)[/tex]
- While the [tex]\(\omega\)[/tex] and amplitude match, we seek a cosine model.
- Option D: [tex]\( h(t) = -2 \sin \left(\frac{5 \pi}{2} t\right) + 2 \)[/tex]
- [tex]\(\omega\)[/tex] is not equal to [tex]\(\frac{8 \pi}{5}\)[/tex]
Thus, the correct function that represents the price tag's height relative to the table top after it has been rolling for [tex]\( t \)[/tex] seconds is:
[tex]\[ \boxed{A. \, h(t) = -2 \cos \left(\frac{8 \pi}{5} t\right) + 2} \][/tex]
Given:
- Diameter of the can: 4 inches
- Time taken to roll: 5 seconds
- Number of rotations: 4
To find the suitable function [tex]\( h(t) \)[/tex] representing the price tag's height relative to the table top, we'll need to determine several key factors:
### 1. Determine the radius and amplitude:
The diameter of the can is 4 inches, so the radius [tex]\( r \)[/tex] is:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{4}{2} = 2 \text{ inches} \][/tex]
The amplitude [tex]\( A \)[/tex] of the function will be equal to the radius:
[tex]\[ A = r = 2 \text{ inches} \][/tex]
### 2. Find the angular frequency ([tex]\(\omega\)[/tex]):
The angular frequency [tex]\(\omega\)[/tex] relates to how many radians per second the can rotates. This is given by:
[tex]\[ \omega = \frac{2 \pi \times (\text{number of rotations})}{\text{time}} \][/tex]
Here, the number of rotations is 4 and the time is 5 seconds:
[tex]\[ \omega = \frac{2 \pi \times 4}{5} = \frac{8 \pi}{5} \text{ radians per second} \][/tex]
### 3. Determine the vertical shift [tex]\( C \)[/tex]:
Given the can rolls on its side level to the table, the center of the can (which is 2 inches above the table) represents the midline about which the height function oscillates. Therefore, the vertical shift [tex]\( C \)[/tex] is equal to the radius:
[tex]\[ C = 2 \text{ inches} \][/tex]
### 4. Construct the possible functions:
Given the information, we want a trigonometric function involving cosine or sine with the above-calculated parameters. Specifically, the correct function should reflect these values:
[tex]\[ h(t) = A \cos(\omega t) + C \quad \text{ or } \quad h(t) = A \sin(\omega t) + C \][/tex]
Hence, combining the values we found:
[tex]\[ A = 2, \quad \omega = \frac{8 \pi}{5}, \quad C = 2 \][/tex]
Now, reviewing the options provided:
- Option A: [tex]\( h(t) = -2 \cos \left(\frac{8 \pi}{5} t\right) + 2 \)[/tex]
- Amplitude: 2
- [tex]\(\omega\)[/tex]: [tex]\( \frac{8 \pi}{5} \)[/tex]
- Vertical shift: 2
This matches our parameters perfectly (ignoring the negative sign on the cosine, which just indicates a phase shift).
- Option B: [tex]\( h(t) = 2 \cos \left(\frac{5 \pi}{2} t\right) + 2 \)[/tex]
- [tex]\(\omega\)[/tex] is not equal to [tex]\(\frac{8 \pi}{5}\)[/tex]
- Option C: [tex]\( h(t) = 2 \sin \left(\frac{8 \pi}{5} t\right) + 2 \)[/tex]
- While the [tex]\(\omega\)[/tex] and amplitude match, we seek a cosine model.
- Option D: [tex]\( h(t) = -2 \sin \left(\frac{5 \pi}{2} t\right) + 2 \)[/tex]
- [tex]\(\omega\)[/tex] is not equal to [tex]\(\frac{8 \pi}{5}\)[/tex]
Thus, the correct function that represents the price tag's height relative to the table top after it has been rolling for [tex]\( t \)[/tex] seconds is:
[tex]\[ \boxed{A. \, h(t) = -2 \cos \left(\frac{8 \pi}{5} t\right) + 2} \][/tex]
