Answer :
Certainly! Let's break down the problem and find the required values step-by-step.
Given the position function:
[tex]\[ s(t) = 8 \sin (5t) \][/tex]
We're asked to find the average velocities over specific intervals and then make a conjecture about the instantaneous velocity at [tex]\( t = \frac{\pi}{2} \)[/tex].
1. Average Velocities
The average velocity over the interval [tex]\([t_1, t_2]\)[/tex] can be calculated using the difference quotient:
[tex]\[ v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \][/tex]
Now, let's compute this for each given interval.
- Interval [tex]\(\left[\frac{\pi}{2}, \pi\right]\)[/tex]:
- [tex]\( t_1 = \frac{\pi}{2} \)[/tex]
- [tex]\( t_2 = \pi \)[/tex]
Using these:
- [tex]\( s(t_1) = 8 \sin \left(5 \cdot \frac{\pi}{2}\right) = 8 \sin \left(\frac{5\pi}{2}\right) = 8 \)[/tex]
- [tex]\( s(t_2) = 8 \sin \left(5 \cdot \pi\right) = 8 \sin (5\pi) = -8 \)[/tex]
Thus, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{-8 - 8}{\pi - \frac{\pi}{2}} = \frac{-16}{\frac{\pi}{2}} = \frac{-16 \cdot 2}{\pi} = -\frac{32}{\pi} \approx -10.1859 \][/tex]
- Interval [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.1\right]\)[/tex]:
- [tex]\( t_2 = \frac{\pi}{2} + 0.1 \)[/tex]
Using these:
- [tex]\( s(t_1) = 8 \)[/tex]
- [tex]\( s(t_2) = 8 \sin \left(5 \left(\frac{\pi}{2} + 0.1\right)\right) \approx -0.792584 \)[/tex]
Thus, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{-0.792584 - 8}{0.1} = -5.092958 \][/tex]
- Interval [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.01\right]\)[/tex]:
- [tex]\( t_2 = \frac{\pi}{2} + 0.01 \)[/tex]
Using these:
- [tex]\( s(t_1) = 8 \)[/tex]
- [tex]\( s(t_2) = 8 \sin \left(5 \left(\frac{\pi}{2} + 0.01\right)\right) \approx -0.079933 \)[/tex]
Thus, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{-0.079933 - 8}{0.01} = -0.999791 \][/tex]
- Interval [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.001\right]\)[/tex]:
- [tex]\( t_2 = \frac{\pi}{2} + 0.001 \)[/tex]
Using these:
- [tex]\( s(t_1) = 8 \)[/tex]
- [tex]\( s(t_2) = 8 \sin \left(5 \left(\frac{\pi}{2} + 0.001\right)\right) \approx -0.007999986 \)[/tex]
Thus, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{-0.007999986 - 8}{0.001} = -0.0999998 \][/tex]
- Interval [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.0001\right]\)[/tex]:
- [tex]\( t_2 = \frac{\pi}{2} + 0.0001 \)[/tex]
Using these:
- [tex]\( s(t_1) = 8 \)[/tex]
- [tex]\( s(t_2) = 8 \sin \left(5 \left(\frac{\pi}{2} + 0.0001\right)\right) \approx -0.0007999998 \)[/tex]
Thus, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{-0.0007999998 - 8}{0.0001} = -0.0099999998 \][/tex]
2. Instantaneous Velocity at [tex]\( t = \frac{\pi}{2} \)[/tex]:
The instantaneous velocity is found by taking the derivative of [tex]\(s(t)\)[/tex] and evaluating it at [tex]\( t = \frac{\pi}{2} \)[/tex].
[tex]\[ v(t) = \frac{ds}{dt} = 40 \cos (5t) \][/tex]
At [tex]\( t = \frac{\pi}{2} \)[/tex]:
[tex]\[ v\left(\frac{\pi}{2}\right) = 40 \cos \left(5 \cdot \frac{\pi}{2}\right) = 40 \cos \left(\frac{5\pi}{2}\right) = 40 \cdot 0 = 0 \][/tex]
However, we note that [tex]\(40 \cos \left(\frac{5\pi}{2}\right) \approx 1.2246467991473532 \times 10^{-14}\)[/tex], which is very close to zero (practically zero).
Summary:
Here's the completed table with the average velocities and conjecture on instantaneous velocity:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline Time \ Interval & \left[\frac{\pi}{2}, \pi\right] & \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.1 \right] & \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.01 \right] & \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.001 \right] & \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.0001 \right] \\ \hline Average \ Velocity & -10.1859 & -5.092958 & -0.999791 & -0.0999998 & -0.0099999998 \\ \hline \end{array} \][/tex]
Conjecture about the instantaneous velocity at [tex]\(t = \frac{\pi}{2}\)[/tex]:
The instantaneous velocity at [tex]\(t = \frac{\pi}{2}\)[/tex] is approximately [tex]\( 1.2246467991473532 \times 10^{-14} \text{ or practically } 0 \)[/tex].
Given the position function:
[tex]\[ s(t) = 8 \sin (5t) \][/tex]
We're asked to find the average velocities over specific intervals and then make a conjecture about the instantaneous velocity at [tex]\( t = \frac{\pi}{2} \)[/tex].
1. Average Velocities
The average velocity over the interval [tex]\([t_1, t_2]\)[/tex] can be calculated using the difference quotient:
[tex]\[ v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \][/tex]
Now, let's compute this for each given interval.
- Interval [tex]\(\left[\frac{\pi}{2}, \pi\right]\)[/tex]:
- [tex]\( t_1 = \frac{\pi}{2} \)[/tex]
- [tex]\( t_2 = \pi \)[/tex]
Using these:
- [tex]\( s(t_1) = 8 \sin \left(5 \cdot \frac{\pi}{2}\right) = 8 \sin \left(\frac{5\pi}{2}\right) = 8 \)[/tex]
- [tex]\( s(t_2) = 8 \sin \left(5 \cdot \pi\right) = 8 \sin (5\pi) = -8 \)[/tex]
Thus, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{-8 - 8}{\pi - \frac{\pi}{2}} = \frac{-16}{\frac{\pi}{2}} = \frac{-16 \cdot 2}{\pi} = -\frac{32}{\pi} \approx -10.1859 \][/tex]
- Interval [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.1\right]\)[/tex]:
- [tex]\( t_2 = \frac{\pi}{2} + 0.1 \)[/tex]
Using these:
- [tex]\( s(t_1) = 8 \)[/tex]
- [tex]\( s(t_2) = 8 \sin \left(5 \left(\frac{\pi}{2} + 0.1\right)\right) \approx -0.792584 \)[/tex]
Thus, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{-0.792584 - 8}{0.1} = -5.092958 \][/tex]
- Interval [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.01\right]\)[/tex]:
- [tex]\( t_2 = \frac{\pi}{2} + 0.01 \)[/tex]
Using these:
- [tex]\( s(t_1) = 8 \)[/tex]
- [tex]\( s(t_2) = 8 \sin \left(5 \left(\frac{\pi}{2} + 0.01\right)\right) \approx -0.079933 \)[/tex]
Thus, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{-0.079933 - 8}{0.01} = -0.999791 \][/tex]
- Interval [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.001\right]\)[/tex]:
- [tex]\( t_2 = \frac{\pi}{2} + 0.001 \)[/tex]
Using these:
- [tex]\( s(t_1) = 8 \)[/tex]
- [tex]\( s(t_2) = 8 \sin \left(5 \left(\frac{\pi}{2} + 0.001\right)\right) \approx -0.007999986 \)[/tex]
Thus, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{-0.007999986 - 8}{0.001} = -0.0999998 \][/tex]
- Interval [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.0001\right]\)[/tex]:
- [tex]\( t_2 = \frac{\pi}{2} + 0.0001 \)[/tex]
Using these:
- [tex]\( s(t_1) = 8 \)[/tex]
- [tex]\( s(t_2) = 8 \sin \left(5 \left(\frac{\pi}{2} + 0.0001\right)\right) \approx -0.0007999998 \)[/tex]
Thus, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{-0.0007999998 - 8}{0.0001} = -0.0099999998 \][/tex]
2. Instantaneous Velocity at [tex]\( t = \frac{\pi}{2} \)[/tex]:
The instantaneous velocity is found by taking the derivative of [tex]\(s(t)\)[/tex] and evaluating it at [tex]\( t = \frac{\pi}{2} \)[/tex].
[tex]\[ v(t) = \frac{ds}{dt} = 40 \cos (5t) \][/tex]
At [tex]\( t = \frac{\pi}{2} \)[/tex]:
[tex]\[ v\left(\frac{\pi}{2}\right) = 40 \cos \left(5 \cdot \frac{\pi}{2}\right) = 40 \cos \left(\frac{5\pi}{2}\right) = 40 \cdot 0 = 0 \][/tex]
However, we note that [tex]\(40 \cos \left(\frac{5\pi}{2}\right) \approx 1.2246467991473532 \times 10^{-14}\)[/tex], which is very close to zero (practically zero).
Summary:
Here's the completed table with the average velocities and conjecture on instantaneous velocity:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline Time \ Interval & \left[\frac{\pi}{2}, \pi\right] & \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.1 \right] & \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.01 \right] & \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.001 \right] & \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.0001 \right] \\ \hline Average \ Velocity & -10.1859 & -5.092958 & -0.999791 & -0.0999998 & -0.0099999998 \\ \hline \end{array} \][/tex]
Conjecture about the instantaneous velocity at [tex]\(t = \frac{\pi}{2}\)[/tex]:
The instantaneous velocity at [tex]\(t = \frac{\pi}{2}\)[/tex] is approximately [tex]\( 1.2246467991473532 \times 10^{-14} \text{ or practically } 0 \)[/tex].
