Let's solve part **b)** step by step: --- ### **Step 1: Set Up the Equation** Given: \[ h(t) = -3\sin\left(\frac{\pi}{12}(t-3)\right) + 4 \] We want \( h(t) = 3 \) in the first 20 seconds. Set \( h(t) = 3 \): \[ 3 = -3\sin\left(\frac{\pi}{12}(t-3)\right) + 4 \] Subtract 4 from both sides: \[ 3 - 4 = -3\sin\left(\frac{\pi}{12}(t-3)\right) \] \[ -1 = -3\sin\left(\frac{\pi}{12}(t-3)\right) \] Divide both sides by -3: \[ \frac{-1}{-3} = \sin\left(\frac{\pi}{12}(t-3)\right) \] \[ \frac{1}{3} = \sin\left(\frac{\pi}{12}(t-3)\right) \] **Explanation:** We isolated the sine function to solve for \( t \). --- ### **Step 2: Solve for \( t \) in the First 20 Seconds** Take the inverse sine: \[ \frac{\pi}{12}(t-3) = \arcsin\left(\frac{1}{3}\right) \] Or, since sine is periodic: \[ \frac{\pi}{12}(t-3) = \arcsin\left(\frac{1}{3}\right) \quad \text{or} \quad \pi - \arcsin\left(\frac{1}{3}\right) \] Let \( \theta = \arcsin\left(\frac{1}{3}\right) \). So: \[ t-3 = \frac{12}{\pi} \arcsin\left(\frac{1}{3}\right) \] \[ t = 3 + \frac{12}{\pi} \arcsin\left(\frac{1}{3}\right) \] And for the other solution: \[ t-3 = \frac{12}{\pi} \left[\pi - \arcsin\left(\frac{1}{3}\right)\right] \] \[ t = 3 + \frac{12}{\pi} \left[\pi - \arcsin\left(\frac{1}{3}\right)\right] \] Since sine is periodic with period \( 2\pi \), add multiples of the period: \[ \text{Period} = \frac{2\pi}{\frac{\pi}{12}} = 24 \] So, general solutions: \[ t_1 = 3 + \frac{12}{\pi} \arcsin\left(\frac{1}{3}\right) + 24n \] \[ t_2 = 3 + \frac{12}{\pi} \left[\pi - \arcsin\left(\frac{1}{3}\right)\right] + 24n \] for integer \( n \). Now, we want \( 0 \leq t \leq 20 \): Calculate \( \arcsin\left(\frac{1}{3}\right) \approx 0.3398 \) radians. So: \[ t_1 = 3 + \frac{12}{\pi}(0.3398) \approx 3 + 1.297 \approx 4.30 \] \[ t_2 = 3 + \frac{12}{\pi}(\pi - 0.3398) = 3 + \frac{12}{\pi}(2.8018) \approx 3 + 10.708 \approx 13.71 \] For \( n = 0 \), \( t_1 \approx 4.30 \), \( t_2 \approx 13.71 \). For \( n = 1 \), \( t_1 \approx 4.30 + 24 = 28.30 \) (out of range), \( t_2 \approx 13.71 + 24 = 37.71 \) (out of range). So, the times in the first 20 seconds are: \[ \boxed{t \approx 4.30\, \text{seconds and}\; 13.71\, \text{seconds}} \] --- ## **Final Answer** **The time(s) in the first 20 seconds that have a height of 3 m are:** \[ \boxed{t \approx 4.3\, \text{seconds and}\; 13.7\, \text{seconds}} \]

Let's proceed step by step to answer the question thoroughly. --- ### Part (a): Sketch a well-labelled graph for at least 2 cycles **Step 1: Understand the given function** Given: \[ h(t) = -3 \sin\left(\frac{\pi}{12}(t - 3)\right) + 4 \] - Amplitude: \( |\text{-}3| = 3 \) - Vertical shift: \( +4 \) - Inside sine (phase): \( \frac{\pi}{12}(t - 3) \) **Explanation:** The amplitude tells us the height variation from the midline. The midline is at \( h = 4 \). The period \( T \) of the sine wave is found by: \[ T = \frac{2\pi}{\text{coefficient of } t} = \frac{2\pi}{\pi/12} = 2\pi \times \frac{12}{\pi} = 24 \text{ seconds} \] **Step 2: Sketch the graph** - The wave has a period of 24 seconds; so two cycles will cover 48 seconds. - At \( t = 3 \), the sine argument is zero, so: \[ h(3) = -3 \sin(0) + 4 = 4 \] - Key points: - Maximum at \( t \) where \( \sin \) is \(-1\): \[ \sin\left(\frac{\pi}{12}(t - 3)\right) = -1 \] which occurs at: \[ \frac{\pi}{12}(t - 3) = \frac{3\pi}{2} \implies t - 3 = \frac{3\pi/2 \times 12}{\pi} = 18 \] \[ t = 3 + 18 = 21 \] At this point: \[ h(21) = -3 \times (-1) + 4 = 3 + 4 = 7 \] - Minimum at \( \sin \) is 1: \[ \frac{\pi}{12}(t - 3) = \frac{\pi}{2} \implies t - 3 = \frac{\pi/2 \times 12}{\pi} = 6 \] \[ t = 3 + 6 = 9 \] At this point: \[ h(9) = -3 \times 1 + 4 = -3 + 4 = 1 \] - Key points: - \( (3, 4) \) (midline) - \( (9, 1) \) (minimum) - \( (21, 7) \) (maximum) - Next cycle will start at \( t = 3 + 24 = 27 \), repeating similar points. **Label these points and sketch the sinusoidal wave over \( t = 0 \) to \( t = 48 \).** --- ### Part (b): Determine the times in the first 20 seconds when height \( h(t) = 3 \) **Step 1: Set \( h(t) = 3 \) and solve** \[ 3 = -3 \sin\left(\frac{\pi}{12}(t - 3)\right) + 4 \] Subtract 4: \[ -1 = -3 \sin\left(\frac{\pi}{12}(t - 3)\right) \] Divide both sides by -3: \[ \frac{1}{3} = \sin\left(\frac{\pi}{12}(t - 3)\right) \] **Explanation:** We isolated the sine term to find \( t \). --- **Step 2: Solve for \( t \)** \[ \frac{\pi}{12}(t - 3) = \arcsin\left(\frac{1}{3}\right) \] or \[ \frac{\pi}{12}(t - 3) = \pi - \arcsin\left(\frac{1}{3}\right) \] Let: \[ \theta = \arcsin\left(\frac{1}{3}\right) \approx 0.3398 \text{ radians} \] Calculate the solutions: \[ t - 3 = \frac{12}{\pi} \times \theta \approx \frac{12}{3.1416} \times 0.3398 \approx 3.82 \times 0.3398 \approx 1.297 \] So: \[ t = 3 + 1.297 \approx 4.30 \text{ seconds} \] Similarly, for the second solution: \[ t - 3 = \frac{12}{\pi} \times (\pi - \theta) = \frac{12}{\pi} \times (3.1416 - 0.3398) \approx 3.82 \times 2.8018 \approx 10.70 \] \[ t = 3 + 10.70 \approx 13.70 \text{ seconds} \] Since the period is 24 seconds and both solutions are within 20 seconds, these are valid. --- ### **Final answer:** The times in the first 20 seconds when the height is 3 meters are approximately: \[ \boxed{ \textbf{t} \approx 4.3 \text{ seconds and } 13.7 \text{ seconds} } \] --- **Summary:** - **Part (a):** The graph is a sine wave oscillating between 1 m and 7 m, centered at 4 m, with a period of 24 seconds, starting at \( t=3 \). - **Part (b):** The times when height is 3 m are approximately at 4.3 s and 13.7 s within the first 20 seconds.