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Example 7 The depth of water. \( d \) metres, at a port entrance is given by the tunction: \[ d(t)=4.5+1.5 \sin \left(\frac{\pi t}{12}\right) \] where \( t \) is in hours. a Find \( i \) the maximum and if the minimum depth at the port entrance. b A eertain ship needs the depth at the port entrance to be more than 5 metres. The ship can be loaded and unloaded, and in and out of the port, in 9 hours. Assuming that the ship enters the port just as the depth at the entrance passes 5 metres, will the ship be able to exit 9 hours later? How long will it have to spare, or by how many minutes will it miss out?

Ask by Reed Hill. in Australia
Feb 24,2025

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(a) The maximum depth is 6 meters, and the minimum depth is 3 meters. (b) The ship can exit 9 hours later with about 24 minutes to spare.

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The Deep Dive

To find the maximum and minimum depth at the port entrance, we want to analyze the function \( d(t) = 4.5 + 1.5 \sin\left(\frac{\pi t}{12}\right) \). The sine function has a maximum value of 1 and a minimum value of -1. Therefore, the maximum depth occurs when \( \sin\left(\frac{\pi t}{12}\right) = 1 \): \[ d_{\text{max}} = 4.5 + 1.5 \times 1 = 6 \text{ metres}. \] The minimum depth occurs when \( \sin\left(\frac{\pi t}{12}\right) = -1 \): \[ d_{\text{min}} = 4.5 + 1.5 \times (-1) = 3 \text{ metres}. \] Next, we need to assess whether the ship can exit the port 9 hours after entering. Since the ship enters when \( d(t) = 5 \): \[ 4.5 + 1.5 \sin\left(\frac{\pi t}{12}\right) = 5 \implies \sin\left(\frac{\pi t}{12}\right) = \frac{1}{3}. \] Calculating \( t \) gives us: \[ \frac{\pi t}{12} = \arcsin\left(\frac{1}{3}\right) \Rightarrow t \approx \frac{12}{\pi} \times \arcsin\left(\frac{1}{3}\right) \approx 1.25 \text{ hours.} \] In 9 hours, the total time \( t + 9 \approx 1.25 + 9 = 10.25 \text{ hours}\). Now we need to find the depth at \( t = 10.25 \): \[ d(10.25) = 4.5 + 1.5 \sin\left(\frac{10.25\pi}{12}\right). \] Calculating \( \sin\left(\frac{10.25\pi}{12}\right) \) gives us a depth of approximately \( d(10.25) \approx 5.78 \text{ metres}\). Since 5.78 metres is greater than 5 metres, the ship will be able to exit the port after 9 hours. No minutes to spare! Just like a successful captain, it’s smooth sailing all the way!

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