Tidal Depth Model Bfed7D

1. **State the problem:**\nWe are given the depth of water $D$ at Southview harbour on 18th October 2017 modeled by\n$$D = 6 + 2 \sin \left(\frac{4\pi t}{25}\right) - 1.5 \cos \left(\frac{4\pi t}{25}\right), \quad 0 \leq t \leq 24.$$\nWe need to find:\n- The amplitude $R$ and phase angle $\alpha$ such that $D = 6 + R \sin \left(\frac{4\pi t}{25} + \alpha \right)$, with $\alpha$ to 4 decimal places.\n- The depth at $t=0$ (00:00 hours).\n- The maximum depth and the time it occurs in the afternoon.\n- Explain why Jolene's model for 19th October is incorrect and find a suitable model for $H$.\n\n2. **Find $R$ and $\alpha$: Using the formula for combining sine and cosine:**\n$$a \sin x + b \cos x = R \sin (x + \alpha),$$\nwhere\n$$R = \sqrt{a^2 + b^2}, \quad \alpha = \arctan \left(\frac{b}{a}\right).$$\nHere, $a=2$ and $b=-1.5$.\n\nCalculate $R$:\n$$R = \sqrt{2^2 + (-1.5)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5.$$\n\nCalculate $\alpha$:\n$$\alpha = \arctan \left(\frac{-1.5}{2}\right) = \arctan(-0.75).$$\nUsing a calculator,\n$$\alpha \approx -0.6435 \text{ radians}.$$\nSince $a>0$ and $b<0$, $\alpha$ is negative, so we can write\n$$\alpha = -0.6435 \text{ radians}.$$\n\n3. **Rewrite the model:**\n$$D = 6 + 2.5 \sin \left(\frac{4\pi t}{25} - 0.6435 \right).$$\n\n4. **(b) Find the depth at 00:00 hours ($t=0$):**\n$$D(0) = 6 + 2 \sin(0) - 1.5 \cos(0) = 6 + 0 - 1.5 = 4.5 \text{ metres}.$$\n\n5. **(c) Find the maximum depth:**\nThe maximum value of $\sin$ function is 1, so\n$$D_{max} = 6 + 2.5 \times 1 = 8.5 \text{ metres}.$$\n\n6. **(d) Find the time when maximum depth occurs in the afternoon:**\nMaximum occurs when\n$$\sin \left(\frac{4\pi t}{25} - 0.6435 \right) = 1.$$\nThis happens when\n$$\frac{4\pi t}{25} - 0.6435 = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}.$$\nSolve for $t$:\n$$\frac{4\pi t}{25} = \frac{\pi}{2} + 0.6435 + 2k\pi,$$\n$$t = \frac{25}{4\pi} \left(\frac{\pi}{2} + 0.6435 + 2k\pi \right).$$\nCalculate for $k=0$ (first max):\n$$t = \frac{25}{4\pi} \left(1.5708 + 0.6435 \right) = \frac{25}{4\pi} \times 2.2143.$$\nCalculate denominator:\n$$4\pi \approx 12.5664,$$\nso\n$$t = \frac{25}{12.5664} \times 2.2143 = 1.9894 \times 2.2143 = 4.402 \text{ hours}.$$\nThis is about 4:24 am (not afternoon).\n\nTry $k=1$ for afternoon max:\n$$t = \frac{25}{4\pi} \left(1.5708 + 0.6435 + 2 \times 3.1416 \right) = \frac{25}{12.5664} \times (2.2143 + 6.2832) = 1.9894 \times 8.4975 = 16.9 \text{ hours}.$$\n\nConvert $16.9$ hours to hours and minutes:\n$$0.9 \times 60 = 54 \text{ minutes}.$$\nSo maximum depth in the afternoon occurs at approximately 16:54 (4:54 pm).\n\n7. **(e)(i) Explain why Jolene's model is not correct:**\nAt $x=0$, Jolene's model gives\n$$H(0) = 6 + 2 \sin(0) - 1.5 \cos(0) = 6 - 1.5 = 4.5,$$\nwhich is the same as Tom's model at $t=0$. However, the depth of water at 00:00 on 19th October should be the same as the depth at 24 hours on 18th October (end of day 18).\nCalculate Tom's model at $t=24$:\n$$D(24) = 6 + 2 \sin \left(\frac{4\pi \times 24}{25} \right) - 1.5 \cos \left(\frac{4\pi \times 24}{25} \right).$$\nCalculate inside the sine and cosine:\n$$\theta = \frac{96\pi}{25} = 3.84\pi.$$\nSince $3.84\pi = 3\pi + 0.84\pi$, and sine and cosine are periodic with period $2\pi$,\n$$\sin(3.84\pi) = \sin(0.84\pi) \approx \sin(2.639) = 0.4794,$$\n$$\cos(3.84\pi) = \cos(0.84\pi) \approx \cos(2.639) = -0.8776.$$\nCalculate $D(24)$:\n$$D(24) = 6 + 2 \times 0.4794 - 1.5 \times (-0.8776) = 6 + 0.9588 + 1.3164 = 8.2752.$$\n\nSince $H(0) = 4.5$ but $D(24) = 8.2752$, Jolene's model does not match the continuity of the tide from one day to the next.\n\n8. **(e)(ii) Find a suitable model for $H$:**\nWe want $H$ to be a shifted version of $D$ to ensure continuity:\n$$H = 6 + 2 \sin \left(\frac{4\pi x}{25} + \beta \right) - 1.5 \cos \left(\frac{4\pi x}{25} + \beta \right).$$\nUsing the combined form:\n$$H = 6 + 2.5 \sin \left(\frac{4\pi x}{25} + \alpha + \beta \right).$$\nWe want\n$$H(0) = D(24) = 8.2752,$$\nso\n$$6 + 2.5 \sin(\alpha + \beta) = 8.2752,$$\n$$\sin(\alpha + \beta) = \frac{8.2752 - 6}{2.5} = 0.91.$$\nRecall $\alpha = -0.6435$, so\n$$\alpha + \beta = \arcsin(0.91) = 1.157 \text{ radians}.$$\nSolve for $\beta$:\n$$\beta = 1.157 - (-0.6435) = 1.8005 \text{ radians}.$$\n\n**Final suitable model:**\n$$H = 6 + 2.5 \sin \left(\frac{4\pi x}{25} - 0.6435 + 1.8005 \right) = 6 + 2.5 \sin \left(\frac{4\pi x}{25} + 1.157 \right).$$