Let's break down the problem step by step. --- ### Given Function \[ H(t) = 4.8 \cos \left( \frac{\pi (t - 8)}{6} \right) + 6.1 \] where \( t \) is in hours, \( 0 \leq t \leq 24 \), and \( t = 0 \) corresponds to 12 midnight. #### **We need to:** (a) Find the rate of change of the water level at 10 A.M. (b) Find the water level at 10 A.M. --- ### **Step 1: Convert 10 A.M. to \( t \)** Since \( t = 0 \) is midnight, 10 A.M. is \( t = 10 \). --- ## **Part (a): Rate of Change at 10 A.M.** We need the derivative \( H'(t) \): \[ H(t) = 4.8 \cos \left( \frac{\pi (t - 8)}{6} \right) + 6.1 \] Derivative using the chain rule: \[ H'(t) = 4.8 \cdot \left[ -\sin \left( \frac{\pi (t - 8)}{6} \right) \right] \cdot \frac{\pi}{6} \] \[ = -4.8 \cdot \frac{\pi}{6} \sin \left( \frac{\pi (t - 8)}{6} \right) \] \[ = -0.8\pi \sin \left( \frac{\pi (t - 8)}{6} \right) \] Plug in \( t = 10 \): \[ H'(10) = -0.8\pi \sin \left( \frac{\pi (10 - 8)}{6} \right) = -0.8\pi \sin \left( \frac{\pi (2)}{6} \right) = -0.8\pi \sin \left( \frac{\pi}{3} \right) \] Recall: \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \) \[ H'(10) = -0.8\pi \cdot \frac{\sqrt{3}}{2} = -0.4\pi\sqrt{3} \] Now, calculate numerically (\( \pi \approx 3.1416, \sqrt{3} \approx 1.7321 \)): \[ H'(10) \approx -0.4 \times 3.1416 \times 1.7321 \] \[ \approx -0.4 \times 5.4414 \] \[ \approx -2.1766 \] **Final Answer for (a):** \[ \boxed{-2.1766 \text{ feet/hour}} \] --- ## **Part (b): Water Level at 10 A.M.** Plug \( t = 10 \) into \( H(t) \): \[ H(10) = 4.8 \cos \left( \frac{\pi (10 - 8)}{6} \right) + 6.1 = 4.8 \cos \left( \frac{\pi (2)}{6} \right) + 6.1 = 4.8 \cos \left( \frac{\pi}{3} \right) + 6.1 \] Recall: \( \cos(\frac{\pi}{3}) = \frac{1}{2} \) \[ H(10) = 4.8 \cdot \frac{1}{2} + 6.1 = 2.4 + 6.1 = 8.5 \] **Final Answer for (b):** \[ \boxed{8.5000 \text{ feet}} \] --- ## **Summary of Answers** **(a)** Rate of change at 10 A.M.: \(\boxed{-2.1766}\) feet/hour **(b)** Water level at 10 A.M.: \(\boxed{8.5000}\) feet