Questions: The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function D(t)=5 cos (π/3 t+2π/3)+4 where t is the number of hours after midnight. Find the rate at which the depth is changing at 2 a.m. Round your answer to 4 decimal places. ft / hr

Solution Steps

The depth of water at a dock is modeled by the function: \[ D(t) = 5 \cos \left( \frac{\pi}{3} t + \frac{2\pi}{3} \right) + 4 \] where \( t \) is the number of hours after midnight.

To find the rate at which the depth is changing, we need to calculate the derivative of \( D(t) \) with respect to \( t \): \[ D'(t) = \frac{d}{dt} \left[ 5 \cos \left( \frac{\pi}{3} t + \frac{2\pi}{3} \right) + 4 \right] \] Using the chain rule, we get: \[ D'(t) = -5 \cdot \frac{\pi}{3} \sin \left( \frac{\pi}{3} t + \frac{2\pi}{3} \right) \] Simplifying, we have: \[ D'(t) = -\frac{5\pi}{3} \sin \left( \frac{\pi}{3} t + \frac{2\pi}{3} \right) \]

To find the rate of change at 2 a.m. (\( t = 2 \)): \[ D'(2) = -\frac{5\pi}{3} \sin \left( \frac{\pi}{3} \cdot 2 + \frac{2\pi}{3} \right) \] Simplifying the argument of the sine function: \[ \frac{\pi}{3} \cdot 2 + \frac{2\pi}{3} = \frac{2\pi}{3} + \frac{2\pi}{3} = \frac{4\pi}{3} \] Thus: \[ D'(2) = -\frac{5\pi}{3} \sin \left( \frac{4\pi}{3} \right) \] Since \(\sin \left( \frac{4\pi}{3} \right) = -\frac{\sqrt{3}}{2}\): \[ D'(2) = -\frac{5\pi}{3} \left( -\frac{\sqrt{3}}{2} \right) = \frac{5\pi \sqrt{3}}{6} \]

Evaluating the expression numerically: \[ D'(2) \approx 4.53449841058554 \] Rounding to four decimal places: \[ D'(2) \approx 4.5345 \]

Final Answer