Question

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)=4cos(πt + 7π/6)+2

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

Explanation:

Step1: Find the derivative of \(D(t)\)

The derivative of \(y = A\cos(Bt + C)+D\) is \(y^\prime=-AB\sin(Bt + C)\).
For \(D(t)=4\cos(\frac{\pi}{6}t+\frac{7\pi}{6}) + 2\), \(A = 4\), \(B=\frac{\pi}{6}\).
So \(D^\prime(t)=4\times(-\frac{\pi}{6})\sin(\frac{\pi}{6}t+\frac{7\pi}{6})\).

Step2: Calculate \(t\) value

At 2 a.m., \(t = 2\) (since \(t\) is the number of hours after midnight).

Step3: Substitute \(t = 2\) into \(D^\prime(t)\)

\(D^\prime(2)=-\frac{2\pi}{3}\sin(\frac{\pi}{6}\times2+\frac{7\pi}{6})\).
First, simplify the argument of the sine function: \(\frac{\pi}{6}\times2+\frac{7\pi}{6}=\frac{2\pi + 7\pi}{6}=\frac{9\pi}{6}=\frac{3\pi}{2}\).
Then \(\sin(\frac{3\pi}{2})=- 1\).
So \(D^\prime(2)=-\frac{2\pi}{3}\times(-1)=\frac{2\pi}{3}\approx - 1.8138\) (using \(\pi\approx3.1416\)).

Answer:

$-1.8138$