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)=3cosleft(\frac{pi}{4}t + \frac{3pi}{4}
ight)+5
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
question help: video

Explanation:

Step1: Differentiate the function

The derivative of $y = A\cos(Bt + C)+D$ with respect to $t$ is $y^\prime=-AB\sin(Bt + C)$. For $D(t)=3\cos(\frac{\pi}{4}t+\frac{3\pi}{4}) + 5$, using the chain - rule, $D^\prime(t)=-3\times\frac{\pi}{4}\sin(\frac{\pi}{4}t+\frac{3\pi}{4})=-\frac{3\pi}{4}\sin(\frac{\pi}{4}t+\frac{3\pi}{4})$.

Step2: Substitute $t = 2$

Since $t$ is the number of hours after midnight and we want to find the rate at 2 a.m., so $t = 2$. Substitute $t = 2$ into $D^\prime(t)$:
\[

$$\begin{align*} D^\prime(2)&=-\frac{3\pi}{4}\sin(\frac{\pi}{4}\times2+\frac{3\pi}{4})\\ &=-\frac{3\pi}{4}\sin(\frac{\pi}{2}+\frac{3\pi}{4})\\ &=-\frac{3\pi}{4}\sin(\frac{2\pi + 3\pi}{4})\\ &=-\frac{3\pi}{4}\sin(\frac{5\pi}{4}) \end{align*}$$

\]
We know that $\sin(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}$. Then $D^\prime(2)=-\frac{3\pi}{4}\times(-\frac{\sqrt{2}}{2})=\frac{3\pi\sqrt{2}}{8}$.

Step3: Calculate the value

\[

$$\begin{align*} \frac{3\pi\sqrt{2}}{8}&=\frac{3\times3.14159\times1.41421}{8}\\ &=\frac{3\times4.44288}{8}\\ &=\frac{13.32864}{8}\\ & = 1.6661 \end{align*}$$

\]

Answer:

$1.6661$