suppose the depth, d, in metres, of the tide in a certain harbour can be modelled by d(t)=-3cos(π/6t)+5…

Explanation:

Step1: Recall cosine - function properties

The cosine function $y = A\cos(Bt - C)+D$ has an amplitude $|A|$, vertical shift $D$. Here $A=- 3$, $B = \frac{\pi}{6}$, $D = 5$. The range of $y=\cos(\frac{\pi}{6}t)$ is $[-1,1]$.

Step2: Find the maximum value of $d(t)$

The maximum value of $d(t)$ occurs when $\cos(\frac{\pi}{6}t)=-1$. [ \begin{align*} d(t)&=-3\times(-1)+5\ &=3 + 5\ &=8 \end{align*} ] Set $\cos(\frac{\pi}{6}t)=-1$. We know that $\cos\theta=-1$ when $\theta=(2k + 1)\pi,k\in\mathbb{Z}$. So $\frac{\pi}{6}t=(2k + 1)\pi$. Solving for $t$ gives $t = 6(2k + 1)$. For $k = 0$, $t = 6$; for $k = 1$, $t=18$; for $k = 2$, $t = 30$ (but considering a 24 - hour day, when $k = 2$, $t=30\equiv6\pmod{24}$), and when $t = 24$, $\cos(\frac{\pi}{6}\times24)=\cos(4\pi)=1$, but $d(24)=-3\times1 + 5=2$ (this is wrong for max). The correct times for max are $t = 6$ and $t = 18$ with $d = 8$.

Step3: Find the minimum value of $d(t)$

The minimum value of $d(t)$ occurs when $\cos(\frac{\pi}{6}t)=1$. [ \begin{align*} d(t)&=-3\times1+5\ &=2 \end{align*} ] Set $\cos(\frac{\pi}{6}t)=1$. We know that $\cos\theta = 1$ when $\theta = 2k\pi,k\in\mathbb{Z}$. So $\frac{\pi}{6}t=2k\pi$. Solving for $t$ gives $t = 12k$. For $k = 1$, $t = 12$.

Answer:

Max. at 8 m at t=6 Max. at 8 m at t=18 Min. at 2 m at t=12