the depth of the water at the end of a pier changes periodically along with the movement of tides. on a…

the depth of the water at the end of a pier changes periodically along with the movement of tides. on a particular day, low tides occur at 12:00 am and 12:30 pm, with a depth of 2.5 m, while high tides occur at 6:15 am and 6:45 pm, with a depth of 5.5 m. let t = 0 be 12:00 am.\n\ngraph the equation (d=-1.5cos(\frac{4pi}{25}t)+4) that models the situation using a graphing calculator and use it to answer the following questions.\n\nhow many times during this day is the depth at the end of the pier equal to 4 meters?\n2 times\n3 times\n4 times\n5 times\n\nat approximately what time on the next day does the depth first reach 4 meters?\n3:00 am\n4:00 am\n5:00 pm\n10:00 am

the depth of the water at the end of a pier changes periodically along with the movement of tides. on a particular day, low tides occur at 12:00 am and 12:30 pm, with a depth of 2.5 m, while high tides occur at 6:15 am and 6:45 pm, with a depth of 5.5 m. let t = 0 be 12:00 am.\n\ngraph the equation (d=-1.5cos(\frac{4pi}{25}t)+4) that models the situation using a graphing calculator and use it to answer the following questions.\n\nhow many times during this day is the depth at the end of the pier equal to 4 meters?\n2 times\n3 times\n4 times\n5 times\n\nat approximately what time on the next day does the depth first reach 4 meters?\n3:00 am\n4:00 am\n5:00 pm\n10:00 am

Answer

Explanation:

Step1: Set up the equation

We are given the equation $d=- 1.5\cos(\frac{4\pi}{25}t)+4$ and we want to find when $d = 4$. So we set up the equation $4=-1.5\cos(\frac{4\pi}{25}t)+4$.

Step2: Simplify the equation

Subtract 4 from both sides: $0=-1.5\cos(\frac{4\pi}{25}t)$. Then divide both sides by - 1.5 to get $\cos(\frac{4\pi}{25}t)=0$.

Step3: Solve for t

We know that $\cos\theta = 0$ when $\theta=(2n + 1)\frac{\pi}{2}$, where $n$ is an integer. So $\frac{4\pi}{25}t=(2n + 1)\frac{\pi}{2}$. Cross - multiply to get $8t = 25(2n + 1)$, then $t=\frac{25(2n + 1)}{8}$.

Step4: Find the number of solutions in one day

The period of the cosine function $y = A\cos(Bt)+C$ is $T=\frac{2\pi}{B}$. Here $B=\frac{4\pi}{25}$, so $T=\frac{2\pi}{\frac{4\pi}{25}}=\frac{25}{2}=12.5$ hours. In a 24 - hour day, we find the number of non - negative values of $t$ for which $\cos(\frac{4\pi}{25}t)=0$. When $n = 0$, $t=\frac{25}{8}=3.125$ hours; when $n = 1$, $t=\frac{25\times3}{8}=9.375$ hours; when $n = 2$, $t=\frac{25\times5}{8}=15.625$ hours; when $n = 3$, $t=\frac{25\times7}{8}=21.875$ hours. So the depth is 4 meters 4 times in a day.

Step5: Find when the depth first reaches 4 meters on the next day

We continue to find values of $t$ for $n$. When $n = 4$, $t=\frac{25\times9}{8}=28.125$ hours. $28.125-24 = 4.125$ hours after the start of the new day. Since 0.125 of an hour is $0.125\times60 = 7.5$ minutes, it is approximately 4:00 am.

Answer:

How many times during this day is the depth at the end of the pier equal to 4 meters? 4 times At approximately what time on the next day does the depth first reach 4 meters? 4:00 am