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?\n\n2 times\n3 times\n4 times\n5 times\ndone

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?\n\n2 times\n3 times\n4 times\n5 times\ndone

Answer

Explanation:

Step1: Set the equation equal to 4

Set $d = 4$ in the equation $d=-1.5\cos(\frac{4\pi}{25}t)+4$. So we get $4=-1.5\cos(\frac{4\pi}{25}t)+4$.

Step2: Solve for $\cos(\frac{4\pi}{25}t)$

Subtract 4 from both sides of the equation: $0=-1.5\cos(\frac{4\pi}{25}t)$. Then divide both sides by - 1.5, we have $\cos(\frac{4\pi}{25}t)=0$.

Step3: Find the general solution for $\frac{4\pi}{25}t$

The general solution for $\cos x = 0$ is $x=(2n + 1)\frac{\pi}{2}$, where $n$ is an integer. So $\frac{4\pi}{25}t=(2n + 1)\frac{\pi}{2}$.

Step4: Solve for $t$

Cross - multiply to get $8t = 25(2n + 1)$, then $t=\frac{25(2n + 1)}{8}$.

Step5: Find the number of solutions in a day

A day has 24 hours. We need to find the number of non - negative integer values of $n$ for which $0\leq t=\frac{25(2n + 1)}{8}\leq24$. When $n = 0$, $t=\frac{25}{8}=3.125$. When $n = 1$, $t=\frac{25\times(2\times1 + 1)}{8}=\frac{75}{8}=9.375$. When $n = 2$, $t=\frac{25\times(2\times2+ 1)}{8}=\frac{125}{8}=15.625$. When $n = 3$, $t=\frac{25\times(2\times3 + 1)}{8}=\frac{175}{8}=21.875$. When $n = 4$, $t=\frac{25\times(2\times4+ 1)}{8}=\frac{225}{8}=28.125>24$. So there are 4 non - negative values of $t$ less than 24.

Answer:

4 times