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=-1.5cosleft(\frac{4pi}{25}t\right)+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=-1.5cosleft(\frac{4pi}{25}t\right)+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 up the equation

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: Simplify the equation

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

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

One day has 24 hours. We need to find the number of non - negative integer values of $n$ for which $0\leq t=\frac{25(2n + 1)}{8}\leq24$. First, solve the left - hand inequality $0\leq\frac{25(2n + 1)}{8}$, which is true for $n\geq0$. Then solve the right - hand inequality $\frac{25(2n + 1)}{8}\leq24$. Cross - multiply: $25(2n + 1)\leq192$. Expand: $50n+25\leq192$. Subtract 25: $50n\leq167$. Then $n\leq\frac{167}{50}=3.34$. Since $n$ is a non - negative integer, $n = 0,1,2,3$.

Answer:

4 times