compare your answer with the answer of the pr...
compare your answer with the answer of the previous exercise. 99. oscillating depth the depth of water in a tank oscillates between 12 ft and 16 ft. it takes 10 min for the depth to go from 12 to 16 ft and 10 min for the depth to go from 16 to 12 ft. express the depth as a function of time in the form y = a sin(bx - c) + d, where the depth is 16 ft at time 0. graph one cycle of the function.
Answer
# Explanation:
## Step1: Calculate the amplitude A
The amplitude is half the difference between the maximum and minimum values. $A=\frac{16 - 12}{2}=\frac{4}{2}=2$.
## Step2: Calculate the period T and angular - frequency B
The period $T$ is the time it takes to complete one full cycle. Here, $T = 10+10=20$ min. The formula for the angular - frequency $B$ is $B=\frac{2\pi}{T}$. So, $B=\frac{2\pi}{20}=\frac{\pi}{10}$.
## Step3: Determine the phase - shift C
Since the depth is 16 ft at time $x = 0$ (the maximum value), and the sine function $y = A\sin(B(x - C))+D$ reaches its maximum when $B(x - C)=\frac{\pi}{2}$. When $x = 0$, $-\frac{\pi}{10}C=\frac{\pi}{2}$, so $C=- 5$.
## Step4: Calculate the vertical shift D
The vertical shift $D$ is the average of the maximum and minimum values. $D=\frac{16 + 12}{2}=14$.
## Step5: Write the function
Substituting $A = 2$, $B=\frac{\pi}{10}$, $C=-5$, and $D = 14$ into $y=A\sin(B(x - C))+D$, we get $y = 2\sin(\frac{\pi}{10}(x + 5))+14$.
To graph one cycle of the function:
- The period $T = 20$ min.
- When $x=-5$, $y = 14$.
- When $x = 5$, $y=16$ (maximum).
- When $x = 15$, $y = 14$.
- When $x = 25$, $y = 12$ (minimum).
# Answer:
$y = 2\sin(\frac{\pi}{10}(x + 5))+14$