The pressure of a fluid flowing through a closed system, , after seconds can be modeled by the function . Answer parts (a) through © below.
(a) In the interval , determine the times at which the pressure of the fluid is 100 mmHg .
The solution set is .
(Type an integer or decimal rounded to two decimal places as needed. Use a comma to separate answers as needed.)
(b) In the interval , determine the times at which the pressure of the fluid is 140 mmHg .
The solution set is .
(Type an integer or decimal rounded to two decimal places as needed. Use a comma to separate answers as needed.)
© In the interval , determine the times at which the pressure of the fluid is between 100 and 105 mmHg .
(Use ascending order. Type an integer or decimal rounded to two decimal places as needed.)

Ask by Marsh Lynch. in the United States

Jan 25,2025

Question

The pressure of a fluid flowing through a closed system, , after seconds can be modeled by the function . Answer parts (a) through © below.
(a) In the interval , determine the times at which the pressure of the fluid is 100 mmHg .
The solution set is .
(Type an integer or decimal rounded to two decimal places as needed. Use a comma to separate answers as needed.)
(b) In the interval , determine the times at which the pressure of the fluid is 140 mmHg .
The solution set is .
(Type an integer or decimal rounded to two decimal places as needed. Use a comma to separate answers as needed.)
© In the interval , determine the times at which the pressure of the fluid is between 100 and 105 mmHg .
(Use ascending order. Type an integer or decimal rounded to two decimal places as needed.)

Ask by Marsh Lynch. in the United States

Jan 25,2025

UpStudy AI · Accepted Answer

To solve the problem, we will analyze the function $$ P(t) = 100 + 40 \sin\left(\frac{12 \pi}{5} t\right) $$ for the specified conditions.

### Part (a)
We need to find the times $$ t $$ in the interval $$ [0, 1] $$ when the pressure $$ P(t) = 100 $$ mmHg.

1. Set the equation:
   $$
   100 + 40 \sin\left(\frac{12 \pi}{5} t\right) = 100
   $$
2. Simplify:
   $$
   40 \sin\left(\frac{12 \pi}{5} t\right) = 0
   $$
3. This implies:
   $$
   \sin\left(\frac{12 \pi}{5} t\right) = 0
   $$
4. The sine function is zero at integer multiples of $$ \pi $$:
   $$
   \frac{12 \pi}{5} t = n\pi \quad \text{for } n \in \mathbb{Z}
   $$
5. Solving for $$ t $$:
   $$
   t = \frac{5n}{12}
   $$
6. We need $$ t $$ in the interval $$ [0, 1] $$:
   - For $$ n = 0 $$: $$ t = 0 $$
   - For $$ n = 1 $$: $$ t = \frac{5}{12} \approx 0.42 $$
   - For $$ n = 2 $$: $$ t = \frac{10}{12} = \frac{5}{6} \approx 0.83 $$

Thus, the solution set is:
$$
\{0, 0.42, 0.83\}
$$

### Part (b)
Now, we need to find the times $$ t $$ in the interval $$ [0, 1] $$ when the pressure $$ P(t) = 140 $$ mmHg.

1. Set the equation:
   $$
   100 + 40 \sin\left(\frac{12 \pi}{5} t\right) = 140
   $$
2. Simplify:
   $$
   40 \sin\left(\frac{12 \pi}{5} t\right) = 40
   $$
3. This implies:
   $$
   \sin\left(\frac{12 \pi}{5} t\right) = 1
   $$
4. The sine function is equal to 1 at:
   $$
   \frac{12 \pi}{5} t = \frac{\pi}{2} + 2k\pi \quad \text{for } k \in \mathbb{Z}
   $$
5. Solving for $$ t $$:
   $$
   t = \frac{5}{12} \left(\frac{1}{2} + 2k\right)
   $$
6. For $$ k = 0 $$:
   $$
   t = \frac{5}{12} \cdot \frac{1}{2} = \frac{5}{24} \approx 0.21
   $$
7. For $$ k = 1 $$:
   $$
   t = \frac{5}{12} \cdot \frac{5}{2} = \frac{25}{24} \quad \text{(not in the interval)}
   $$

Thus, the solution set is:
$$
\{0.21\}
$$

### Part (c)
Now, we need to find the times $$ t $$ in the interval $$ [0, 1] $$ when the pressure $$ P(t) $$ is between 100 and 105 mmHg.

1. Set the inequality:
   $$
   100 < 100 + 40 \sin\left(\frac{12 \pi}{5} t\right) < 105
   $$
2. Simplifying gives:
   $$
   0 < 40 \sin\left(\frac{12 \pi}{5} t\right) < 5
   $$
3. Dividing by 40:
   $$
   0 < \sin\left(\frac{12 \pi}{5} t\right) < \frac{1}{8}
   $$
4. The sine function is positive in the intervals $$ (0, \pi) $$ and $$ (2\pi, 3\pi) $$, etc. We need to find the values of $$ t $$ that satisfy this condition.

5. We will find the angles where $$ \sin(x) = \frac{1}{8} $$:
   $$
   x = \arcsin\left(\frac{1}{8}\right) \quad \text{and} \quad x = \pi - \arcsin\left(\frac{1}{8}\right)
   $$
6. Calculate $$ t $$:
   $$
   t_1 = \frac{5}{12} \cdot \frac{\arcsin\left(\frac{1}{8}\right)}{\pi} \quad \text{and} \quad t_2 = \frac{5}{12} \cdot \frac{\pi - \arcsin\left(\frac{1}{8}\right)}{\pi}
   $$

Let's calculate these values.

First, we find $$ \arcsin\left(\frac{1}{8}\right) $$:
$$
\arcsin\left(\frac{1}{8}\right) \approx 0.124 \text{ radians}
$$

Now, calculate $$ t_1 $$ and $$ t_2 $$:
$$
t_1 \approx \frac{5}{12} \cdot \frac{0.124}{\pi} \approx 0.05
$$
$$
t_2 \approx \frac{5}{12} \cdot \frac{\pi - 0.124}{\pi} \approx \frac{5}{12} \cdot \left(1 - \frac{0.124}{\pi}\right) \approx 0.95
$$

Thus, the solution set for part (c) is:
$$
\{0.05, 0.95\}
$$

### Final Answers
(a) $$ \{0, 0.42, 0.83\} $$

(b) $$ \{0.21\} $$

(c) $$ \{0.05, 0.95\} $$
(a) Times when pressure is 100 mmHg: 0, 0.42, 0.83 seconds. (b) Time when pressure is 140 mmHg: 0.21 seconds. (c) Times when pressure is between 100 and 105 mmHg: 0.05, 0.95 seconds.

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