https://r2.erweima.ai/mathbot/image/1726636506723oy5yr15e.jpg

The problem provided involves modeling the pressure \( P(t) \) in a mechanical system that operates with LP gas, following a sinusoidal behavior. The general form of the function is given by:

\[
P(t) = k + A \sin(\omega t)
\]

Where:

- \( P(t) \) is the pressure in PSI at time \( t \),
- \( A \) is the amplitude of oscillation,
- \( \omega \) is the angular frequency,
- \( k \) is the mean pressure value,
- The pressure oscillates between 80 PSI and 120 PSI, with a mean pressure of 110 PSI after 10 minutes.

Let's go through the steps:

### Part (a) - Model the function \( P(t) \)

We know:

- The maximum pressure is 120 PSI and the minimum is 80 PSI. Hence, the amplitude \( A \) is:

\[
A = \frac{120 - 80}{2} = 20
\]

- The mean pressure \( k \) is the midpoint between 80 and 120 PSI:

\[
k = \frac{120 + 80}{2} = 100
\]

- The pressure reaches 110 PSI after 10 minutes. We can assume that this is one-quarter of the way through a full oscillation (a quarter period), meaning \( \omega t = \frac{\pi}{2} \) at \( t = 10 \). This gives us:

\[
\omega(10) = \frac{\pi}{2} \implies \omega = \frac{\pi}{20}
\]

Thus, the model for \( P(t) \) is:

\[
P(t) = 100 + 20 \sin\left(\frac{\pi}{20} t\right)
\]

### Part (b) - Graph the function and determine domain and range

- **Domain**: The time \( t \) is given as \( 0 \leq t \leq 120 \), so the domain is:
  \[
  [0, 120]
  \]

- **Range**: The pressure oscillates between 80 PSI and 120 PSI, so the range is:
  \[
  [80, 120]
  \]

### Part (c) - Find the times when the pressure is 90 PSI

We need to solve for \( t \) when \( P(t) = 90 \):

\[
100 + 20 \sin\left(\frac{\pi}{20} t\right) = 90
\]

Simplifying this equation:

\[
20 \sin\left(\frac{\pi}{20} t\right) = -10
\]
\[
\sin\left(\frac{\pi}{20} t\right) = -\frac{1}{2}
\]

The sine function equals \( -\frac{1}{2} \) at \( \frac{7\pi}{6} \) and \( \frac{11\pi}{6} \). Solving for \( t \):

1. \( \frac{\pi}{20} t = \frac{7\pi}{6} \implies t = \frac{7 \times 20}{6} = \frac{140}{6} \approx 23.33 \)
2. \( \frac{\pi}{20} t = \frac{11\pi}{6} \implies t = \frac{11 \times 20}{6} \approx 36.67 \)

So, the pressure reaches 90 PSI at \( t \approx 23.33 \) minutes and \( t \approx 36.67 \) minutes.

---

Would you like a further breakdown of any part of this solution? Here are some related questions to explore:

1. How do we determine the period of a sinusoidal function?
2. What would happen to the function \( P(t) \) if the maximum pressure changed to 130 PSI?
3. Could we model this scenario with a cosine function instead of sine? How?
4. How does the angular frequency \( \omega \) affect the graph of the function?
5. How would the graph change if the time limit was extended beyond 120 minutes?

**Tip**: When solving sinusoidal equations, always consider the periodic nature of sine, which gives multiple solutions over time.

Model a sinusoidal function P(t) for pressure oscillating between 80 and 120 PSI in a mechanical system, given that the pressure reaches 110 PSI after 10 minutes.

Learn how to model the sinusoidal pressure function P(t) for a machine with LP gas. The problem involves pressure oscillations between 80 PSI and 120 PSI, reaching 110 PSI after 10 minutes. Explore the graph of the function, its domain and range, and calculate when the pressure hits 90 PSI. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation