91Ó°ÊÓ

When understanding the concept of the **Air Pressure Equation** used in the given exercise, it is important to know the relationship between air pressure and altitude. The equation in focus here is:\[ p(h) = 14.7 e^{-0.0000385 h} \] This equation indicates that as altitude, represented by \( h \) (in feet), increases, the air pressure \( p(h) \) decreases exponentially. The value \( 14.7 \) represents the air pressure at sea level in pounds per square inch (psi). The constant \( -0.0000385 \) determines how rapidly the pressure decreases with altitude. Using this equation, you can calculate the pressure at various altitudes, which is crucial for different engineering and scientific applications.- **Variable Definitions**: - \( p(h) \): Air pressure at altitude \( h \) - \( e \): Base of the natural logarithm, approximately 2.71828 - \( h \): Altitude above sea level in feet- **Exponential Decay**: This particular equation describes a natural exponential decay, where the pressure decreases as altitude increases.

In solving the exercise, particularly in the steps for determining altitude, **Natural Logarithms** (ln) play a crucial role. Natural logarithms are the inverse of exponential functions, which makes them useful for solving equations where the variable is in the exponent, like our air pressure equation. For instance, if we know that:\[ e^x = a \]we can find \( x \) by taking the natural logarithm:\[ x = \ln(a) \]In this problem, to solve for \( h \) when the air pressure at a specific altitude needs to be determined, you rearrange the air pressure equation:- Apply the ln function to both sides of the equation after isolating the exponential term:\[ \ln\left(\frac{p}{14.7}\right) = -0.0000385 h \]- This allows you to solve for \( h \) by isolating it:\[ h = \frac{\ln\left(\frac{p}{14.7}\right)}{-0.0000385} \]The use of natural logarithms simplifies the process of manipulating equations involving exponential functions and helps you determine unknown variables, like altitude in this context.

**Altitude Calculation** is the key objective in the exercise, specifically finding the height at which a certain air pressure is achieved. We use the air pressure formula, manipulate it using natural logarithms, and solve for the altitude \( h \).- **Part (a)** requires finding \( h \) when the pressure \( p(h) = 10 \, \text{lb/in}^2 \): - Set the equation: \( 10 = 14.7 e^{-0.0000385 h} \) - Divide by 14.7 and use natural logarithms: - Solve for \( h \): \[ h = \frac{\ln\left(\frac{10}{14.7}\right)}{-0.0000385} \] - **Part (b)** involves finding \( h \) where pressure is half of its sea-level value: - Here, \( p(h) = 7.35 \): - Set up: \( 7.35 = 14.7 e^{-0.0000385 h} \), divide and use natural logarithms: - Solve for \( h \) similarly: \[ h = \frac{\ln\left(\frac{1}{2}\right)}{-0.0000385} \]By solving these expressions, you are calculating how high you need to go until the air pressure drops to given levels. Understanding these procedures enables you to determine altitudes based on air pressure quickly and is particularly valuable in fields like aviation and meteorology.