Problem 45 - Algebra A Combined Function

Question

The formula \(P=14.7 e^{-0.21 x}\) gives the average atmospheric pressure \(P\), in pounds per square inch, at an altitude \(x\), in miles above sea level. Use this formula to solve. Round to the nearest tenth. Find the elevation of a Delta jet if the atmospheric pressure outside the jet is 7.5 pounds per square inch.

Step-by-Step Solution

Answer

The elevation of the Delta jet is approximately 3.2 miles.

Step 1: Understand the Problem

We have the formula for atmospheric pressure as a function of altitude: \( P = 14.7 e^{-0.21 x} \). We need to find the altitude \( x \) when the pressure \( P \) is 7.5 pounds per square inch.

Step 2: Set Up the Equation

Substitute the given pressure \( P = 7.5 \) into the equation: \[ 7.5 = 14.7 e^{-0.21 x} \] We need to solve this equation for \( x \).

Step 3: Isolate the Exponential Term

Divide both sides of the equation by 14.7 to isolate the exponential term:\[ \frac{7.5}{14.7} = e^{-0.21 x} \] Calculating the left side gives approximately 0.5102.

Step 4: Solve for the Exponent

Use the natural logarithm to solve for \( x \). Take the natural log of both sides:\[ \ln(0.5102) = \ln(e^{-0.21 x}) \] Since \( \ln(e^{a}) = a \), we simplify to:\[ \ln(0.5102) = -0.21 x \]

Step 5: Compute the Natural Logarithm

Calculate \( \ln(0.5102) \) which is approximately -0.6722. So the equation becomes:\[ -0.6722 = -0.21 x \]

Step 6: Solve for x

Divide both sides by -0.21 to find \( x \):\[ x = \frac{-0.6722}{-0.21} \] This calculation gives approximately \( x = 3.2 \).

Step 7: Round the Result

The altitude \( x \) is found to be approximately 3.2 miles. Since we are asked to round to the nearest tenth, the final answer is 3.2 miles.

Key Concepts

Understanding Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of air above a certain point, typically measured in pounds per square inch (psi) or pascal. It is highest at sea level where the air is dense and decreases as altitude increases due to the thinning of the air.
This exercise uses a particular formula, \( P = 14.7 e^{-0.21 x} \), to express atmospheric pressure at different altitudes. Here, \( P \) represents atmospheric pressure and \( x \) is the altitude in miles.
Lower atmospheric pressure at higher altitudes is why breathing becomes more challenging on mountains and why aircraft cabins need to be pressurized. As you ascend, the molecules of air spread out, resulting in fewer molecules above you, leading to decreased pressure.

The Role of Natural Logarithms

Natural logarithms, denoted as \( \ln \), are logarithms to the base \( e \), where \( e \) is an irrational number approximately equal to 2.718. They are essential in solving equations involving exponential functions.
In this exercise, we used the natural logarithm to solve for altitude by isolating the variable \( x \) in the exponential equation. When both sides of the equation \( e^{-0.21 x} = 0.5102 \) are passed through the natural logarithm, it simplifies due to the property \( \ln(e^{a}) = a \).

• This allows for easy manipulation and solving of exponential equations.
• Understanding how to convert exponentials using logarithms is a valuable tool in mathematics, physics, and engineering applications.

Using natural logarithms simplifies the process and provides clear solutions to what initially appear as complex exponential equations.

Techniques of Altitude Calculation

Altitude calculation using atmospheric pressure relies on understanding the relationship between pressure and elevation. In our given formula, \( P = 14.7 e^{-0.21 x} \), the variable \( x \) represents the altitude.
When given an atmospheric pressure, as in the exercise question where \( P = 7.5 \), reasoning and computation are done step-by-step to find \( x \). Here’s how it unfolds:

• Direct substitution of \( P = 7.5 \) into the formula sets the scene for further calculation.
• Dividing both sides by 14.7 isolates the exponential term, simplifying the equation to make it easier to handle logarithmic operations.

Calculating \( \ln(0.5102) \) involves using a calculator, resulting in a manageable linear equation that lets us solve for \( x \).
This approach emphasizes the interplay between algebra and calculus in deriving meaningful data from mathematical models.