Problem 76 - College Algebra

Question

Atmospheric pressure \(P\) in pounds per square inch is represented by the formula \(P=14.7 e^{-0.21 x},\) where \(x\) is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.369 pounds per square inch? (Hint: there are 5280 feet in a mile)

Step-by-Step Solution

Answer

The mountain peak is approximately 18153 feet high.

Step 1: Identify Given Values

The problem provides the formula for atmospheric pressure: \(P = 14.7 e^{-0.21x}\), and the pressure at the mountain peak is given as 8.369 psi. We need to find the height in feet where this pressure occurs.

Step 2: Rearrange the Formula

Start with the equation \(8.369 = 14.7 e^{-0.21x}\). To solve for \(x\), first, divide both sides by 14.7: \(\frac{8.369}{14.7} = e^{-0.21x}\).

Step 3: Take the Natural Logarithm

Apply the natural logarithm (ln) to both sides to isolate \(x\): \(\ln\left(\frac{8.369}{14.7}\right) = \ln(e^{-0.21x})\). Using the property of logarithms, this simplifies to \(-0.21x = \ln\left(\frac{8.369}{14.7}\right)\).

Step 4: Solve for x

Now, solve for \(x\) by dividing both sides by \(-0.21\): \(x = \frac{\ln(\frac{8.369}{14.7})}{-0.21}\). Evaluate this using a calculator.

Step 5: Convert Miles to Feet

Calculate the value: \(x \approx 3.438\) miles. To convert miles to feet, multiply by the number of feet per mile: \(3.438 \times 5280 = 18152.64\) feet. Round to the nearest foot.

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a fundamental mathematical concept used to solve exponential equations. It is particularly useful when dealing with exponential functions involving the mathematical constant \( e \), which is approximately equal to 2.718. The natural logarithm is the inverse operation of exponentiation with base \( e \), meaning it helps to undo exponential effects.

• For instance, if you have an equation like \( e^y = x \), applying the natural logarithm to both sides gives \( \ln(x) = y \).
• This property allows you to solve for variables that are exponents in equations containing \( e \).

In the context of the given problem, we use \( \ln \) to isolate \( x \) in the atmospheric pressure formula \( 8.369 = 14.7 e^{-0.21x} \). By taking the natural logarithm of each side, we can solve for \( x \), which represents the height in miles of a mountain based on atmospheric pressure.

Atmospheric Pressure Formula

The atmospheric pressure formula is an exponential function that describes how pressure decreases with altitude. It is a key part of understanding how the atmosphere works at different heights. The formula used to describe this relationship is generally given by:
\[P = 14.7 e^{-0.21 x}\]Where:

• \( P \) is the atmospheric pressure at a certain height in pounds per square inch (psi).
• \( e \) is the base of the natural logarithm, approximately equal to 2.718.
• \( x \) is the height above sea level in miles.

The base pressure at sea level is approximately 14.7 psi. This value decreases exponentially as elevation increases. In the original exercise, this formula helps predict atmospheric pressure at different altitudes by solving for \( x \), which provides the mountain's height given the pressure measurement.

Unit Conversion

Unit conversion is an essential skill in solving problems involving different units of measurement. It ensures that our results are consistent and accurate. When calculating physical quantities like distance, converting between units like miles and feet is often necessary because these units measure the same dimension (length) but on different scales.

• There are 5280 feet in one mile.
• To convert miles to feet, multiply the number of miles by 5280.
• Conversely, to convert feet to miles, divide the number of feet by 5280.

In the original problem, once we determined the height \( x \) in miles using the atmospheric pressure formula, we then converted this value into feet to specify the mountain's height more precisely. This conversion allows us to express the answer in a unit more familiar and practical for describing elevation in everyday terms.