Problem 44 - 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 average atmospheric pressure of Pikes Peak, which is 2.7 miles above sea level.

Step-by-Step Solution

Answer

The average atmospheric pressure of Pikes Peak is approximately 8.3 psi.

Step 1: Understanding the Problem

We need to find the average atmospheric pressure at 2.7 miles above sea level using the formula \( P = 14.7 e^{-0.21 x} \). Here, \( x \) is the altitude in miles, and for Pikes Peak, \( x = 2.7 \).

Step 2: Substituting the Value of x

Substitute \( x = 2.7 \) into the formula to get: \[ P = 14.7 e^{-0.21 \times 2.7} \]

Step 3: Calculating the Exponent

Calculate the product \(-0.21 \times 2.7\): \(-0.21 \times 2.7 = -0.567\).So the expression becomes:\[ P = 14.7 e^{-0.567} \]

Step 4: Evaluating the Exponential Function

Calculate \( e^{-0.567} \) using a scientific calculator or an online calculator. The approximate value is: \[ e^{-0.567} \approx 0.567 \]

Step 5: Multiplying to Find Pressure

Multiply \( 14.7 \) by \( 0.567 \): \[ P = 14.7 \times 0.567 \approx 8.3 \]

Step 6: Rounding the Result

After multiplication, the result is \( 8.3309 \), which rounded to the nearest tenth is \( 8.3 \). Therefore, the average atmospheric pressure of Pikes Peak is approximately \( 8.3 \) pounds per square inch.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to the power of a variable exponent. In the formula for atmospheric pressure, \( P = 14.7 e^{-0.21x} \), the base of the exponential function is the mathematical constant \( e \), which is approximately 2.71828. This constant often appears in real-world phenomena, like decay or growth processes.
Exponential functions like this one describe how the pressure decreases exponentially with increasing altitude. At every mile above sea level, the atmospheric pressure is reduced by a factor depending on the value of the exponent \(-0.21x\).

• A negative exponent indicates a decay function, meaning the value decreases as \( x \) increases.
• The function's curve for atmospheric pressure is not a straight line. Instead, it slopes downward more steeply as the altitude increases.

Understanding how exponential functions work is critical for accurately modeling natural processes, such as calculating atmospheric conditions at various altitudes.

Substitution in Formulas

Substitution is a crucial step in many mathematical calculations. It involves replacing a variable with a specific value to find an unknown quantity. In the problem regarding Pikes Peak's pressure, substitution is used to replace \( x \) with 2.7, the given altitude in miles.
This process transforms the general formula into a specific one that allows us to calculate the pressure at this height:
\[ P = 14.7 e^{-0.21 \times 2.7} \]
Here are a few tips to ensure correct substitution:

• Ensure you understand what each symbol represents in the formula before making a substitution.
• Perform arithmetic operations step-by-step, as these interim calculations are important for accuracy.

Substitution converts abstract formulas into concrete computations that are easier to evaluate.

Rounding Numbers

Rounding is an important skill in mathematics, especially when dealing with results from calculations that produce many decimal places. In most practical scenarios, only a certain degree of precision is necessary or manageable. In this exercise, rounding is applied to the final calculated value of the atmospheric pressure.
This is necessary because:

• Real-world measurements often do not require extreme precision.
• It simplifies the number, making it easier to understand and discuss.

To round to the nearest tenth, examine the digit in the hundredths place. If it is 5 or higher, increase the tenths digit by one. From our previous steps, \( P = 8.3309 \) would round to \( 8.3 \) because the hundredths place is 3, which does not necessitate rounding up.
Successful rounding isn't just about shortening numbers; it’s about presenting a realistic representation of calculated results.