Problem 18 - Calculus with Concepts in Calculus

Question

Assume that the air pressure \(p(x)\) in pounds per square foot at \(x\) feet above sea level is given by $$ p(x) \approx 2140 e^{-0.000035 x} \quad \text { for } x \geq 0 $$ and that an airplane is losing altitude at the rate of 20 miles per hour. At what rate is the air pressure just outside the plane increasing when the plane is 2 miles above sea level?

Step-by-Step Solution

Answer

The air pressure outside the plane is increasing at approximately 5470.15 pounds per square foot per hour when at 2 miles above sea level.

Step 1: Convert Altitude Rate to Feet per Hour

First, convert the rate at which the airplane is losing altitude from miles per hour to feet per hour. Since there are 5280 feet in a mile, the rate is given by:\[ 20 \text{ miles/hour} \times 5280 \text{ feet/mile} = 105600 \text{ feet/hour}. \]

Step 2: Differentiate Pressure Function with Respect to Altitude

The air pressure is given by the function: \[ p(x) = 2140 e^{-0.000035 x}. \]To find how pressure changes with respect to altitude, we need to find \( \frac{dp}{dx} \): \[ \frac{dp}{dx} = 2140 \cdot (e^{-0.000035 x}) \cdot (-0.000035), \]which simplifies to:\[ \frac{dp}{dx} = -0.0749 e^{-0.000035 x}. \]

Step 3: Apply the Chain Rule for Derivatives

We want to find how the pressure changes over time, \( \frac{dp}{dt} \). By using the chain rule:\[ \frac{dp}{dt} = \frac{dp}{dx} \cdot \frac{dx}{dt}. \]Substitute \(\frac{dp}{dx}\) from the previous step and \(\frac{dx}{dt} = -105600 \text{ ft/hr} \) (since the plane is losing altitude):\[ \frac{dp}{dt} = -0.0749 e^{-0.000035 x} \cdot (-105600). \]

Step 4: Evaluate the Expression at 2 miles Above Sea Level

Substitute \(x = 2 \text{ miles} = 2 \times 5280 \text{ feet} = 10560 \text{ feet}\) into the expression found in Step 3:\[ \frac{dp}{dt} = -0.0749 e^{-0.000035 \times 10560} \cdot (-105600). \]Calculate the exponential term first using a calculator:\[ e^{-0.000035 \times 10560} \approx e^{-0.3696} \approx 0.6908. \]Then, evaluate the entire expression:\[ \frac{dp}{dt} = -0.0749 \times 0.6908 \times (-105600), \]which results in:\[ \frac{dp}{dt} \approx 5470.150688 \text{ pounds/ft}^2/\text{hr}. \]

Step 5: Interpret the Result

The positive result for \( \frac{dp}{dt} \) means that the air pressure outside the airplane is increasing at a rate of approximately 5470.15 pounds per square foot per hour when the airplane is at an altitude of 2 miles.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding how one quantity changes with respect to another. In the context of our problem, we want to determine how air pressure changes as a function of altitude. This means we are finding the derivative of the pressure function with respect to altitude.

The function given is \( p(x) = 2140 e^{-0.000035 x} \), which represents the air pressure dependent on altitude \(x\). Differentiation allows us to compute \( \frac{dp}{dx} \), the rate of change of pressure with respect to altitude. The differentiation process finds the slope or the rate of change of this function at any point \(x\).

To differentiate \( p(x) \), we apply differentiation rules specific to exponential functions. The derivative of \( e^{kx} \) is \( ke^{kx} \). So, applying this in our function, we get:

• First, multiply the constant 2140 with the derivative of \( e^{-0.000035 x} \).
• The derivative of \( e^{-0.000035 x} \) is \(-0.000035 imes e^{-0.000035 x}\).
• Thus, \( \frac{dp}{dx} = 2140 imes (e^{-0.000035 x}) imes (-0.000035) \).

This derivative tells us how air pressure decreases as altitude increases.

Chain Rule

The chain rule is a crucial method in calculus for finding the derivative of a composition of functions. It is especially useful when a variable depends on another, and that one depends on a third. In our scenario, the pressure changes with altitude and altitude changes over time, due to the airplane's movement.

To find the rate of pressure change with time, \( \frac{dp}{dt} \), when both pressure depends on altitude \(x\) and \(x\) depends on time \(t\), the chain rule helps to make this connection:

• First, identify \( \frac{dp}{dx} \), which we derived already.
• Second, identify \( \frac{dx}{dt} \), the rate of altitude change, given as \(-105600 \) feet/hour.
• Then, apply the chain rule: \( \frac{dp}{dt} = \frac{dp}{dx} \cdot \frac{dx}{dt} \).

This formula incorporates the rate at which the airplane descends, connecting it to how air pressure changes with time. Using the chain rule effectively allows us to understand changes in complex systems by breaking them down into individual parts.

Altitude Conversion

Before utilizing our calculus tools, we often need to convert units to ensure consistency throughout the calculations. This problem involves converting altitude from miles to feet, a crucial step due to the given rate of the aircraft's descent.

Initially, we know:

• 1 mile equals 5280 feet.
• The airplane's altitude change rate is 20 miles per hour.
• To convert this into feet per hour, multiply by 5280, resulting in \(20 \times 5280 = 105600 \text{ feet/hour} \).

Similarly, when determining the altitude of 2 miles in feet, which is \(2 \times 5280 = 10560 \text{ feet}\), it ensures that our calculations for \( \frac{dp}{dt} \) are based on uniform measurements. Proper conversion helps maintain accuracy and straightforward application of calculus formulas for real-world problems.