Problem 34 - Calculus

Question

As the altitude of an object increases, the pressure of the earth's atmosphere on the object decreases. Let \(P=f(h)\) denote the atmospheric pressure, in kilopascals (kPa), at an altitude of \(h\) thousand meters above sea level. Interpret each of the following statements in terms of altitude and atmospheric pressure. (a) \(\quad f(1)=88\) (b) \(\quad f^{\prime}(1)=-11.5\) (c) \(\quad f^{-1}(10)=17.6\) (d) \(\quad\left(f^{-1}\right)^{\prime}(10)=-0.76\)

Step-by-Step Solution

Answer

(a) Pressure is 88 kPa at 1 km. (b) Pressure decreases by 11.5 kPa per km increase at 1 km. (c) Altitude is 17.6 km at 10 kPa. (d) Altitude increases by 0.76 km per kPa decrease at 10 kPa.

Step 1: Interpret f(1)=88

The notation \( f(1) = 88 \) tells us that when the altitude \( h \) is 1 thousand meters above sea level, the atmospheric pressure \( P \) is 88 kilopascals (kPa). So, at 1 km altitude, the pressure is 88 kPa.

Step 2: Interpret f'(1)=-11.5

The derivative \( f'(1) = -11.5 \) indicates the rate of change of atmospheric pressure at an altitude of 1 thousand meters. Specifically, it means that for every 1 thousand meters increase in altitude, the atmospheric pressure decreases by 11.5 kPa.

Step 3: Interpret f^{-1}(10)=17.6

The inverse function \( f^{-1}(10) = 17.6 \) implies that at a pressure of 10 kPa, the altitude is 17.6 thousand meters above sea level. So, 10 kPa is the atmospheric pressure at an altitude of 17.6 km.

Step 4: Interpret (f^{-1})'(10)=-0.76

The derivative of the inverse function \( (f^{-1})'(10) = -0.76 \) represents the rate of change of altitude with respect to pressure at 10 kPa. This means that for every decrease of 1 kPa in atmospheric pressure, the altitude increases by 0.76 thousand meters.

Key Concepts

Altitude

Altitude refers to the height of an object or point in relation to sea level. In the context of this exercise, altitude is expressed in thousands of meters. Atmospheric pressure changes with altitude. As you move higher above sea level, the thickness of the air layer above you decreases. This thinner layer means less air pressing down, resulting in lower pressure.

• At lower altitudes, there are more air molecules above, leading to higher pressure.
• At higher altitudes, there are fewer air molecules above, resulting in lower pressure.

Understanding this concept helps explain why mountain climbers often need oxygen tanks. At high altitudes, the air is too thin to supply enough oxygen for the body's needs.

Derivative

The derivative is a key concept in calculus that describes how a function changes as its input changes. It essentially gives us the rate of change of the function. In the context of our exercise, we look at the derivative of the pressure with respect to altitude.

• The derivative, denoted as \( f'(h) \), tells us how quickly atmospheric pressure \( P \) changes as the altitude \( h \) changes.
• For example, if \( f'(1) = -11.5 \), it indicates that for every 1 km increase in altitude, the atmospheric pressure decreases by 11.5 kPa.

Calculating and understanding derivatives allow us to predict and understand physical phenomena, such as how rapidly pressure drops as you climb a mountain.

Inverse function

An inverse function essentially reverses the role of inputs and outputs. For a function \( f \), the inverse function \( f^{-1} \) will give you back the original input when applied to the output of \( f \). In this exercise, the inverse function helps us understand altitude based on atmospheric pressure.

• Given \( f^{-1}(10) = 17.6 \), it means that when the atmospheric pressure is 10 kPa, the altitude is 17.6 thousand meters.
• This allows us to compute how high we need to be to experience a specific pressure level.

Inverse functions are powerful in scenarios where you want to control for output levels, like setting altitudes where pressure conditions need to be met.

Rate of change

The rate of change is a measure of how one quantity varies in relation to another quantity. It is a fundamental concept used to describe the relationship between various mathematical and physical phenomena. In our exercise, we look at two rates of change:

• The rate at which pressure decreases with altitude, represented by the derivative \( f'(h) \).
• The rate at which altitude increases as pressure decreases, seen in the derivative of the inverse function \( (f^{-1})'(10) \), which is -0.76. This means for every decrease of 1 kPa in pressure, altitude increases by 0.76 km.

This concept helps describe how alterations in one factor, like altitude, directly affect another, such as pressure. Understanding the rate of change is crucial for predicting future values in dynamic systems.