Question

Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in kilometers, the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}\) a. Compute the pressure at the summit of Mt. Everest, which has an elevation of roughly \(10 \mathrm{km} .\) Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\) d. Does \(p^{\prime}(z)\) increase or decrease with \(z ?\) Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)

Short answer

**Question:** Compute the pressure at the summit of Mt. Everest (10 km) and compare it to sea-level pressure using the given formula \(p(z) = 1000 e^{-z / 10}\). **Short Answer:** The pressure at the summit of Mt. Everest is calculated as \(1000 e^{-1}\), which is about 36.8% of the sea-level pressure. **Question:** Compute the average change in pressure in the first 5 km above Earth's surface. **Short answer:** The average change in pressure in the first 5 km above Earth's surface is \(\frac{1000 e^{-1/2} - 1000}{5}\). **Question:** Compute the rate of change of the pressure at an elevation of 5 km. **Short answer:** The rate of change of pressure at an elevation of 5 km is \(-100 e^{-1/2}\). **Question:** Analyze whether \(p'(z)\) increases or decreases with \(z\). **Short answer:** The derivative \(p'(z)\) decreases with respect to \(z\). **Question:** Interpret the meaning of \(\lim_{z \rightarrow \infty} p(z) = 0\). **Short answer:** The limit means that as we go infinitely far away from Earth's surface (increase the altitude), the atmospheric pressure approaches zero, indicating virtually no air molecules at extremely high altitudes.

Step-by-step solution

Evaluate the function at 10 km

To find the atmospheric pressure at the summit of Mt. Everest (10 km above sea level), we need to plug in \(z = 10\) in the given function: $$p(10) = 1000 e^{-10 / 10} = 1000 e^{-1}$$

Compare pressure at the summit of Mt. Everest and sea level

Using the given formula, we can compute the atmospheric pressure at sea level (z = 0): $$p(0) = 1000 e^{-0 / 10} = 1000 e^0 = 1000$$ To compare the pressure on Mt. Everest to the pressure at sea level, we find the ratio of the two pressures: $$\frac{p(10)}{p(0)} = \frac{1000 e^{-1}}{1000} = e^{-1} \approx 0.368$$ Thus, the atmospheric pressure at the summit of Mt. Everest is about 36.8% of the pressure at sea level. #b. Average change in pressure in the first 5 km#

Calculate the difference between pressures at 5 km and sea level

To find the average change in pressure within the first 5 km above Earth's surface, we first need to find the pressure at 5 km: $$p(5) = 1000 e^{-5 / 10} = 1000 e^{-1/2}$$ Then, compute the difference between the pressure at 5 km and sea level: $$\Delta p = p(5) - p(0) = 1000 e^{-1/2} - 1000$$

Compute the average change in pressure

Now, divide the pressure difference by the distance (5 km) to find the average change in pressure: $$\bar{p} = \frac{\Delta p}{5} = \frac{1000 e^{-1/2} - 1000}{5}$$ #c. Rate of change at an elevation of 5 km#

Compute the derivative of the function

To find the rate of change of pressure, we need to differentiate the atmospheric pressure function: $$p'(z) = -\frac{1}{10} \cdot 1000 e^{-z / 10} = -100 e^{-z / 10}$$

Evaluate the derivative at 5 km

Now, we can plug in \(z = 5\) to find the rate of change of atmospheric pressure at an elevation of 5 km: $$p'(5) = -100 e^{-5 / 10} = -100 e^{-1/2}$$ #d. Behavior of \(p'(z)\) with respect to \(z\)#

Investigate whether the derivative increases or decreases with \(z\)

The derivative of the atmospheric pressure function is given by: $$p'(z) = -100 e^{-z / 10}$$ Since the exponential function is always positive, and we have a negative sign in front of it, the derivative is always negative. Therefore, \(p'(z)\) decreases with respect to \(z\). #e. Interpretation of the limit as \(z\) approaches infinity#

Evaluate the limit

We are asked to interpret the meaning of the limit: $$\lim_{z \rightarrow \infty} p(z) = \lim_{z \rightarrow \infty} (1000 e^{-z / 10})$$ As \(z\) approaches infinity, the exponent \(-z / 10\) approaches \(-\infty\). Therefore, the exponential term approaches zero. Thus: $$\lim_{z \rightarrow \infty} p(z) = 1000 \cdot 0 = 0$$

Interpret the meaning of the limit

The limit represents the atmospheric pressure as we go infinitely far away from Earth's surface (i.e., as we keep increasing the altitude). In this case, the limit of the atmospheric pressure function as \(z\) goes to infinity is 0, which means that the atmospheric pressure approaches zero as we move farther away from Earth's surface. This makes sense since there would be virtually no air molecules at extremely high altitudes.

Key concepts

Exponential Decay

Exponential decay is a mathematical concept where quantities decrease at a rate proportional to their current value. In our atmospheric pressure model, this is represented by the equation: \[ p(z) = 1000 e^{-z / 10} \] Here, the atmospheric pressure decreases as altitude \(z\) increases. The factor \(-z / 10\) in the exponent indicates the rate of decay. The base of the exponential, \(e\), is a constant approximately equal to 2.718, which is the natural base for logarithms and is widely used to describe growth and decay processes.

• As \(z\) increases, \(e^{-z/10}\) decreases, indicating that pressure decreases exponentially with altitude.
• The exponential function never touches zero, but gets infinitely close to it as \(z\) grows.

Exponential decay appears in many natural processes beyond atmospheric pressure, such as radioactive decay, cooling of objects, and the discharge of capacitors.

Altitude

Altitude refers to the height above Earth's surface, usually measured in kilometers or meters. In the context of atmospheric pressure, altitude is a crucial variable because it directly influences pressure levels. The higher you climb in altitude:

• The lower the atmospheric pressure becomes, as demonstrated in the exponential decay model.
• This relationship shows that climbing to higher altitudes results in a significant drop in pressure, which can affect breathing and other physical processes.

The model \(p(z) = 1000 e^{-z / 10}\) is particularly used to understand how atmospheric pressure changes with altitude, providing insights into variations that affect aviation, weather forecasts, and climbing expeditions. Altitude factors into a range of scientific and engineering fields that study or utilize Earth's atmosphere.

Derivative

The derivative of a function gives us the rate at which the function's value changes. For the atmospheric pressure function \(p(z) = 1000 e^{-z / 10}\), the derivative is: \[ p'(z) = -100 e^{-z/10} \] This indicates how rapidly the atmospheric pressure is changing with altitude. The negative sign means the pressure is decreasing as altitude increases.

• The magnitude \(100 e^{-z/10}\) tells us how steep the decline in pressure is.
• At different altitudes, the derivative helps us understand the rate of pressure change, which is essential for weather predictions and altitudinal studies.

Fundamentally, derivatives are invaluable for optimization and predicting dynamic behaviors in mathematical modeling and real-world applications.

Limit

Limits help us understand the behavior of functions as they approach certain points, often infinity. In this exercise, the limit is examined as altitude tends to infinity: \[ \lim_{z \rightarrow \infty} p(z) = 0 \] This mathematical expression shows that, as we ascend to infinitely large altitudes, the atmospheric pressure approaches zero, though it never exactly becomes zero. Here's what this entails:

• It reflects the reality that at extremely high altitudes, such as in outer space, the density of air molecules becomes negligible.
• Understanding limits is crucial when dealing with functions that describe natural phenomena, as it allows us to predict the ultimate behavior of these functions.

The concept of a limit is foundational in calculus and is essential for analyzing the asymptotic behavior of functions in various scientific and engineering tasks.