/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Atmospheric Pressure Earth's atm... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 41

Atmospheric Pressure Earth's atmospheric pressure \(p\) is often modeled by assuming that the rate \(d p / d h\) at which \(p\) changes with the altititude \(h\) above sea level is proportional to \(p\) . Suppose that the pressure at sea level is 1013 millibars (about 14.7 lb/in \(^{2} )\) and that the pressure at an altitude of 20 \(\mathrm{km}\) is 90 millibars. (a) Solve the initial value problem $$\begin{array}{ll}{\text { Differential equation: }} & {\frac{d p}{d h}=k p} \\\ {\text { Initial condition: }} & {p=p_{0} \text { when } h=0}\end{array}$$ to express \(p\) in terms of \(h .\) Determine the values of \(p_{0}\) and \(k\) from the given altitude-pressure data. (b) What is the atmospheric pressure at \(h=50 \mathrm{km} ?\) (c) At what altitude does the pressure equal 900 millibars?

Short Answer

Expert verified
The atmospheric pressure at an altitude of 50 km is approximately 30.789 millibars. The altitude at which the pressure equals 900 millibars is approximately 116.3 km.

Step by step solution

01

Finding the general solution of the differential equation

Our differential equation is given by \( \frac{dp}{dh} = kp \). This is a first-order linear differential equation. Separating the variables and integrating, we obtain the general solution: \( p(h) = p_0 e^{kh} \). Where \( p_0 \) is the initial pressure (pressure at sea level), \( k \) is a constant, and \( h \) is altitude.
02

Calculating the constant k

To get the value for \(k\), we use the provided atmospheric pressure at the altitude of 20 km. We solve the equations \( 90 = p_0 * e^{20k} \) and \( 1013 = p_0 \), which gives \( k = \frac{1}{20} \ln \left( \frac{90}{1013} \right) \). Now we have our full model \( p(h) = 1013 * e^{ \frac{h}{20} \ln \left( \frac{90}{1013} \right) } \).
03

Finding the pressure at an altitude of 50 km

We can directly substitute \( h = 50 \) km into our model equation to give: \( p(50) = 1013 * e^{ \frac{50}{20} \ln \left( \frac{90}{1013} \right) } \).
04

Determining the altitude at which the pressure equals 900 millibars

We find the altitude \( h \) for when \( p = 900 \) mb by setting our model equal to 900 and solving for \( h \): \( 900 = 1013 * e^{ \frac{h}{20} \ln \left( \frac{90}{1013} \right) } \). Then, isolate \( h \) to give the altitude: \( h = 20 \frac{\ln \left(\frac{900}{1013}\right)}{\ln \left(\frac{90}{1013}\right)} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Understanding differential equations is crucial when we discuss atmospheric pressure variability with altitude. In this problem, the atmospheric pressure change with respect to altitude is described using a differential equation. Specifically, the rate of change of pressure \((dp/dh)\) is proportional to the current pressure \(p\). This simplifies into an equation of the form \(\frac{dp}{dh} = kp\), where \(k\) is a proportionality constant. This type of equation is a first-order linear differential equation, which is particularly significant in fields like physics and engineering for modeling natural processes.

To solve this, we separate the variables, integrate, and form a general solution. By isolating \(dp/p\) on one side and \(dh\) on the other, we integrate both sides. This results in an equation involving the natural exponential function, typically written as \(p(h) = p_0 e^{kh}\), where \(p_0\) is the initial condition. This form beautifully links to exponential functions, indicating exponential decay or growth behavior, common in nature.
Exponential Functions
Exponential functions are pervasive in calculus, especially when dealing with atmospheric pressure. The relationship \(p(h) = p_0 e^{kh}\) for atmospheric pressure is a classic model that relies on exponential function properties. The nature of this function essentially captures how quantities change at a rate proportional to their current value, which is indicative of either exponential decay or growth.

The exponential function \(e^{kh}\) in the equation dynamically alters the pressure \(p\) as altitude \(h\) increases or decreases. A negative \(k\) in our differential model reflects the real-world observation that pressure decreases with increasing altitude, aptly showing exponential decay. As altitude rises, the air gets thinner, leading to a decrease in atmospheric pressure, and these changes can be mathematically represented and predicted using these exponential functions.
Rate of Change
Rate of change symbolizes how a quantity varies with another in calculus. Here, it refers to how atmospheric pressure changes with altitude. In calculus, rate of change is mathematically expressed as a derivative, denoted as \(\frac{dp}{dh}\), which represents how pressure \(p\) changes for a tiny change in altitude \(h\) in our scenario.

One important insight from this is that the rate of change being proportional to a current value often implies that the change is consistent across similar conditions. In atmospheric contexts, this implies that changes in altitude steadily affect the pressure in a predictable manner. Discerning these rates and patterns helps in crafting models like the one we've seen, aiding in weather predictions, aviation, and more. Thus, learning about this rate allows us to better understand and predict how atmospheric pressure might behave as you ascend through different altitudes.

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Most popular questions from this chapter

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Trigonometric Substitution Suppose \(u=\tan ^{-1} x\) (a) Use the substitution \(x=\tan u, d x=\sec ^{2} u d u\) to show that \(\int_{0}^{\sqrt{3}} \frac{d x}{\sqrt{1+x^{2}}}=\int_{0}^{\pi / 3} \sec u d u\) (b) Use the hint in Exercise 45 to evaluate the definite integral without a calculator.

Limited Growth Equation Another differential equation that models limited growth of a population \(P\) in an environment with carrying capacity \(M\) is \(d P / d t=k(M-P)\) (where \(k>0\) and \(M>0\) ). (a) Show that \(P=M-A e^{-k t},\) where \(A\) is a constant determined by an appropriate initial condition. (b) What is lim \(P(t) ? ~ M\) (c) For what time \(t \geqslant 0\) is the population growing the fastest? (d) Writing to Learn How does the growth curve in this model differ from the growth curve in the logistic model? See answ

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