8.6: Integrable Functions. Convergence Theorems
I. Some important theorems apply to integrable functions.
If \(f, g : S \rightarrow E^{*}\) are integrable on a set \(A \in \mathcal{M}\) in \((S, \mathcal{M}, m),\) so is
\[p f+q g\]
for any \(p, q \in E^{1},\) and
\[\int_{A}(p f+q g)=p \int_{A} f+q \int_{A} g;\]
in particular,
\[\int_{A}(f \pm g)=\int_{A} f \pm \int_{A} g.\]
- Proof
-
By Problem 5 in §5,
\[\overline{\int}_{A} f+\overline{\int}_{A} g \geq \overline{\int}_{A}(f+g) \geq \underline{\int}_{A}(f+g) \geq \underline{\int}_{A} f + \underline{\int}_{A} g.\]
(Here
\[\overline{\int}_{A} f, \underline{\int}_{A} f, \overline{\int}_{A} g, \text { and } \underline{\int}_{A} g\]
are finite by integrability; so all is orthodox.)
As
\[\overline{\int}_{A} f=\underline{\int}_{A} f \text { and } \overline{\int}_{A} g=\underline{\int}_{A} g,\]
the inequalities turn into equalities, so that
\[\int_{A} f+\int_{A} g=\overline{\int}_{A}(f+g)=\underline{\int}_{A}(f+g).\]
Using also Theorem 1(e)(e') from §5, we obtain the desired result for any \(p, q \in E^{1}. \quad \square\)
A function \(f : S \rightarrow E^{*}\) is integrable on \(A\) in \((S, \mathcal{M}, m)\) iff
(i) it is \(m\)-measurable on \(A,\) and
(ii) \(\overline{\int}_{A} f\) (equivalently \(\overline{\int}_{A}|f|\)) is finite.
- Proof
-
If these conditions hold, \(f\) is integrable on \(A\) by Theorem 3 of §5.
Conversely, let
\[\overline{\int}_{A} f=\underline{\int}_{A} f \neq \pm \infty.\]
Using Lemma 2 in §5, fix measurable maps \(g\) and \(h\) (\(g \leq f \leq h\)) on \(A,\) with
\[\int_{A} g=\int_{A} f=\int_{A} h \neq \pm \infty.\]
By Theorem 3 in §5, \(g\) and \(h\) are integrable on \(A;\) so by Theorem 1,
\[\int_{A}(h-g)=\int_{A} h-\int_{A} g=0.\]
As
\[h-g \geq h-f \geq 0,\]
we get
\[\int_{A}(h-f)=0,\]
and so by Theorem 1(h) of §5, \(h-f=0\) a.e. on \(A\).
Hence \(f\) is almost measurable on \(A,\) and
\[\int_{A} f \neq \pm \infty\]
by assumption. From formula (1), we then get
\[\int_{A} f^{+} \text { and } \int_{A} f^{-}<\infty,\]
and hence
\[\int_{A}|f|=\int_{A}\left(f^{+}+f^{-}\right)=\int_{A} f^{+}+\int_{A} f^{-}<\infty\]
by Theorem 1 and by Theorem 2 of §2. Thus all is proved.\(\quad \square\)
Simultaneously, we also obtain the following corollary.
A function \(f : S \rightarrow E^{*}\) is integrable on \(A\) iff \(f^{+}\) and \(f^{-}\) are.
If \(f, g : S \rightarrow E^{*}\) are integrable on \(A,\) so also are
\[f \vee g, f \wedge g,|f|, \text { and } k f \text { for } k \in E^{1},\]
with
\[\int_{A} k f=k \int_{A} f.\]
Exercise!
For products \(f g,\) this holds if \(f\) or \(g\) is bounded. In fact, we have the following theorem.
Let \(f\) be \(m\)-measurable and bounded on A. Set
\[p=\inf f[A] \text { and } q=\sup f[A].\]
Then if \(g\) is \(m\)-integrable on \(A,\) so is \(f g,\) and
\[\int_{A} f|g|=c \int_{A}|g|\]
for some \(c \in[p, q]\).
If, further, \(f\) also has the Darboux property on A (Chapter 4, §9), then \(c=f(x_{0})\) for some \(x_{0} \in A.\)
- Proof
-
By assumption,
\[\left(\exists k \in E^{1}\right) \quad|f| \leq k\]
on \(A.\) Hence if \(\int_{A}|g|=0\),
\[
\left|\int_{A} f|g|\right| \leq \int_{A}|f g| \leq k \int_{A}|g|=0;\]so any \(c \in[p, q]\) yields
\[\int_{A} f|g|=c \int_{A}|g|=0.\]
If, however, \(\int_{A}|g| \neq 0,\) the number
\[c=\left(\int_{A} f|g|\right) / \int_{A}|g|\]
is the required constant.
Moreover, as \(f\) and \(g\) are \(m\)-measurable on \(A,\) so is \(f g;\) and as
\[\left|\int_{A} f g\right| \leq|c| \int_{A}|g|<\infty,\]
\(f g\) is integrable on \(A\) by Theorem 2.
Finally, if \(f\) has the Darboux property and if \(p<c<q\) (with \(p, q\) as above), then
\[f(x)<c<f(y)\]
for some \(x, y \in A\) (why?); hence by the Darboux property, \(f\left(x_{0}\right)=c\) for some \(x_{0} \in A.\)
If, however,
\[c \leq \inf f[A]=p,\]
then
\[(f-c)|g| \geq 0\]
and
\[\int_{A}(f-c)|g|=\int_{A} f|g|-c \int_{A}|g|=0 \text { (why?);}\]
so by Theorem 1(h) in §5, \(f-c=0\) a.e. on \(A.\) Then surely \(f\left(x_{0}\right)=c\) for some \(x_{0} \in A\) (except the trivial case \(m A=0\)). This also implies \(c \in f[A] \in[p, q]\).
Proceed similarly in the case \(c \geq q. \quad \square\)
If \(f\) is integrable on \(A \in \mathcal{M},\) it is so on any \(B \subseteq A(B \in \mathcal{M})\).
- Proof
-
Apply Theorem 1(f) in §5, and Theorem 3 of §5, to \(f^{+}\) and \(f^{-}. \quad \square\)
II. Convergence Theorems. If \(f_{n} \rightarrow f\) on \(A\) (pointwise, a.e., or uniformly), does it follow that
\[\int_{A} f_{n} \rightarrow \int_{A} f ?\]
To give some answers, we need a lemma.
If \(f \geq 0\) on \(A \in \mathcal{M}\) and if
\[\underline{\int}_{A} f>p \in E^{*},\]
there is an elementary and nonnegative map \(g\) on \(A\) such that
\[\int_{A} g>p,\]
and \(g<f\) on \(A\) except only at those \(x \in A\) (if any) at which
\[f(x)=g(x)=0.\]
(We then briefly write \(g \subset f\) on \(A.\))
- Proof
-
By Lemma 1 in §5, there is an elementary and nonnegative map \(G \leq f\) on \(A,\) with
\[\underline{\int}_{A} f \geq \int_{A} G>p.\]
For the rest, proceed as in Problem 7 of §4, replacing \(f\) by \(G\) there.\(\quad \square\)
If \(0 \leq f_{n} \nearrow f\) (a.e.) on \(A \in \mathcal{M},\) i.e.,
\[0 \leq f_{n} \leq f_{n+1} \quad (\forall n),\]
and \(f_{n} \rightarrow f\) (a.e.) on \(A,\) then
\[\overline{\int}_{A} f_{n} \nearrow \overline{\int}_{A} f.\]
- Proof for \(\mathcal{M}\)-measurable \(f_{n}\) and \(f\) on \(A.\)
-
By Corollary 2 in §5, we may assume that \(f_{n} \nearrow f\) (pointwise) on \(A\) (otherwise, drop a null set).
By Theorem 1(c) of §5, \(0 \leq f_{n} \nearrow f\) implies
\[0 \leq \int_{A} f_{n} \leq \int_{A} f,\]
and so
\[\lim _{n \rightarrow \infty} \int_{A} f_{n} \leq \int_{A} f.\]
The limit, call it \(p,\) exists in \(E^{*},\) as \(\left\{\int_{A} f_{n}\right\} \uparrow.\) It remains to show that
\[p \geq \overline{\int}_{A} f=\underline{\int}_{A} f.\]
(We know that
\[\overline{\int}_{A} f=\underline{\int}_{A} f,\]
by the assumed measurability of \(f;\) see Theorem 3 in §5.)
Suppose
\[\underline{\int}_{A} f>p.\]
Then Lemma 1 yields an elementary and nonnegative map \(g \subset f\) on \(A,\) with
\[p<\int_{A} g.\]
Let
\[A_{n}=A\left(f_{n} \geq g\right), \quad n=1,2, \ldots.\]
Then \(A_{n} \in \mathcal{M}\) and
\[A_{n} \nearrow A=\bigcup_{n=1}^{\infty} A_{n}.\]
For if \(f(x)=0,\) then \(x \in A_{1},\) and if \(f(x)>0,\) then \(f(x)>g(x),\) so that \(f_{n}(x)>g(x)\) for large \(n;\) hence \(x \in A_{n}.\)
By Note 4 in §5, the set function \(s=\int g\) is a measure, hence continuous by Theorem 2 in Chapter 7, §4. Thus
\[\int_{A} g=s A=\lim _{n \rightarrow \infty} s A_{n}=\lim _{n \rightarrow \infty} \int_{A_{n}} g.\]
But as \(g \leq f_{n}\) on \(A_{n},\) we have
\[\int_{A_{n}} g \leq \int_{A_{n}} f_{n} \leq \int_{A} f_{n}.\]
Hence
\[\int_{A} g=\lim \int_{A_{n}} g \leq \lim \int_{A} f_{n}=p,\]
contrary to \(p<\int_{A} g.\) This contradiction completes the proof.\(\quad \square\)
If \(f_{n} \geq 0\) on \(A \in \mathcal{M}(n=1,2, \ldots),\) then
\[\overline{\int}_{A} \underline{\lim} f_{n} \leq \underline{\lim } \overline{\int}_{A} f_{n}.\]
- Proof
-
Let
\[g_{n}=\inf _{k \geq n} f_{k}, \quad n=1,2, \ldots;\]
so \(f_{n} \geq g_{n} \geq 0\) and \(\left\{g_{n}\right\} \uparrow\) on \(A.\) Thus by Theorem 4,
\[\overline{\int}_{A} \lim g_{n} = \lim \overline{\int}_{A} g_{n} = \underline{\lim} \overline{\int}_{A} g_{n} \leq \underline{\lim} \overline{\int}_{A} f_{n}.\]
But
\[\lim _{n \rightarrow \infty} g_{n}=\sup _{n} g_{n}=\sup _{n} \inf _{k \geq n} f_{k}=\lim _{n} f_{n}.\]
Hence
\[\overline{\int}_{A} \underline{\lim} f_{n} = \overline{\int}_{A} \lim g_{n} \leq \underline{\lim} \overline{\int}_{A} f_{n},\]
as claimed.\(\quad \square\)
Let \(f_{n} : S \rightarrow E\) be \(m\)-measurable on \(A \in \mathcal{M} (n=1,2,\ldots).\) Let
\[f_{n} \rightarrow f \text{ (a.e.) on } A.\]
Then
\[\lim _{n \rightarrow \infty} \int_{A}\left|f_{n}-f\right|=0,\]
provided that there is a map \(g : S \rightarrow E^{1}\) such that
\[\int_{A} g<\infty\]
and
\[(\forall n) \quad\left|f_{n}\right| \leq g \text { a.e. on } A.\]
- Proof
-
Neglecting null sets, we may assume that
\[\left|f_{n}\right| \leq g<\infty\]
on \(A\) and \(f_{n} \rightarrow f\) (pointwise) on \(A;\) so \(|f| \leq g\) and
\[\left|f_{n}-f\right| \leq\left|f_{n}\right|+|f| \leq 2 g\]
on \(A.\) As \(|f|<\infty,\) we have
\[\left|f_{n}-f\right| \rightarrow 0\]
on \(A.\) Hence, setting
\[h_{n}=2 g-\left|f_{n}-f\right| \geq 0,\]
we get
\[2 g=\lim _{n \rightarrow \infty} h_{n}=\underline{\lim} h_{n}.\]
We may also assume that \(g\) is measurable on \(A.\) (If not, replace it by a measurable \(G \geq g,\) with
\[\int_{A} G=\int_{A} g<\infty,\]
by Lemma 2 in §5.) Then all
\[h_{n}=2 g-\left|f_{n}-f\right|\]
are measurable (even integrable) on \(A\).
Thus by Lemma 2,
\begin{aligned} \int_{A} 2 g = \int_{A} \underline{\lim} h_{n} & \leq \underline{\lim} \int_{A}\left(2 g-\left|f_{n}-f\right|\right) \\ &=\underline{\lim}\left(\int_{A} 2 g+\int_{A}\left(-\left|f_{n}-f\right|\right)\right) \\ &= \int_{A} 2 g+\underline{\lim}\left(-\int_{A}\left|f_{n}-f\right|\right) \\ &= \int_{A} 2 g-\overline{\lim} \int_{A}\left|f_{n}-f\right|. \end{aligned}
(See Problems 5 and 8 in Chapter 2, §13.)Canceling \(\int_{A} 2 g\) (finite!), we have
\[0 \leq-\overline{\lim } \int_{A}\left|f_{n}-f\right|.\]
Hence
\[0 \geq \overline{\lim } \int_{A}\left|f_{n}-f\right| \geq \underline{\lim }\int_{A}\left|f_{n}-f\right| \geq 0,\]
as \(\left|f_{n}-f\right| \geq 0.\) This yields
\[0=\overline{\lim } \int_{A}\left|f_{n}-f\right|=\underline{\lim} \int_{A}\left|f_{n}-f\right|=\lim \int_{A}\left|f_{n}-f\right|,\]
as required.\(\quad \square\)
Note 1. Theorem 5 holds also for complex and vector-valued functions (for \(\left|f_{n}-f\right|\) is real).
In the extended-real case, Theorems 1(g) in §5 and Theorems 1 and 2 in §6 yield
\[\left|\int_{A} f_{n}-\int_{A} f\right|=\left|\int_{A}\left(f_{n}-f\right)\right| \leq \int_{A}\left|f_{n}-f\right| \rightarrow 0,\]
i.e.,
\[\int_{A} f_{n} \rightarrow \int_{A} f.\]
Moreover, \(f\) is integrable on \(A,\) being measurable (why?), with
\[\int_{A}|f| \leq \int_{A} g<\infty.\]
For complex and vector-valued functions, this will follow from §7. Observe that Theorem 5, unlike Theorem 4, requires the \(m\)-measurability of the \(f_{n}\).
Note 2. Theorem 5 fails if there is no "dominating"
\[g \geq\left|f_{n}\right| \text { with } \int_{A} g<\infty,\]
even if \(f\) and the \(f_{n}\) are integrable.
Let \(m\) be Lebesgue measure in \(A=E^{1}, f=0,\) and
\[f_{n}=\left\{\begin{array}{ll}{1} & {\text { on }[n, n+1],} \\ {0} & {\text { elsewhere. }}\end{array}\right.\]
Then \(f_{n} \rightarrow f\) and \(\int_{A} f_{n}=1;\) so
\[\lim _{n \rightarrow \infty} \int_{A} f_{n}=1 \neq 0=\int_{A} f.\]
The trouble is that any
\[g \geq f_{n} \quad(n=1,2, \ldots)\]
would have to be \(\geq 1\) on \(B=[1, \infty);\) so
\[\int_{A} g \geq \int_{B} g=1 \cdot m B=\infty,\]
instead of \(\int_{A} g<\infty\).
This example also shows that \(f_{n} \rightarrow f\) alone does not imply
\[\int_{A} f_{n} \rightarrow \int_{A} f.\]
Given \(f : S \rightarrow E\) with
\[\overline{\int}_{A}|f|<\infty\]
and \(\varepsilon>0,\) there is \(\delta>0\) such that
\[\overline{\int}_{X}|f|<\varepsilon\]
whenever
\[m X<\delta \quad(A \supseteq X, X \in \mathcal{M}).\]
- Proof
-
By Lemma 2 in §5, fix \(h \geq|f|,\) measurable on \(A,\) with
\[\int_{A} h=\overline{\int}_{A}|f|<\infty.\]
Neglecting a null set, we assume that \(|h|<\infty\) on \(A\) (Corollary 1 of §5). Now, \((\forall n)\) set
\[g_{n}(x)=\left\{\begin{array}{ll}{h(x),} & {x \in A_{n}=A(h<n),} \\ {0,} & {x \in-A_{n}.}\end{array}\right.\]
Then \(g_{n} \leq n\) and \(g_{n}\) is measurable on \(A\). (Why?)
Also, \(g_{n} \geq 0\) and \(g_{n} \rightarrow h\) (pointwise) on \(A\).
For let \(\varepsilon>0,\) fix \(x \in A,\) and find \(k>h(x).\) Then
\[(\forall n \geq k) \quad h(x) \leq n \text { and } g_{n}(x)=h(x).\]
So
\[(\forall n \geq k) \quad\left|g_{n}(x)-h(x)\right|=0<\varepsilon.\]
Clearly, \(g_{n} \leq h.\) Hence by Theorem 5
\[\lim _{n \rightarrow \infty} \int_{A}\left|h-g_{n}\right|=0.\]
Thus we can fix \(n\) so large that
\[\int_{A}\left(h-g_{n}\right)<\frac{1}{2} \varepsilon.\]
For that \(n,\) let
\[\delta=\frac{\varepsilon}{2 n}\]
and take any \(X \subseteq A(X \in \mathcal{M}),\) with \(m X<\delta\).
As \(g_{n} \leq n\) (see above), Theorem 1(c) in §5 yields
\[\int_{X} g_{n} \leq \int_{X}(n)=n \cdot m X<n \delta=\frac{1}{2} \varepsilon.\]
Hence as \(|f| \leq h\) and
\[\int_{X}\left(h-g_{n}\right) \leq \int_{A}\left(h-g_{n}\right)<\frac{1}{2} \varepsilon\]
(Theorem 1(f) of §5), we obtain
\[\overline{\int}_{X}|f| \leq \int_{X} h=\int_{X}\left(h-g_{n}\right)+\int_{X} g_{n}<\frac{1}{2} \varepsilon+\frac{1}{2} \varepsilon=\varepsilon,\]
as required.\(\quad \square\)


14.5: Summary
14.1 Types of Resources
Determining which resources a venture needs is essential to its success. Assets are types of property that benefit the person or company in possession of them. They can be tangible or intangible. Tangible resources can be seen, touched, and felt. They may include raw materials, land, facilities, buildings, machinery, computers, supplies, furniture, information technology, and vehicles.
Intangible resources cannot be seen, touched, or felt. They include intellectual property such as designs, formulas, art, written work, brands, and inventions that can be protected by patents, trademarks, and copyrights.
Entrepreneurs oftentimes have novel ideas about how to provide a service or how to create a better product. These ideas are important to safeguard so other competitors don’t copy those same characteristics of a product, process, machine, piece of writing, or any other of the works cited earlier. A way to protect them is to ensure that you have a patent, trademark, or copyright so others cannot benefit from your work. It is always beneficial to do a check before applying for one of the protections and make sure that the invention hasn’t been created before. Searching the USPTO website and hiring an attorney will ensure you are the first one to register your idea and help you through the lengthy and often expensive process of protecting a new idea.
Funding sources are essential to starting and scaling a business. These include an entrepreneur’s own savings, bank loans, venture capitalists, angel investors, crowdfunding, and friends and family. It is important to consider the pros and cons of your financing options in relation to your resource needs in order to plan for short- and long-term needs.
14.2 Using the PEST Framework to Assess Resource Needs
Business owners need to know what resources they need before launching. The PEST framework can help you become aware of outside forces that affect the procurement of resources you need to succeed. PEST looks at the political, economic, sociocultural, and technological factors that affect resource costs and availability. Different businesses require different resources, so going through a checklist of basic needs, finding the associated costs, splitting them into one-time costs and ongoing costs, and adding them up will ensure the entrepreneur is aware and ready to document for an investor or bank the resources needed.
14.3 Managing Resources over the Venture Life Cycle
The life cycle of the business requires resource allocation and planning at each stage. These stages include the startup phase, growth phase, maturity phase, and decline/rebirth phase. There are instances where a business has failed because the owner failed to re-assess shifting resource needs caused by changes in the marketplace and the overall environment. Research is an ongoing process, and keeping an eye on the external environment allows a business to be able to shift on time.
Resource dependence theory (RDT) is a model that explores creating networks with other companies through mergers, vertical integration, and joint ventures. RDT can help counteract the effects of competing with each other by optimizing collaboration.
Human resources include the labor that produces a product or service, and provides administrative support and customer service. Having good employees adds value to the enterprise because they help generate sales and profit. Education in marketing, management, and leadership are important topics to engage in as a business owner as well as having personal support from mentors.


40.0: Prelude to the Circulatory System
Most animals are complex multicellular organisms that require a mechanism for transporting nutrients throughout their bodies and removing waste products. The circulatory system has evolved over time from simple diffusion through cells in the early evolution of animals to a complex network of blood vessels that reach all parts of the human body. This extensive network supplies the cells, tissues, and organs with oxygen and nutrients, and removes carbon dioxide and waste, which are byproducts of respiration.
At the core of the human circulatory system is the heart. The size of a clenched fist, the human heart is protected beneath the rib cage. Made of specialized and unique cardiac muscle, it pumps blood throughout the body and to the heart itself. Heart contractions are driven by intrinsic electrical impulses that the brain and endocrine hormones help to regulate. Understanding the heart’s basic anatomy and function is important to understanding the body’s circulatory and respiratory systems.
Gas exchange is one essential function of the circulatory system. A circulatory system is not needed in organisms with no specialized respiratory organs because oxygen and carbon dioxide diffuse directly between their body tissues and the external environment. However, in organisms that possess lungs and gills, oxygen must be transported from these specialized respiratory organs to the body tissues via a circulatory system. Therefore, circulatory systems have had to evolve to accommodate the great diversity of body sizes and body types present among animals.


2.3: Gaseous Exchange Mechanism
- Compare counter- and crosscurrent exchange using specific respiratory system examples.
Use the following gif animation as you are reviewing counter-current gas exchange.
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License (Eleanor Lutz).
How would you describe gas exchange and breathing cycle in each of the examples you see in the gif animation? What are their similarities? What are their differences
Which one do you think is an example of counter-current and which one is an example of crosscurrent exchange
Pulmonary ventilation is the act of breathing, which can be described as the movement of air into and out of the lungs. The major mechanisms that drive pulmonary ventilation are atmospheric pressure ( P atm ); the air pressure within the alveoli, called intra-alveolar pressure ( P alv ); and the pressure within the pleural cavity, called intrapleural pressure ( P ip ).
Mechanisms of breathing
The intra-alveolar and intrapleural pressures are dependent on certain physical features of the lung. However, the ability to breathe—to have air enter the lungs during inspiration and air leave the lungs during expiration—is dependent on the air pressure of the atmosphere and the air pressure within the lungs.
- Pressure relationships
Inspiration (or inhalation) and expiration (or exhalation) are dependent on the differences in pressure between the atmosphere and the lungs. In a gas, the pressure is a force created by the movement of gas molecules that are confined. For example, a certain number of gas molecules in a two-liter container has more room than the same number of gas molecules in a one-liter container (Figure 2.11). In this case, the force exerted by the movement of the gas molecules against the walls of the two-liter container is lower than the force exerted by the gas molecules in the one-liter container. Therefore, the pressure is lower in the two-liter container and higher in the one-liter container. At a constant temperature, changing the volume occupied by the gas changes the pressure, as does changing the number of gas molecules. Boyle’s law describes the relationship between volume and pressure in a gas at a constant temperature. Boyle discovered that the pressure of a gas is inversely proportional to its volume: If volume increases, pressure decreases. Likewise, if volume decreases, pressure increases. Pressure and volume are inversely related (P = k/V). Therefore, the pressure in the one-liter container (one-half the volume of the two-liter container) would be twice the pressure in the two-liter container. Boyle’s law is expressed by the following formula:
[P 1 V 1 =P 2 V 2 ]
In this formula, P 1 represents the initial pressure and V 1 represents the initial volume, whereas the final pressure and volume are represented by P 2 and V 2 , respectively. If the two- and one-liter containers were connected by a tube and the volume of one of the containers were changed, then the gases would move from higher pressure (lower volume) to lower pressure (higher volume).
Pulmonary ventilation is dependent on three types of pressure: atmospheric, intra-alveolar, and intrapleural. Atmospheric pressure is the amount of force that is exerted by gases in the air surrounding any given surface, such as the body. Atmospheric pressure can be expressed in terms of the unit atmosphere, abbreviated atm, or in millimeters of mercury (mm Hg). One atm is equal to 760 mm Hg, which is the atmospheric pressure at sea level. Typically, for respiration, other pressure values are discussed in relation to atmospheric pressure. Therefore, negative pressure is pressure lower than the atmospheric pressure, whereas positive pressure is the pressure that it is greater than the atmospheric pressure. A pressure that is equal to the atmospheric pressure is expressed as zero.
Intra-alveolar pressure (intrapulmonary pressure) is the pressure of the air within the alveoli, which changes during the different phases of breathing (Figure 2.12). Because the alveoli are connected to the atmosphere via the tubing of the airways (similar to the two- and one-liter containers in the example above), the intrapulmonary pressure of the alveoli always equalizes with the atmospheric pressure.
Intrapleural pressure is the pressure of the air within the pleural cavity, between the visceral and parietal pleurae. Similar to intra-alveolar pressure, intrapleural pressure also changes during the different phases of breathing. However, due to certain characteristics of the lungs, the intrapleural pressure is always lower than, or negative to, the intra-alveolar pressure (and therefore also to atmospheric pressure). Although it fluctuates during inspiration and expiration, intrapleural pressure remains approximately –4 mm Hg throughout the breathing cycle.
Competing forces within the thorax cause the formation of the negative intrapleural pressure. One of these forces relates to the elasticity of the lungs themselves—elastic tissue pulls the lungs inward, away from the thoracic wall. Surface tension of the alveolar fluid, which is mostly water, also creates an inward pull of the lung tissue. This inward tension from the lungs is countered by opposing forces from the pleural fluid and thoracic wall. Surface tension within the pleural cavity pulls the lungs outward. Too much or too little pleural fluid would hinder the creation of the negative intrapleural pressure; therefore, the level must be closely monitored by the mesothelial cells and drained by the lymphatic system. Since the parietal pleura is attached to the thoracic wall, the natural elasticity of the chest wall opposes the inward pull of the lungs. Ultimately, the outward pull is slightly greater than the inward pull, creating the –4 mm Hg intrapleural pressure relative to the intra-alveolar pressure. Transpulmonary pressure is the difference between the intrapleural and intra-alveolar pressures, and it determines the size of the lungs. A higher transpulmonary pressure corresponds to a larger lung.
Physical factors affecting ventilation
In addition to the differences in pressures, breathing is also dependent upon the contraction and relaxation of muscle fibers of both the diaphragm and thorax. The lungs themselves are passive during breathing, meaning they are not involved in creating the movement that helps inspiration and expiration. This is because of the adhesive nature of the pleural fluid, which allows the lungs to be pulled outward when the thoracic wall moves during inspiration. The recoil of the thoracic wall during expiration causes compression of the lungs. Contraction and relaxation of the diaphragm and intercostals muscles (found between the ribs) cause most of the pressure changes that result in inspiration and expiration. These muscle movements and subsequent pressure changes cause air to either rush in or be forced out of the lungs.
Other characteristics of the lungs influence the effort that must be expended to ventilate. Resistance is a force that slows motion, in this case, the flow of gases. The size of the airway is the primary factor affecting resistance. A small tubular diameter forces air through a smaller space, causing more collisions of air molecules with the walls of the airways. The following formula helps to describe the relationship between airway resistance and pressure changes:
[F=∆P/RF=∆P/R]
As noted earlier, there is surface tension within the alveoli caused by water present in the lining of the alveoli. This surface tension tends to inhibit expansion of the alveoli. However, pulmonary surfactant secreted by type II alveolar cells mixes with that water and helps reduce this surface tension. Without pulmonary surfactant, the alveoli would collapse during expiration.
Thoracic wall compliance is the ability of the thoracic wall to stretch while under pressure. This can also affect the effort expended in the process of breathing. In order for inspiration to occur, the thoracic cavity must expand. The expansion of the thoracic cavity directly influences the capacity of the lungs to expand. If the tissues of the thoracic wall are not very compliant, it will be difficult to expand the thorax to increase the size of the lungs.
Gas laws and air composition
Gas molecules exert a force on the surfaces with which they are in contact; this force is called pressure. In natural systems, gases are normally present as a mixture of different types of molecules. For example, the atmosphere consists of oxygen, nitrogen, carbon dioxide, and other gaseous molecules, and this gaseous mixture exerts a certain pressure referred to as atmospheric pressure (Table 2.1). Partial pressure ( P x ) is the pressure of a single type of gas in a mixture of gases. For example, in the atmosphere, oxygen exerts a partial pressure, and nitrogen exerts another partial pressure, independent of the partial pressure of oxygen (Figure 2.13). Total pressure is the sum of all the partial pressures of a gaseous mixture. Dalton’s law describes the behavior of nonreactive gases in a gaseous mixture and states that a specific gas type in a mixture exerts its own pressure; thus, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the gases in the mixture.
| Gas | Percent of total composition | Partial pressure (mm Hg) |
|---|---|---|
| Nitrogen (N 2 ) | 78.6 | 597.4 |
| Oxygen (O 2 ) | 20.9 | 158.8 |
| Water (H 2 O) | 0.4 | 3.0 |
| Carbon dioxide (CO 2 ) | 0.04 | 0.3 |
| Others | 0.06 | 0.5 |
| Total composition/total atmospheric pressure | 100% | 760.0 |
Partial pressure is extremely important in predicting the movement of gases. Recall that gases tend to equalize their pressure in two regions that are connected. A gas will move from an area where its partial pressure is higher to an area where its partial pressure is lower. In addition, the greater the partial pressure difference between the two areas, the more rapid is the movement of gases.
Solubility of gases in liquids
Henry’s law describes the behavior of gases when they come into contact with a liquid, such as blood. Henry’s law states that the concentration of gas in a liquid is directly proportional to the solubility and partial pressure of that gas. The greater the partial pressure of the gas, the greater the number of gas molecules that will dissolve in the liquid. The concentration of the gas in a liquid is also dependent on the solubility of the gas in the liquid. For example, although nitrogen is present in the atmosphere, very little nitrogen dissolves into the blood because the solubility of nitrogen in blood is very low. The exception to this occurs in scuba divers; the composition of the compressed air that divers breathe causes nitrogen to have a higher partial pressure than normal, causing it to dissolve in the blood in greater amounts than normal. Too much nitrogen in the bloodstream results in a serious condition that can be fatal if not corrected. Gas molecules establish an equilibrium between those molecules dissolved in liquid and those in the air.
The composition of air in the atmosphere and in the alveoli differs. In both cases, the relative concentration of gases is nitrogen > oxygen > water vapor > carbon dioxide. The amount of water vapor present in the alveolar air is greater than that in atmospheric air (Table 2.2). Recall that the respiratory system works to humidify incoming air, thereby causing the air present in the alveoli to have a greater amount of water vapor than atmospheric air. In addition, alveolar air contains a greater amount of carbon dioxide and less oxygen than atmospheric air. This is no surprise, as gas exchange removes oxygen from and adds carbon dioxide to the alveolar air. Both deep and forced breathing cause the alveolar air composition to be changed more rapidly than during quiet breathing. As a result, the partial pressures of oxygen and carbon dioxide change, affecting the diffusion process that moves these materials across the membrane. This will cause oxygen to enter and carbon dioxide to leave the blood more quickly.
| Gas | Percent of total composition | Partial pressure (mm Hg) |
|---|---|---|
| Nitrogen (N 2 ) | 74.9 | 569 |
| Oxygen (O 2 ) | 13.7 | 104 |
| Water (H 2 O) | 6.2 | 40 |
| Carbon dioxide (CO 2 ) | 5.2 | 47 |
| Total composition/total alveolar pressure | 100% | 760.0 |
Ventilation and perfusion
Two important aspects of gas exchange in the lung are ventilation and perfusion. Ventilation is the movement of air into and out of the lungs, and perfusion is the flow of blood in the pulmonary capillaries. For gas exchange to be efficient, the volumes involved in ventilation and perfusion should be compatible. However, factors such as regional gravity effects on the blood, blocked alveolar ducts, or disease can cause ventilation and perfusion to be unbalanced.
The partial pressure of oxygen in the alveolar air is about 104 mm Hg, whereas the partial pressure of oxygenated blood in pulmonary veins is about 100 mm Hg. When ventilation is sufficient, oxygen enters the alveoli at a high rate, and the partial pressure of oxygen in the alveoli remains high. In contrast, when ventilation is insufficient, the partial pressure of oxygen in the alveoli drops. Without the large difference in partial pressure between the alveoli and the blood, oxygen does not diffuse efficiently across the respiratory membrane. The body has mechanisms that counteract this problem. In cases when ventilation is not sufficient for an alveolus, the body redirects blood flow to alveoli that are receiving sufficient ventilation. This is achieved by constricting the pulmonary arterioles that serve the dysfunctional alveolus, which redirects blood to other alveoli that have sufficient ventilation. At the same time, the pulmonary arterioles that serve alveoli receiving sufficient ventilation vasodilate, which brings in greater blood flow. Factors such as carbon dioxide, oxygen, and pH levels can all serve as stimuli for adjusting blood flow in the capillary networks associated with the alveoli.
Ventilation is regulated by the diameter of the airways, whereas perfusion is regulated by the diameter of the blood vessels. The diameter of the bronchioles is sensitive to the partial pressure of carbon dioxide in the alveoli. A greater partial pressure of carbon dioxide in the alveoli causes the bronchioles to increase their diameter as will a decreased level of oxygen in the blood supply, allowing carbon dioxide to be exhaled from the body at a greater rate. As mentioned above, a greater partial pressure of oxygen in the alveoli causes the pulmonary arterioles to dilate, increasing blood flow.
Gas exchange
Gas exchange occurs at two sites in the body: in the lungs, where oxygen is picked up and carbon dioxide is released at the respiratory membrane, and at the tissues, where oxygen is released and carbon dioxide is picked up. External respiration is the exchange of gases with the external environment and occurs in the alveoli of the lungs. Internal respiration is the exchange of gases with the internal environment and occurs in the tissues. The actual exchange of gases occurs due to simple diffusion. Energy is not required to move oxygen or carbon dioxide across membranes. Instead, these gases follow pressure gradients that allow them to diffuse. The anatomy of the lung maximizes the diffusion of gases: The respiratory membrane is highly permeable to gases; the respiratory and blood capillary membranes are very thin, and there is a large surface area throughout the lungs.
External respiration
The pulmonary artery carries deoxygenated blood into the lungs from the heart, where it branches and eventually becomes the capillary network composed of pulmonary capillaries. These pulmonary capillaries create the respiratory membrane with the alveoli (Figure 2.14). As the blood is pumped through this capillary network, gas exchange occurs. Although a small amount of the oxygen is able to dissolve directly into the plasma from the alveoli, most of the oxygen is picked up by erythrocytes (red blood cells) and binds to a protein called hemoglobin, a process described later in this chapter. Oxygenated hemoglobin is red, causing the overall appearance of bright red oxygenated blood, which returns to the heart through the pulmonary veins. Carbon dioxide is released in the opposite direction of oxygen, from the blood to the alveoli. Some of the carbon dioxide is returned on hemoglobin, but can also be dissolved in plasma or is present as a converted form, also explained in greater detail later in this chapter.
External respiration occurs as a function of partial pressure differences in oxygen and carbon dioxide between the alveoli and the blood in the pulmonary capillaries.
External Respiration
Figure 2.14. In external respiration, oxygen diffuses across the respiratory membrane from the alveolus to the capillary, whereas carbon dioxide diffuses out of the capillary into the alveolus.
Although the solubility of oxygen in the blood is not high, there is a drastic difference in the partial pressure of oxygen in the alveoli versus in the blood of the pulmonary capillaries. This difference is about 64 mm Hg: The partial pressure of oxygen in the alveoli is about 104 mm Hg, whereas its partial pressure in the blood of the capillary is about 40 mm Hg. This large difference in partial pressure creates a very strong pressure gradient that causes oxygen to rapidly cross the respiratory membrane from the alveoli into the blood.
The partial pressure of carbon dioxide is also different between the alveolar air and the blood of the capillary. However, the partial pressure difference is less than that of oxygen, about 5 mm Hg. The partial pressure of carbon dioxide in the blood of the capillary is about 45 mm Hg, whereas its partial pressure in the alveoli is about 40 mm Hg. However, the solubility of carbon dioxide is much greater than that of oxygen—by a factor of about 20—in both blood and alveolar fluids. As a result, the relative concentrations of oxygen and carbon dioxide that diffuse across the respiratory membrane are similar.
Internal respiration
Internal respiration is the gas exchange that occurs at the level of body tissues (Figure 2.15). Similar to external respiration, internal respiration also occurs as simple diffusion due to a partial pressure gradient. However, the partial pressure gradients are the opposite of those present at the respiratory membrane. The partial pressure of oxygen in tissues is low, about 40 mm Hg, because oxygen is continuously used for cellular respiration. In contrast, the partial pressure of oxygen in the blood is about 100 mm Hg. This creates a pressure gradient that causes oxygen to dissociate from hemoglobin, diffuse out of the blood, cross the interstitial space, and enter the tissue. Hemoglobin that has little oxygen bound to it loses much of its brightness, so that blood returning to the heart is more burgundy in color.
Considering that cellular respiration continuously produces carbon dioxide, the partial pressure of carbon dioxide is lower in the blood than it is in the tissue, causing carbon dioxide to diffuse out of the tissue, cross the interstitial fluid, and enter the blood. It is then carried back to the lungs either bound to hemoglobin, dissolved in plasma, or in a converted form. By the time blood returns to the heart, the partial pressure of oxygen has returned to about 40 mm Hg, and the partial pressure of carbon dioxide has returned to about 45 mm Hg. The blood is then pumped back to the lungs to be oxygenated once again during external respiration.
A hyperbaric chamber is a unit that can be sealed and expose a patient to either 100 percent oxygen with increased pressure or a mixture of gases that includes a higher concentration of oxygen than normal atmospheric air, also at a higher partial pressure than the atmosphere (Figure 2.15). Hyperbaric chamber treatment is based on the behaviour of gases. As you recall, gases move from a region of higher partial pressure to a region of lower partial pressure. In a hyperbaric chamber, the atmospheric pressure is increased, causing a greater amount of oxygen than normal to diffuse into the bloodstream of the patient. Hyperbaric chamber therapy is used to treat a variety of medical problems, such as wound and graft healing, anaerobic bacterial infections, and carbon monoxide poisoning. Exposure to and poisoning by carbon monoxide is difficult to reverse, because hemoglobin’s affinity for carbon monoxide is much stronger than its affinity for oxygen, causing carbon monoxide to replace oxygen in the blood. Hyperbaric chamber therapy can treat carbon monoxide poisoning because the increased atmospheric pressure causes more oxygen to diffuse into the bloodstream. At this increased pressure and increased concentration of oxygen, carbon monoxide is displaced from hemoglobin. Another example is the treatment of anaerobic bacterial infections, which are created by bacteria that cannot or prefer not to live in the presence of oxygen. An increase in blood and tissue levels of oxygen helps to kill the anaerobic bacteria that are responsible for the infection, as oxygen is toxic to anaerobic bacteria. For wounds and grafts, the chamber stimulates the healing process by increasing energy production needed for repair. Increasing oxygen transport allows cells to ramp up cellular respiration and thus ATP production, the energy needed to build new structures.
Gas moves from an area of ________ partial pressure to an area of ________ partial pressure.
a. low; high
b. low; low
c. high; high
d. high; low
When ventilation is not sufficient, which of the following occurs?
a. The capillary constricts.
b. The capillary dilates.
c. The partial pressure of oxygen in the affected alveolus increases.
d. The bronchioles dilate
Gas exchange that occurs at the level of the tissues is called ________.
a. external respiration
b. interpulmonary respiration
c. internal respiration
d. pulmonary ventilation
The partial pressure of carbon dioxide is 45 mm Hg in the blood and 40 mm Hg in the alveoli. What happens to the carbon dioxide?
a. It diffuses into the blood.
b. It diffuses into the alveoli.
c. The gradient is too small for carbon dioxide to diffuse.
d. It decomposes into carbon and oxygen
Compare and contrast Dalton’s law and Henry’s law
A smoker develops damage to several alveoli that then can no longer function. How does this affect gas exchange?


4.7: §31. The Legacy of Latin- III. Modern English
Just as Middle English greatly surpassed Old English in its adoption of words with Latin origins, so has Modern English far exceeded Middle English in this regard. When one considers the time frame, that is not surprising, since the term MODERN ENGLISH is applied to the entire period from 1450 or 1500 to the present day. This era is sometimes subdivided into “Early Modern English,” from 1500 to 1800, and “Present-Day English,” the language of the l9th and 20th centuries. The period as a whole is an era when English has become stabilized as a language that would be mutually intelligible to speakers—or at least to readers—from any time within those five hundred years.
If Middle English was heralded by the Norman invasion of AD 1066, the advent of Modern English was signalled by two not unrelated events, the invention of the printing press (by Johannes Gutenberg, a Mainz goldsmith, in 1450) [1] and the great intellectual and cultural movement known as the Renaissance. Arising in Italy in the 15th century and sweeping into northern Europe with full force in the 16th, this new way of looking at the world sparked a passionate reawakening of interest in the classical civilizations of ancient Greece and Rome. The movement was largely caused, of course, by the rediscovery of classical authors and the so-called “Revival of Learning”—a label that is perhaps a little unfair to the excellent Latin scholars of the late Middle Ages, who were the forerunners of the Renaissance. Great works of literature that had previously been accessible only to a learned few were now being keenly studied by a growing circle of literate readers, were being widely disseminated by the new technology of the printed word, and were being translated into English from the original Greek and Latin. The Protestant Reformation caused an upheaval in Christian worship and secular education, with consequences that had a profound effect on the language. During the reign of Elizabeth I, England enjoyed a period unprecedented in its intellectual energy and creativity. William Shakespeare and his fellow poets of the late 16th and early 17th centuries were riding a high crest of linguistic experiment and innovation.
As we observed in Chapter 3 (§23), there had been a certain amount of direct borrowing from Latin during the Middle English period. After 1500, however, the trickle of Latin words became a flood. Poets and scholars vied with each other in coining new English vocabulary that was inspired by their knowledge of the classics—Greek, as well as Latin, though the knowledge of Greek was still rather limited. Some of the results were solid and useful words that have survived to the present day; others were fantastic and preposterous inventions that deserved the early death that was their fate. Students of Elizabethan literature will be aware of the controversy that raged during the 16th century on the propriety of these bizarre and outlandish coinages (“inkhorn terms,” as they were scornfully known). Because scholars and poets on the continent were equally bold and inventive, it was not uncommon for an English writer to borrow a Latin or Greek adaptation that had recently become current in French or Italian usage. If you have access to a dictionary that gives the date of a word’s first attested use in English—for example, the current edition of Webster’s New Collegiate or any of the larger Oxford dictionaries—notice how many Latin derivatives can be dated to the period between 1560 and 1620. That is roughly Shakespeare’s lifetime, but Shakespeare was merely the most conspicuous representative of a widespread general trend.
The extraordinary number of Latin DOUBLETS in modern English can be explained by this activity during the high Renaissance. Though English already had words like chamber and choir , derived through French from Latin camera and chorus , the well-educated writers of the l6th and l7th centuries would go back to the original tongues to adopt forms like camera and chorus as new English words. More often than not, the new word became a naturalized adaptation, rather than an exact adoption: the Latin adjective fragilis , for example, was Anglicized as fragile , now coexisting with its earlier doublet frail . Whenever you are confronted with such a pair of English doublets, it is usually safe to assume that the form that looks less like the Latin original is the older English word, since it probably evolved through French in the Middle English period. The more Latinate form is almost certain to be the learned creation of some Modern English borrower.
In the next few chapters, we will be studying patterns of resemblance between relatively complex Latin words and their corresponding English derivatives. For the most part, our English examples will be drawn from these more learned adaptations of the Modern English period—vocabulary that is unmistakably Latin, once you have had some experience with Latin. There are thousands of words like this in our language, and they continue to be coined at an astonishing rate. The flood of classical neologisms subsided a little in the l7th and 18th centuries, but it began to swell again in the l9th century, as Latin and Greek both began to be exploited for scientific and technical terminology.
Because the English language is so thoroughly stocked with borrowed Latin roots and affixes, speakers of English are now able to invent new combinations of Latin word elements without actually knowing any Latin. Occasionally, a new coinage may cause the scholar some distress, if it violates the standard rules of Latin word formation.  No one is startled any longer by the word humanitarian (1831), though it, too, attaches an historically inappropriate Latin suffix to an abstract Latin noun. The great virtue of English is its flexibility and its receptivity to change. We can take Latin words and add Germanic suffixes, creating hybrids like graceful , masterful , brutish , and scholarly . It is an astonishing fact that there are more Latin derivatives in English today than the total number of Latin words in the largest dictionary of classical Latin. We have conquered the Romans on their own linguistic battlefield.
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The celebrated Gutenberg 42-line Bible was printed in 1456; in 1476, William Caxton, who had learned his craft on the European continent, established the first printing press in England. ↵


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blended family
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parents and children who are not biologically related to the other parent, usually created after a death or divorce but also include those who have never married; also called a binuclear or stepfamily
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collectivism
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culture in which people believe in interconnectedness with others and respect relationships
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cultural competence
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lifelong process of applying evidence-based nursing in agreement with the cultural values, beliefs, worldview, and practices of patients to produce improved patient outcomes
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culturally responsive care
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occurs when a person’s cultural beliefs are integrated into their health care; culturally responsive care is required for a trusting, effective relationship with the patient and their family
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culture
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set of beliefs, attitudes, and practices shared by a group of people or community that is accepted, followed, and passed down to other members of the group
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Culture Care Theory
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states that health care cannot be effectively provided without considering the person's cultural background
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diversity
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differences in social and ethnic backgrounds, gender, and sexual orientation within a community
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dual-career family
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household in which both parents work
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extended family
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grandparents, aunts, uncles, and cousins
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Four Cs of Culture model
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example of a quick cultural assessment tool that asks questions about what the patient
C
onsiders to be a problem, the
C
ause of the problem, how they are
C
oping with the problem, and how
C
oncerned they are about the problem
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gender-affirming language
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language that describes one’s gender identity
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Healthy People 2030
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national objectives aimed at improving health and well-being over the next 10 years
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historical trauma
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cumulative trauma associated with a specific cultural, racial, marginalized, or ethnic group
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individualism
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culture in which people respect independence and view themselves as separate from others, with personal ideals and goals
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single-career family
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household in which one parent works and the other stays home
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social determinants of health
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nonmedical factors that influence health outcomes, including conditions in which people are born, grow, work, live, and age, and the wider sets of forces and systems shaping the conditions of daily life
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traditional healing
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various medicines and healing practices around the world that differ from the modern, Western health-care system
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transcultural nursing
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incorporates cultural beliefs and practices of people to help them maintain and regain health or to face death in a meaningful way
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trauma-informed care
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care that acknowledges all past and present parts of a person’s life situation, including any trauma they have endured (Center for Health Care Strategies, 2021) in an effort to provide treatment that supports the patient’s autonomy, strength, and control over making health-care decisions


6.8: Measuring Supply Chain Performance
Inventory Turnover
Key Performance Indicators (KPI) are measurements used to evaluate supply chain performance.
One of the ways to evaluate the supply chain performance is to calculate inventory turnover (inventory turns):
“Average aggregate inventory value” is a term used to describe all of the inventory held in stock, which includes raw materials, work in process and finished goods, all valued at cost.
Inventory turnover is an indicator of the policies and practices of an organization. It represents their ability to purchase materials, produce and sell their products in a timely manner. A higher value for the inventory turnover means that the organization has been capable of replenishing and selling its inventory more number of times in any particular amount of time, and as a result, have a better cash flow.
It is important to keep in mind that high or low value of inventory turnover for each company is relative to its own industry. For example, dairy (milk) manufacturing has an annual inventory turnover of around 23, while this number is 14.7 for the grocery supermarkets, and 4.8 for the automotive industry. [9] Industries with higher volume, but lower margin, usually have the highest inventory turnovers.
Example
NED’s Food Supply is a supplier to restaurants and institutions for frozen foods, meats, fish, canned and fresh fruits and vegetables. Here is an analysis from the past two years regarding their inventory management. In which year was their supply chain performance better?
| Last year | Two years ago | |
|---|---|---|
| Cost of goods sold | 17,550,000 | 16,255,000 |
| Average aggregate inventory value | $1,650,000 | $1,763,350 |
Solution
Inventory turns for last year = 17,550,000 / 1,650,000 = 10.64 turns
Inventory turns for two years ago = 16,255,000 / 1,763,350 = 9.22 turns
Last year, their inventory turnover was faster. If customer service was equivalent in both years, then their performance was better last year than it was two years ago. This may have resulted in customers receiving fresher foods as well.
Days of Supply
Another related performance measure is days of supply :
Example
J’s Custom Automotive Finishing has calculated that his annual cost of goods sold at 45,000,000. His average inventory value in 2019 is:
| Production components | 2,350,000 |
|---|---|
| Production supplies | 450,000 |
| Finished goods | 225,600 |
| Total aggregate inventory value: | 3,025,600 |
Solution
Days of supply = (3,025,600 / 45,000,000) x 365 = 24.54
This measure can be thought of as how much inventory is sitting in the building at any one time. In terms of measuring the efficiency of the inventory, a lower number is better. It would imply that goods are purchased more frequently and spend less time in the facility before being converted into sales.
There are other ways to measure supply chain performance as well. In a warehouse or distribution setting, fill rate is an important measure. It is the percentage of customer orders that are filled from on-hand stock. In a manufacturing setting, a measure such as the percentage of orders delivered on time is an important indicator of customer service level.
The following 3:11 video provides additional insights into important measurements throughout the supply chain process:
"Supply Chain" Introduction to Operations Management Copyright © by Hamid Faramarzi and Mary Drane is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.


6.4: Kirchhoff's Rules and Resistor Circuits
By the end of the section, you will be able to:
- State Kirchhoff’s junction rule
- State Kirchhoff’s loop rule
- Analyze complex circuits using Kirchhoff’s rules
We have just seen that some circuits may be analyzed by reducing a circuit to a single voltage source and an equivalent resistance. Many complex circuits cannot be analyzed with the series-parallel techniques developed in the preceding sections. In this section, we elaborate on the use of Kirchhoff’s rules to analyze more complex circuits. For example, the circuit in Figure \(\PageIndex{1}\) is known as a multi-loop circuit , which consists of junctions. A junction, also known as a node, is a connection of three or more wires. In this circuit, the previous methods cannot be used, because not all the resistors are in clear series or parallel configurations that can be reduced. Give it a try. The resistors \(R_1\) and \(R_2\) are in series and can be reduced to an equivalent resistance. The same is true of resistors \(R_4\) and \(R_5\). But what do you do then?
Even though this circuit cannot be analyzed using the methods already learned, two circuit analysis rules can be used to analyze any circuit, simple or complex. The rules are known as Kirchhoff’s rules , after their inventor Gustav Kirchhoff (1824–1887).
- Kirchhoff’s first rule—the junction rule. The sum of all currents entering a junction must equal the sum of all currents leaving the junction: \[\sum I_{in} = \sum I_{out}.\]
- Kirchhoff’s second rule—the loop rule. The algebraic sum of changes in potential around any closed circuit path (loop) must be zero: \[\sum V = 0.\]
We now provide explanations of these two rules, followed by problem-solving hints for applying them and a worked example that uses them.
Kirchhoff’s Junction Rule
Kirchhoff’s j unction rule applies to the charge entering and leaving a junction (Figure \(\PageIndex{2}\)). As stated earlier, a junction, or node, is a connection of three or more wires. Current is the flow of charge, and charge is conserved; thus, whatever charge flows into the junction must flow out.
Although it is an over-simplification, an analogy can be made with water pipes connected in a plumbing junction. If the wires in Figure \(\PageIndex{2}\) were replaced by water pipes, and the water was assumed to be incompressible, the volume of water flowing into the junction must equal the volume of water flowing out of the junction.
Kirchhoff’s Loop Rule
Kirchhoff’s loop rule applies to potential differences. The loop rule is stated in terms of potential \(V\) rather than potential energy, but the two are related since \(U = qV\). In a closed loop, whatever energy is supplied by a voltage source, the energy must be transferred into other forms by the devices in the loop, since there are no other ways in which energy can be transferred into or out of the circuit. Kirchhoff’s loop rule states that the algebraic sum of potential differences, including voltage supplied by the voltage sources and resistive elements, in any loop must be equal to zero. For example, consider a simple loop with no junctions, as in Figure \(\PageIndex{3}\).
The circuit consists of a voltage source and three external load resistors. The labels a , b , c , and d serve as references, and have no other significance. The usefulness of these labels will become apparent soon. The loop is designated as Loop abcda , and the labels help keep track of the voltage differences as we travel around the circuit. Start at point a and travel to point b . The voltage of the voltage source is added to the equation and the potential drop of the resistor \(R_1\) is subtracted. From point b to c , the potential drop across \(R_2\) is subtracted. From c to d , the potential drop across \(R_3\) is subtracted. From points d to a , nothing is done because there are no components.
Figure \(\PageIndex{4}\) shows a graph of the voltage as we travel around the loop. Voltage increases as we cross the battery, whereas voltage decreases as we travel across a resistor. The potential drop , or change in the electric potential, is equal to the current through the resistor times the resistance of the resistor. Since the wires have negligible resistance, the voltage remains constant as we cross the wires connecting the components.
Then Kirchhoff’s loop rule states
\[V - IR_1 - IR_2 - IR_3 = 0.\]
The loop equation can be used to find the current through the loop:
\[I = \frac{V}{R_1 +R_2 +R_3} = \frac{12.00 \, V}{1.00 \, \Omega + 2.00 \, \Omega + 3.00 \, \Omega} = 2.00 \, A.\]
This loop could have been analyzed using the previous methods, but we will demonstrate the power of Kirchhoff’s method in the next section.
Applying Kirchhoff’s Rules
By applying Kirchhoff’s rules, we generate a set of linear equations that allow us to find the unknown values in circuits. These may be currents, voltages, or resistances. Each time a rule is applied, it produces an equation. If there are as many independent equations as unknowns, then the problem can be solved.
Using Kirchhoff’s method of analysis requires several steps, as listed in the following procedure.
- Label points in the circuit diagram using lowercase letters a , b , c , …. These labels simply help with orientation.
- Locate the junctions in the circuit. The junctions are points where three or more wires connect. Label each junction with the currents and directions into and out of it. Make sure at least one current points into the junction and at least one current points out of the junction.
- Choose the loops in the circuit. Every component must be contained in at least one loop, but a component may be contained in more than one loop.
- Apply the junction rule. Again, some junctions should not be included in the analysis. You need only use enough nodes to include every current.
- Apply the loop rule. Use the map in Figure \(\PageIndex{5}\).
Let’s examine some steps in this procedure more closely. When locating the junctions in the circuit, do not be concerned about the direction of the currents. If the direction of current flow is not obvious, choosing any direction is sufficient as long as at least one current points into the junction and at least one current points out of the junction. If the arrow is in the opposite direction of the conventional current flow, the result for the current in question will be negative but the answer will still be correct.
The number of nodes depends on the circuit. Each current should be included in a node and thus included in at least one junction equation. Do not include nodes that are not linearly independent, meaning nodes that contain the same information.
Consider Figure \(\PageIndex{6}\). There are two junctions in this circuit: Junction b and Junction e . Points a , c , d , and f are not junctions, because a junction must have three or more connections. The equation for Junction b is \(I_1 = I_2 + I_3\), and the equation for Junction e is \(I_2 + I_3 = I_1\). These are equivalent equations, so it is necessary to keep only one of them.
When choosing the loops in the circuit, you need enough loops so that each component is covered once, without repeating loops. Figure \(\PageIndex{7}\) shows four choices for loops to solve a sample circuit; choices (a), (b), and (c) have a sufficient amount of loops to solve the circuit completely. Option (d) reflects more loops than necessary to solve the circuit.
Consider the circuit in Figure \(\PageIndex{8a}\). Let us analyze this circuit to find the current through each resistor. First, label the circuit as shown in part (b).
Next, determine the junctions. In this circuit, points b and e each have three wires connected, making them junctions. Start to apply Kirchhoff’s junction rule \(\left(\sum I_{in} = \sum I_{out}\right)\) by drawing arrows representing the currents and labeling each arrow, as shown in Figure \(\PageIndex{9}\). Junction b shows that \(I_1 = I_2 + I_3\) and Junction e shows that \(I_2 + I_3 = I_1\). Since Junction e gives the same information of Junction b , it can be disregarded. This circuit has three unknowns, so we need three linearly independent equations to analyze it.
Next we need to choose the loops. In Figure \(\PageIndex{10}\), Loop abefa includes the voltage source \(V_1\) and resistors \(R_1\) and \(R_2\). The loop starts at point a , then travels through points b , e , and f , and then back to point a . The second loop, Loop ebcde , starts at point e and includes resistors \(R_2\) and \(R_3\), and the voltage source \(V_2\).
Now we can apply Kirchhoff’s loop rule, using the map in Figure \(\PageIndex{5}\). Starting at point a and moving to point b , the resistor \(R_1\) is crossed in the same direction as the current flow \(I_1\), so the potential drop \(I_1R_1\) is subtracted. Moving from point b to point e , the resistor \(R_2\) is crossed in the same direction as the current flow \(I_2\) so the potential drop \(I_2R_2\) is subtracted. Moving from point e to point f , the voltage source \(V_1\) is crossed from the negative terminal to the positive terminal, so \(V_1\) is added. There are no components between points f and a. The sum of the voltage differences must equal zero:
\[Loop \, abefa: \, -I_1R_1 - I_2R_2 + V_1 = 0 \, or \, V_1 = I_1R_1 + I_2R_2.\]
Finally, we check loop ebcde . We start at point e and move to point b , crossing \(R_2\) in the opposite direction as the current flow \(I_2\). The potential drop \(I_2R_2\) is added. Next, we cross \(R_3\) and \(R_4\) in the same direction as the current flow \(I_3\) and subtract the potential drops \(I_3R_3\) and \(I_3R_4\). Note that the current is the same through resistors \(R_3\) and \(R_4\),because they are connected in series. Finally, the voltage source is crossed from the positive terminal to the negative terminal, and the voltage source \(V_2\) is subtracted. The sum of these voltage differences equals zero and yields the loop equation
\[Loop \, ebcde: \, I_2R_2 - I_3(R_3 + R_4) - V_2 = 0.\]
We now have three equations, which we can solve for the three unknowns.
\[\text{Junction b:} \, I_1 - I_2 - I_3 = 0. \label{eq1}\]
\[\text{Loop abefa:} \, I_1R_1 + I_2R_2 = V_1. \label{eq2}\]
\[\text{Loop ebcde:} \, I_2R_2 - I_3(R_3 + R_4) = V_2. \label{eq3}\]
To solve the three equations for the three unknown currents, start by eliminating current \(I_2\). First add Equation \ref{eq1} times \(R_2\) to Equation \ref{eq2}. The result is Equation \ref{eq4}:
\[(R_1 + R_2) I_1 - R_2I_3 = V_1.\]
\[6 \, \Omega I_1 - 3 \Omega I_3 = 24 \, V. \label{eq4}\]
Next, subtract Equation \ref{eq3} from Equation \ref{eq2}. The result is Equation \ref{eq5}:
\[I_1R_1 + I_3(R_3 + R_4) = V_1 - V_2.\]
\[3 \Omega I_1 + 7 \Omega I_3 = -5 \, V. \label{eq5}\]
We can solve Equations \ref{eq4} and \ref{eq5} for current \(I_1\). Adding seven times Equation \ref{eq4} and three times Equation \ref{eq5} results in \(51 \, \Omega I_1 = 153 \, V\), or \(I_1 = 3.00 \, A\). Using Equation \ref{eq4} results in \(I_3 = -2.00 \, A\). Finally, Equation \ref{eq1} yields \(I_2 = I_1 - I_3 = 5.00 \, A\). One way to check that the solutions are consistent is to check the power supplied by the voltage sources and the power dissipated by the resistors:
\[P_{in} = I_1V_1 + I_3V_2 = 130 \, W, \nonumber\]
\[P_{out} = I_1^2R_1 + I_2^2R_2 + I_3^2R_3 + I_3^2R_4 = 130 \, W. \nonumber\]
Note that the solution for the current \(I_3\) is negative. This is the correct answer, but suggests that the arrow originally drawn in the junction analysis is the direction opposite of conventional current flow. The power supplied by the second voltage source is 58 W and not −58 W.
Find the currents flowing in the circuit in Figure \(\PageIndex{11}\).
Strategy
This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the series-parallel techniques—it is necessary to use Kirchhoff’s rules. Currents have been labeled \(I_1, \, I_2\), and \(I_3\) in the figure, and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h . In the solution, we apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents.
Solution
Applying the junction and loop rules yields the following three equations. We have three unknowns, so three equations are required.
\[Junction \, c: \, I_1 + I_2 = I_3.\]
\[Loop \, abcdefa: \, I_1(R_1 + R_4) - I_2(R_2 + R_5 + R_6) = V_1 - V_3.\]
\[Loop \, cdefc: \, I_2(R_2 + R_5 + R_6) + I_3R_3 = V_2 + V_3.\]
Simplify the equations by placing the unknowns on one side of the equations.
\[Junction \, c: \, I_1 + I_2 - I_3 = 0.\]
\[Loop \, abcdefa: \, I_1 (3 \Omega) - I_2(8 \Omega) = 0.5 \, V - 2.30 \, V.\]
\[Loop \, cdefc: \, I_2 (8 \Omega) + I_3 (1 \Omega) = 0.6 \, V + 2.30 \, V.\]
Simplify the equations. The first loop equation can be simplified by dividing both sides by 3.00. The second loop equation can be simplified by dividing both sides by 6.00.
\[Junction \, c: \, I_1 + I_2 - I_3 = 0.\]
\[Loop \, abcdefa: \, I_1 (3 \Omega) - I_2(8 \Omega) = - 1.8 \, V.\]
\[Loop \, cdefc: \, I_2 (8 \Omega) + I_3 (1 \Omega) = 2.90 \, V.\]
The results are
\[I_1 = 0.20 \, A, \, I_2 = 0.30 \, A, \, I_3 = 0.50 \, A.\]
Significance
A method to check the calculations is to compute the power dissipated by the resistors and the power supplied by the voltage sources:
\[P_{R_1} = I_1^2R_1 = 0.04 \, W.\]
\[P_{R_2} = I_2^2R_2 = 0.45 \, W.\]
\[P_{R_3} = I_3^2R_3 = 0.25 \, W.\]
\[P_{R_4} = I_1^2R_4 = 0.08 \, W.\]
\[P_{R_5} = I_2^2R_5 = 0.09 \, W.\]
\[P_{R_6} = I_2^2R_1 = 0.18 \, W.\]
\[P_{dissipated} = 1.09 \, W.\]
\[P_{source} = I_1V_1 + I_2V_3 + I_3V_2 = 0.10 \, + 0.69 \, W + 0.30 \, W = 1.09 \, W.\]
The power supplied equals the power dissipated by the resistors.
In considering the following schematic and the power supplied and consumed by a circuit, will a voltage source always provide power to the circuit, or can a voltage source consume power?
- Answer
-
The circuit can be analyzed using Kirchhoff’s loop rule. The first voltage source supplies power: \(P_{in} = IV_1 = 7.20 \, mW\). The second voltage source consumes power: \(P_{out} = IV_2 + I^2R_1 + I^2R_2 = 7.2 \, mW.\)
Find the current flowing in the circuit in Figure \(\PageIndex{12}\).
Strategy
This circuit can be analyzed using Kirchhoff’s rules. There is only one loop and no nodes. Choose the direction of current flow. For this example, we will use the clockwise direction from point a to point b . Consider Loop abcda and use Figure \(\PageIndex{5}\) to write the loop equation. Note that according to Figure \(\PageIndex{5}\), battery \(V_1\) will be added and battery \(V_2\) will be subtracted.
Solution
Applying the junction rule yields the following three equations. We have one unknown, so one equation is required:
\[Loop \, abcda : \, -IR_1 -V_1 -IR_2 +V_2 -IR_3 = 0.\]
Simplify the equations by placing the unknowns on one side of the equations. Use the values given in the figure.
\[I(R_1 + R_2 + R_3) = V_2 - V_1.\]
\[I = \frac {V_2 - V_1}{R_1 + R_2 + R_3} = \frac{24 \, V - 12 \, V}{10.0 \, \Omega + 30.0 \, \Omega + 10.0 \, \Omega} = 0.20 \, A.\]
Significance
The power dissipated or consumed by the circuit equals the power supplied to the circuit, but notice that the current in the battery \(V_1\) is flowing through the battery from the positive terminal to the negative terminal and consumes power.
\[P_{R_1} = I^2R_1 = 0.40 \, W\]
\[P_{R_2} = I^2R_2 = 1.20 \, W\]
\[P_{R_3} = I^2R_3 = 0.80 \, W\]
\[P_{V_1} = IV_1 = 2.40 \, W\]
\[P_{dissipated} = 4.80 \, W\]
\[P_{source} = IV_2 = 4.80 \, W\]
The power supplied equals the power dissipated by the resistors and consumed by the battery \(V_1\).
When using Kirchhoff’s laws, you need to decide which loops to use and the direction of current flow through each loop. In analyzing the circuit in Example \(\PageIndex{2}\), the direction of current flow was chosen to be clockwise, from point a to point b . How would the results change if the direction of the current was chosen to be counterclockwise, from point b to point a ?
- Answer
-
The current calculated would be equal to \(I = -0.20 \, A\) instead of \(I = 0.20 \, A\). The sum of the power dissipated and the power consumed would still equal the power supplied.
Multiple Voltage Sources
Many devices require more than one battery. Multiple voltage sources, such as batteries, can be connected in series configurations, parallel configurations, or a combination of the two.
In series, the positive terminal of one battery is connected to the negative terminal of another battery. Any number of voltage sources, including batteries, can be connected in series. Two batteries connected in series are shown in Figure \(\PageIndex{13}\). Using Kirchhoff’s loop rule for the circuit in part (b) gives the result
\[\varepsilon_1 - Ir_1 + \varepsilon_2 - Ir_2 - IR = 0,\]
\[[(\varepsilon_1 + \varepsilon_2) - I(r_1 + r_2)] - IR = 0.\]
When voltage sources are in series, their internal resistances can be added together and their source voltages can be added together to get the total values. Series connections of voltage sources are common—for example, in flashlights, toys, and other appliances. Usually, the cells are in series in order to produce a larger total source voltage. In Figure \(\PageIndex{13}\), the terminal voltage is
\[V_{terminal} = (\varepsilon_1 - Ir_1) + (\varepsilon_2 - Ir_2) = [(\varepsilon_1 + \varepsilon_2) - I(r_1 + r_2) - I(r_1 + r_2)] = (\varepsilon_1 + \varepsilon_2) + Ir_{eq}.\]
Note that the same current I is found in each battery because they are connected in series. The disadvantage of series connections of cells is that their internal resistances are additive.
Batteries are connected in series to increase the voltage supplied to the circuit. For instance, an LED flashlight may have two AAA cell batteries, each with a terminal voltage of 1.5 V, to provide 3.0 V to the flashlight.
Any number of batteries can be connected in series. For N batteries in series, the terminal voltage is equal to
\[V_{terminal} = (\varepsilon_1 + \varepsilon_2 + . . . +\varepsilon_{N-1} + \varepsilon_N) - I(r_1 + r_2 + . . . + r_{N-1} + r_N) = \sum_{i=1}^N \varepsilon_i - Ir_{eq}\]
where the equivalent resistance is \[r_{eq} = \sum_{i=1}^N r_i\]
When a load is placed across voltage sources in series, as in Figure \(\PageIndex{14}\), we can find the current:
\[(\varepsilon_1 - Ir_1) + (\varepsilon_2 - Ir_2) = IR,\]
\[Ir_1 + Ir_2 + IR = \varepsilon_1 + \varepsilon_2,\]
\[I = \frac{\varepsilon_1 + \varepsilon_2}{r_1 + r_2 + R}.\]
As expected, the internal resistances increase the equivalent resistance.
Voltage sources, such as batteries, can also be connected in parallel. Figure \(\PageIndex{15}\) shows two batteries with identical source voltages in parallel and connected to a load resistance. When the batteries are connected in parallel, the positive terminals are connected together and the negative terminals are connected together, and the load resistance is connected to the positive and negative terminals. Normally, voltage sources in parallel have identical source voltages. In this simple case, since the voltage sources are in parallel, the total source voltage is the same as the individual source voltages of each battery.
Consider the Kirchhoff analysis of the circuit in Figure \(\PageIndex{15b}\). There are two loops and a node at point b and \(\varepsilon = \varepsilon_1 = \varepsilon_2\).
Node b: \(I_1 + I_2 - I = 0\).
Loop abcfa: \(\varepsilon_2 - I_1r_1 + I_2r_2 - \varepsilon = 0, \, I_1r_1 = I_2r_2\).
Loop fcdef: \(\varepsilon_2 - I_2r_2 - IR = 0,\) \(\varepsilon - I_2r_2 - IR = 0.\)
Solving for the current through the load resistor results in \(I = \frac{\varepsilon}{r_{eq} +R}\), where \(r_{eq} = \left(\frac{1}{r_1} + \frac{1}{r_2}\right)^{-1}\). The terminal voltage is equal to the potential drop across the load resistor \(IR = \left(\frac{\varepsilon}{r_{eq} + R}\right)\).
The parallel connection reduces the internal resistance and thus can produce a larger current.
Any number of batteries can be connected in parallel. For N batteries in parallel, the terminal voltage is equal to
\[V_{terminal} = \varepsilon - I \left(\frac{1}{r_1} + \frac{1}{r_2} + . . . + \frac{1}{r_{N-1}} + \frac{1}{r_N}\right)^{-1} = \varepsilon - Ir_{eq}\]
where the equivalent resistance is \[r_{eq} = \left(\sum_{i=1}^N \frac{1}{r_i} \right)^{-1}\]
As an example, some diesel trucks use two 12-V batteries in parallel; they produce a total source voltage of 12 V but can deliver the larger current needed to start a diesel engine.
In summary, the terminal voltage of batteries in series is equal to the sum of the individual source voltages minus the sum of the internal resistances times the current. When batteries are connected in parallel, they usually have equal source voltages and the terminal voltage is equal to the source voltage minus the equivalent internal resistance times the current, where the equivalent internal resistance is smaller than the individual internal resistances. Batteries are connected in series to increase the terminal voltage to the load. Batteries are connected in parallel to increase the current to the load.
Solar Cell Arrays
Another example dealing with multiple voltage sources is that of combinations of solar cells - wired in both series and parallel combinations to yield a desired voltage and current. Photovoltaic generation, which is the conversion of sunlight directly into electricity, is based upon the photoelectric effect. The photoelectric effect is beyond the scope of this chapter and is covered in Photons and Matter Waves , but in general, photons hitting the surface of a solar cell create an electric current in the cell.
Most solar cells are made from pure silicon. Most single cells have a voltage output of about 0.5 V, while the current output is a function of the amount of sunlight falling on the cell (the incident solar radiation known as the insolation). Under bright noon sunlight, a current per unit area of about \(100 \, mA/cm^2\) of cell surface area is produced by typical single-crystal cells.
Individual solar cells are connected electrically in modules to meet electrical energy needs. They can be wired together in series or in parallel - connected like the batteries discussed earlier. A solar-cell array or module usually consists of between 36 and 72 cells, with a power output of 50 W to 140 W.
Solar cells, like batteries, provide a direct current (dc) voltage. Current from a dc voltage source is unidirectional. Most household appliances need an alternating current (ac) voltage.
Contributors and Attributions
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a
Creative Commons Attribution License (by 4.0)
.


3.E: Biopsychology (Exercises)
3.1: Human Genetics
Psychological researchers study genetics in order to better understand the biological basis that contributes to certain behaviors. While all humans share certain biological mechanisms, we are each unique. And while our bodies have many of the same parts—brains and hormones and cells with genetic codes—these are expressed in a wide variety of behaviors, thoughts, and reactions.
Review Questions
Q1
A(n) ________ is a sudden, permanent change in a sequence of DNA.
- allele
- chromosome
- epigenetic
- mutation
Q2
________ refers to a person’s genetic makeup, while ________ refers to a person’s physical characteristics.
- Phenotype; genotype
- Genotype; phenotype
- DNA; gene
- Gene; DNA
Q3
________ is the field of study that focuses on genes and their expression.
- Social psychology
- Evolutionary psychology
- Epigenetics
- Behavioral neuroscience
Q4
Humans have ________ pairs of chromosomes.
- \(15\)
- \(23\)
- \(46\)
- \(78\)
Critical Thinking Questions
Q5
The theory of evolution by natural selection requires variability of a given trait. Why is variability necessary and where does it come from?
Personal Application Questions
Q6
You share half of your genetic makeup with each of your parents, but you are no doubt very different from both of them. Spend a few minutes jotting down the similarities and differences between you and your parents. How do you think your unique environment and experiences have contributed to some of the differences you see?
Solution
S1
D
S2
B
S3
C
S4
B
S5
Variability is essential for natural selection to work. If all individuals are the same on a given trait, there will be no relative difference in their reproductive success because everyone will be equally adapted to their environments on that trait. Mutations are one source of variability, but sexual reproduction is another important source of variation given that individuals inherit half of their genetic makeup from each of their parents.
3.2: Cells of the Nervous System
Learning how the cells and organs (like the brain) function, help us understand the biological basis behind human psychology. The nervous system is composed of two basic cell types: glial cells (also known as glia) and neurons. Glial cells, which outnumber neurons ten to one, are traditionally thought to play a supportive role to neurons, both physically and metabolically.
Review Questions
Q1
The ________ receive(s) incoming signals from other neurons.
- soma
- terminal buttons
- myelin sheath
- dendrites
Q2
A(n) ________ facilitates or mimics the activity of a given neurotransmitter system.
- axon
- SSRI
- agonist
- antagonist
Q3
Multiple sclerosis involves a breakdown of the ________.
- soma
- myelin sheath
- synaptic vesicles
- dendrites
Q4
An action potential involves \(Na^+\) moving ________ the cell and \(K^+\) moving ________ the cell.
- inside; outside
- outside; inside
- inside; inside
- outside; outside
Critical Thinking Questions
Q5
Cocaine has two effects on synaptic transmission: it impairs reuptake of dopamine and it causes more dopamine to be released into the synapse. Would cocaine be classified as an agonist or antagonist? Why?
Q6
Drugs such as lidocaine and novocaine act as \(Na^+\) channel blockers. In other words, they prevent sodium from moving across the neuronal membrane. Why would this particular effect make these drugs such effective local anesthetics?
Personal Application Question
Q7
Have you or someone you know ever been prescribed a psychotropic medication? If so, what side effects were associated with the treatment?
Solution
S1
D
S2
C
S3
B
S4
A
S5
As a reuptake inhibitor, cocaine blocks the normal activity of dopamine at the receptor. The function causing more dopamine to be released into the synapse is agonist because it mimics and strengthens the effect of the neurotransmitter. Cocaine would be considered an agonist because by preventing the enzymatic degradation of the neurotransmitters, it increases the potential time that these neurotransmitters might be active in the synapse.
S6
The action potential is initiated by an influx of Na+ into the neuron. If this process is prevented, then no action potentials in neurons in a given area will occur. Therefore, any painful stimuli would not result in action potentials carrying that information to the brain.
3.3: Parts of the Nervous System
The nervous system can be divided into two major subdivisions: the central nervous system (CNS) and the peripheral nervous system (PNS), shown in the figure below. The CNS is comprised of the brain and spinal cord; the PNS connects the CNS to the rest of the body. In this section, we focus on the peripheral nervous system; later, we look at the brain and spinal cord.
Review Questions
Q1
Our ability to make our legs move as we walk across the room is controlled by the ________ nervous system.
- autonomic
- somatic
- sympathetic
- parasympathetic
Q2
If your ________ is activated, you will feel relatively at ease.
- somatic nervous system
- sympathetic nervous system
- parasympathetic nervous system
- spinal cord
Q3
The central nervous system is comprised of ________.
- sympathetic and parasympathetic nervous systems
- organs and glands
- somatic and autonomic nervous systems
- brain and spinal cord
Q4
Sympathetic activation is associated with ________.
- pupil dilation
- storage of glucose in the liver
- increased heart rate
- both A and C
Critical Thinking Questions
Q5
What are the implications of compromised immune function as a result of exposure to chronic stress?
Q6
Examine Fig. 3.3.2, illustrating the effects of sympathetic nervous system activation. How would all of these things play into the fight or flight response?
Personal Application Questions
Q7
Hopefully, you do not face real physical threats from potential predators on a daily basis. However, you probably have your fair share of stress. What situations are your most common sources of stress? What can you do to try to minimize the negative consequences of these particular stressors in your life?
Solution
S1
B
S2
C
S3
D
S4
D
S5
Chronic stress can lead to increased susceptibility to bacterial and viral infections, and potentially an increased risk of cancer. Ultimately, this could be a vicious cycle with stress leading to increased risk of disease, disease states leading to increased stress and so on.
S6
Most of these effects directly impact energy availability and redistribution of key resources and heightened sensory capacity. The individual experiencing these effects would be better prepared to fight or flee.
3.4: The Brain and Spinal Cord
The brain is a remarkably complex organ comprised of billions of interconnected neurons and glia. It is a bilateral, or two-sided, structure that can be separated into distinct lobes. Each lobe is associated with certain types of functions, but, ultimately, all of the areas of the brain interact with one another to provide the foundation for our thoughts and behaviors. In this section, we discuss the overall organization of the brain and the functions associated with different brain areas.
Review Questions
Q1
The ________ is a sensory relay station where all sensory information, except for smell, goes before being sent to other areas of the brain for further processing.
- amygdala
- hippocampus
- hypothalamus
- thalamus
Q2
Damage to the ________ disrupts one’s ability to comprehend language, but it leaves one’s ability to produce words intact.
- amygdala
- Broca’s Area
- Wernicke’s Area
- occipital lobe
Q3
A(n) ________ uses magnetic fields to create pictures of a given tissue.
- EEG
- MRI
- PET scan
- CT scan
Q4
Which of the following is not a structure of the forebrain?
- thalamus
- hippocampus
- amygdala
- substantia nigra
Critical Thinking Questions
Q5
Before the advent of modern imaging techniques, scientists and clinicians relied on autopsies of people who suffered brain injury with resultant change in behavior to determine how different areas of the brain were affected. What are some of the limitations associated with this kind of approach?
Q6
Which of the techniques discussed would be viable options for you to determine how activity in the reticular formation is related to sleep and wakefulness? Why?
Personal Application Questions
Q7
You read about H. M.’s memory deficits following the bilateral removal of his hippocampus and amygdala. Have you encountered a character in a book, television program, or movie that suffered memory deficits? How was that character similar to and different from H. M.?
Solution
S1
D
S2
C
S3
B
S4
D
S5
The same limitations associated with any case study would apply here. In addition, it is possible that the damage caused changes in other areas of the brain, which might contribute to the behavioral deficits. Such changes would not necessarily be obvious to someone performing an autopsy, as they may be functional in nature, rather than structural.
S6
The most viable techniques are fMRI and PET because of their ability to provide information about brain activity and structure simultaneously.
3.5: The Endocrine System
The endocrine system consists of a series of glands that produce chemical substances known as hormones. Like neurotransmitters, hormones are chemical messengers that must bind to a receptor in order to send their signal. However, unlike neurotransmitters, which are released in close proximity to cells with their receptors, hormones are secreted into the bloodstream and travel throughout the body, affecting any cells that contain receptors for them.
Review Questions
Q1
The two major hormones secreted from the pancreas are:
- estrogen and progesterone
- norepinephrine and epinephrine
- thyroxine and oxytocin
- glucagon and insulin
Q2
The ________ secretes messenger hormones that direct the function of the rest of the endocrine glands.
- ovary
- thyroid
- pituitary
- pancreas
Q3
The ________ gland secretes epinephrine.
- adrenal
- thyroid
- pituitary
- master
Q4
The ________ secretes hormones that regulate the body’s fluid levels.
- adrenal
- pituitary
- testis
- thyroid
Critical Thinking Questions
Q5
Hormone secretion is often regulated through a negative feedback mechanism, which means that once a hormone is secreted it will cause the hypothalamus and pituitary to shut down the production of signals necessary to secrete the hormone in the first place. Most oral contraceptives are made of small doses of estrogen and/or progesterone. Why would this be an effective means of contraception?
Q6
Chemical messengers are used in both the nervous system and the endocrine system. What properties do these two systems share? What properties are different? Which one would be faster? Which one would result in long-lasting changes?
Personal Application Questions
Q7
Given the negative health consequences associated with the use of anabolic steroids, what kinds of considerations might be involved in a person’s decision to use them?
Solution
S1
D
S2
C
S3
A
S4
B
S5
The introduction of relatively low, yet constant, levels of gonadal hormones places the hypothalamus and pituitary under inhibition via negative feedback mechanisms. This prevents the alterations in both estrogen and progesterone concentrations that are necessary for successful ovulation and implantation.
S6
Both systems involve chemical messengers that must interact with receptors in order to have an effect. The relative proximity of the release site and target tissue varies dramatically between the two systems. In neurotransmission, reuptake and enzymatic breakdown immediately clear the synapse. Metabolism of hormones must occur in the liver. Therefore, while neurotransmission is much more rapid in signaling information, hormonal signaling can persist for quite some time as the concentrations of the hormone in the bloodstream vary gradually over time.
Contributor
-
Rose M. Spielman with many significant contributors. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the creative commons license and may not be reproduced without the prior and express written consent of Rice University. For questions regarding this license, please contact<EMAIL_ADDRESS>.Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>.


2.5: Solve Equations with Fractions or Decimals
-
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Learning Objectives
By the end of this section, you will be able to:
- Solve equations with fraction coefficients
- Solve equations with decimal coefficients
Note
Before you get started, take this readiness quiz.
-
Multiply: \(8\cdot 38\).
If you missed this problem, review Exercise 1.6.16 . -
Find the LCD of \(\frac{5}{6}\) and \(\frac{1}{4}\).
If you missed this problem, review Exercise 1.7.16 . -
Multiply 4.78 by 100.
If you missed this problem, review Exercise 1.8.22 .
Solve Equations with Fraction Coefficients
Let’s use the general strategy for solving linear equations introduced earlier to solve the equation, \(\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}\).
| To isolate the x term, subtract \(\frac{1}{2}\) from both sides. | |
| Simplify the left side. | |
| Change the constants to equivalent fractions with the LCD. | |
| Subtract. | |
| Multiply both sides by the reciprocal of \(\frac{1}{8}\). | |
| Simplify. |
This method worked fine, but many students do not feel very confident when they see all those fractions. So, we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.
We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but without fractions. This process is called “clearing” the equation of fractions.
Let’s solve a similar equation, but this time use the method that eliminates the fractions.
Example \(\PageIndex{1}\): How to Solve Equations with Fraction Coefficients
Solve: \(\frac{1}{6}y - \frac{1}{3} = \frac{5}{6}\)
Solution
Try It \(\PageIndex{2}\)
Solve: \(\frac{1}{4}x + \frac{1}{2} = \frac{5}{8}\)
- Answer
-
\(x= \frac{1}{2}\)
Try It \(\PageIndex{3}\)
Solve: \(\frac{1}{8}x + \frac{1}{2} = \frac{1}{4}\)
- Answer
-
\(x = -2\)
Notice in Exercise \(\PageIndex{1}\), once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve! We then used the General Strategy for Solving Linear Equations.
STRATEGY TO SOLVE EQUATIONS WITH FRACTION COEFFICIENTS.
- Find the least common denominator of all the fractions in the equation.
- Multiply both sides of the equation by that LCD. This clears the fractions.
- Solve using the General Strategy for Solving Linear Equations.
Example \(\PageIndex{4}\)
Solve: \(6 = \frac{1}{2}v + \frac{2}{5}v - \frac{3}{4}v\)
Solution
We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.
| Find the LCD of all fractions in the equation. | ||
| The LCD is 20. | ||
| Multiply both sides of the equation by 20. | ||
| Distribute. | ||
| Simplify—notice, no more fractions! | ||
| Combine like terms. | ||
| Divide by 3. | ||
| Simplify. | ||
| Check: | ||
| Let v=40. | ||
Try It \(\PageIndex{5}\)
Solve: \(7 = \frac{1}{2}x + \frac{3}{4}x - \frac{2}{3}x\)
- Answer
-
\(x = 12\)
Try It \(\PageIndex{6}\)
Solve: \(-1 = \frac{1}{2}u + \frac{1}{4}u - \frac{2}{3}u\)
- Answer
-
\(u = -12\)
In the next example, we again have variables on both sides of the equation.
Example \(\PageIndex{7}\)
Solve: \(a + \frac{3}{4} = \frac{3}{8}a - \frac{1}{2}\)
Solution
|
Find the LCD of all fractions in the equation.
The LCD is 8. |
||
| Multiply both sides by the LCD. | ||
| Distribute. | ||
| Simplify—no more fractions. | ||
| Subtract 3a3a from both sides. | ||
| Simplify. | ||
| Subtract 6 from both sides. | ||
| Simplify. | ||
| Divide by 5. | ||
| Simplify. | ||
| Check: | ||
| Let a=−2. | ||
Try It \(\PageIndex{8}\)
Solve: \(x + \frac{1}{3} = \frac{1}{6}x - \frac{1}{2}\)
- Answer
-
\(x = -1\)
Try It \(\PageIndex{9}\)
Solve: \(c + \frac{3}{4} = \frac{1}{2}c - \frac{1}{4}\)
- Answer
-
\(c = -2\)
In the next example, we start by using the Distributive Property. This step clears the fractions right away.
Example \(\PageIndex{10}\)
Solve: \(-5 = \frac{1}{4}(8x + 4)\)
Solution
| Distribute. | ||
|
Simplify.
Now there are no fractions. |
||
| Subtract 1 from both sides. | ||
| Simplify. | ||
| Divide by 2. | ||
| Simplify. | ||
| Check: | ||
| Let x=−3. | ||
Try It \(\PageIndex{11}\)
Solve: \(-11 = \frac{1}{2}(6p + 2)\)
- Answer
-
\(p = -4\)
Try It \(\PageIndex{12}\)
Solve: \(8 = \frac{1}{3}(9q + 6)\)
- Answer
-
\(q = 2\)
In the next example, even after distributing, we still have fractions to clear.
Example \(\PageIndex{13}\)
Solve: \(\frac{1}{2}(y - 5) = \frac{1}{4}(y - 1)\)
Solution
| Distribute. | ||
| Simplify. | ||
| Multiply by the LCD, 4. | ||
| Distribute. | ||
| Simplify. | ||
| Collect the variables to the left. | ||
| Simplify. | ||
| Collect the constants to the right. | ||
| Simplify. | ||
| Check: | ||
| Let y=9. | ||
| Finish the check on your own. |
Try It \(\PageIndex{14}\)
Solve: \(\frac{1}{5}(n + 3) = \frac{1}{4}(n + 2)\)
- Answer
-
\(n = 2\)
Try It \(\PageIndex{15}\)
Solve: \(\frac{1}{2}(m - 3) = \frac{1}{4}(m - 7)\)
- Answer
-
\(m = -1\)
Example \(\PageIndex{16}\)
Solve: \(\frac{5x - 3}{4} = \frac{x}{2}\)
Solution
| Multiply by the LCD, 4. | ||
| Simplify. | ||
| Collect the variables to the right. | ||
| Simplify. | ||
| Divide. | ||
| Simplify. | ||
| Check: | ||
| Let x=1. | ||
Try It \(\PageIndex{17}\)
Solve: \(\frac{4y - 7}{3} = \frac{y}{6}\)
- Answer
-
\(y = 2\)
Try It \(\PageIndex{18}\)
Solve: \(\frac{-2z - 5}{4} = \frac{z}{8}\)
- Answer
-
\(z = -2\)
Example \(\PageIndex{19}\)
Solve: \(\frac{a}{6} + 2 = \frac{a}{4} + 3\)
Solution
| Multiply by the LCD, 12. | ||
| Distribute. | ||
| Simplify. | ||
| Collect the variables to the right. | ||
| Simplify. | ||
| Collect the constants to the left. | ||
| Simplify. | ||
| Check: | ||
| Let a=−12. | ||
Try It \(\PageIndex{20}\)
Solve: \(\frac{b}{10} + 2 = \frac{b}{4} + 5\)
- Answer
-
\(b = -20\)
Try It \(\PageIndex{21}\)
Solve: \(\frac{c}{6} + 3 = \frac{c}{3} + 4\)
- Answer
-
\(c= -6\)
Example \(\PageIndex{22}\)
Solve: \(\frac{4q + 3}{2}+ 6 = \frac{3q + 5}{4}\)
Solution
| Multiply by the LCD, 4. | ||
| Distribute. | ||
| Simplify. |
|
|
| Collect the variables to the left. | ||
| Simplify. | ||
| Collect the constants to the right. | ||
| Simplify. | ||
| Divide by 5. | ||
| Simplify. | ||
| Check: | ||
| Let q=−5. | ||
| Finish the check on your own. |
Try It \(\PageIndex{23}\)
Solve: \(\frac{3r + 5}{6}+ 1 = \frac{4r + 3}{3}\)
- Answer
-
\(r = 1\)
Try It \(\PageIndex{24}\)
Solve: \(\frac{2s + 3}{2}+ 1 = \frac{3s + 2}{4}\)
- Answer
-
\(s = -8\)
Solve Equations with Decimal Coefficients
Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions. For example, \(0.3 = \frac{3}{10}\) and \(0.17 = \frac{17}{100}\). So, with an equation with decimals, we can use the same method we used to clear fractions—multiply both sides of the equation by the least common denominator.
Example \(\PageIndex{25}\)
Solve: \(0.06x + 0.02 = 0.25x - 1.5\)
Solution
Look at the decimals and think of the equivalent fractions.
\(0.06 = \frac { 6 } { 100 } \quad 0.02 = \frac { 2 } { 100 } \quad 0.25 = \frac { 25 } { 100 } \quad 1.5 = 1 \frac { 5 } { 10 }\)
Notice, the LCD is 100.
By multiplying by the LCD, we will clear the decimals from the equation.
| Multiply both side by 100. | |
| Distribute. | |
| Multiply, and now we have no more decimals. | |
| Collect the variables to the right. | |
| Simplify. | |
| Collect the variables to the right. | |
| Simplify. | |
| Divide by 19. | |
| Simplify. | |
|
Check: Let x=8
|
Try It \(\PageIndex{26}\)
Solve: \(0.14h + 0.12 = 0.35h - 2.4\)
- Answer
-
\(h = 12\)
Try It \(\PageIndex{27}\)
Solve: \(0.65k - 0.1 = 0.4k - 0.35\)
- Answer
-
\(k = -1\)
The next example uses an equation that is typical of the money applications in the next chapter. Notice that we distribute the decimal before we clear all the decimals.
Example \(\PageIndex{28}\)
Solve: \(0.25x + 0.05(x + 3) = 2.85\)
Solution
| Distribute first. | |
| Combine like terms. | |
| To clear decimals, multiply by 100. | |
| Distribute. | |
| Subtract 15 from both sides. | |
| Simplify. | |
| Divide by 30. | |
| Simplify. | |
| Check it yourself by substituting x=9 into the original equation. |
Try It \(\PageIndex{29}\)
Solve: \(0.25n + 0.05(n + 5) = 2.95\)
- Answer
-
\(n = 9\)
Try It \(\PageIndex{30}\)
Solve: \(0.10d + 0.05(d -5) = 2.15\)
- Answer
-
\(d = 16\)
Key Concepts
-
Strategy to Solve an Equation with Fraction Coefficients
- Find the least common denominator of all the fractions in the equation.
- Multiply both sides of the equation by that LCD. This clears the fractions.
- Solve using the General Strategy for Solving Linear Equations.