1.2.7: The Greatest Common Factor and Factor by Grouping
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We begin this section with definition of a factor of a term. Because \(24 = 2\cdot 12\), both \(2\) and \(12\) are factors of \(24\). Note that \(2\) is also a divisor of \(24\), because when you divide \(24\) by \(2\) you get \(12\), with a remainder of zero. Similarly, \(12\) is also a divisor of \(24\), because when you divide \(24\) by \(12\) you get \(2\), with a remainder of zero.
Factors of a Term
Suppose \(m\) and \(n\) are integers. Then \(m\) is a factor of a term, \(n\), if and only if there exists another integer \(k\) so that \(n=m \cdot k\).
A factor is also known as a divisor of a term since the term can be divided evenly by the factor, \(\dfrac{n}{m}=k\).
Example \(\PageIndex{1}\)
List the factors of \(24\).
Solution
First, list all possible ways that we can express \(24\) as a product of two positive integers:
\[24=1 \cdot 24 \quad \text { or } \quad 24=2 \cdot 12 \quad \text { or } \quad 24=3 \cdot 8 \quad \text { or } \quad 24=4 \cdot 6 \nonumber \]
Therefore, the factors of \(24\) are \(1,2,3,4,6,8,\) and \(24\).
You Try \(\PageIndex{1}\)
List the factors of \(18\).
- Answer
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\(1,2,3,6,9,\) and \(18\)
Example \(\PageIndex{2}\)
List the factors that \(36\) and \(48\) have in common.
Solution
First, list all factors of \(36\) and \(48\) separately, then determine the factors that are in common.
Factors of \(36\) are: \(1, 2, 3, 4, 6, 9, 12, 18, 36\)
Factorss of \(48\) are: \(1, 2, 3, 4, 6, 8, 12, 16, 24, 48\)
Therefore, the common factors of \(36\) and \(48\) are \(1, 2, 3, 4, 6,\) and \(12\).
You Try \(\PageIndex{2}\)
List the factors that \(40\) and \(60\) have in common.
- Answer
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\(1,2,4,5,10,\) and \(20\)
Greatest Common Factor
The greatest common factor of \(a\) and \(b\) is the largest positive number that divides evenly (with no remainder) both \(a\) and \(b\). The greatest common factor of \(a\) and \(b\) is denoted by the symbolism \(\operatorname{GCF}(a, b)\).
Since factors are also divisors, the abbreviation \(\operatorname{GCD}(a, b)\) also represents the greatest common factor of \(a\) and \(b\).
In Example \(\PageIndex{2}\), we listed the common factors of \(36\) and \(48\). The largest of these common factor was \(12\). Hence, the greatest common factor of \(36\) and \(48\) is \(12\), written \(\operatorname{GCF}(36, 48)=12\).
With larger numbers, it can be more difficult to identify the greatest common factor. Prime factorization can be helpful in this situation!
Example \(\PageIndex{3}\)
Find the greatest common factor of \(360\) and \(756\).
Solution
Prime factor \(360\) and \(756\), writing your answer in exponential form.
Thus:
\[\begin{array}{l}{360=2^{3} \cdot 3^{2} \cdot 5} \\ {756=2^{2} \cdot 3^{3} \cdot 7}\end{array} \nonumber \]
To find the greatest common factor, list each factor that appears in common and the highest power that appears in common amongst the terms.
In this case, the factors \(2\) and \(3\) appear in common, with \(2\) being the highest power of \(2\) and \(2\) being the highest power of \(3\) that appear in common. Therefore, the greatest common factor of \(360\) and \(756\) is:
\[\begin{aligned} \mathrm{GCF}(360,756) &=2^{2} \cdot 3^{2} \\ &=4 \cdot 9 \\ &=36 \end{aligned} \nonumber \]
Therefore, the greatest common factor is \(\mathrm{GCF}(360,756)=36\).
Note what happens when we write each of the given numbers in the last example as a product of the greatest common factor and a second factor:
\[\begin{array}{l}{360={\color {Red} 36} \cdot 10} \\ {756={\color {Red} 36} \cdot 21}\end{array} \nonumber \]
In each case, note how the second second factors (\(10\) and \(21\)) contain no additional common factors.
You Try \(\PageIndex{3}\)
Find the greatest common factor of \(120\) and \(450\).
- Answer
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\(30\)
Finding the Greatest Common Factor of Monomials
Example \(\PageIndex{3}\) reveals the technique used to find the greatest common factor of two or more monomials.
Finding the Greatest Common Factor (GCF) of Two or More Monomials
To find the greatest common factor of two or more monomials:
- Find the greatest common factor of the coefficients of the given monomials. Use prime factorization if necessary.
- List each variable that appears in common in the given monomials.
- Raise each variable factor that appears in common to the highest power that appears in common among the given monomials.
Example \(\PageIndex{4}\)
Find the greatest common factor of \(6 x^{3} y^{3}\) and \(9 x^{2} y^{5}\).
Solution
To find the \(\mathrm{GCF}\) of \(6 x^{3} y^{3}\) and \(9 x^{2} y^{5}\), we note that:
- The greatest common factor of \(6\) and \(9\) is \(3\).
- The monomials \(6x^3y^3\) and \(9x^2y^5\) have the variables \(x\) and \(y\) in common.
- The highest power of \(x\) in common is \(x^2\). The highest power of \(y\) in common is \(y^3\).
Thus, the greatest common factor is \(\mathrm{GCF}\left(6 x^{3} y^{3}, 9 x^{2} y^{5}\right)=3 x^{2} y^{3}\). Note what happens when we write each of the given monomials as a product of the greatest common factor and a second monomial:
\[\begin{array}{l}{6 x^{3} y^{3}={\color {Red} 3 x^{2} y^{3}} \cdot 2 x} \\ {9 x^{2} y^{5}={\color {Red} 3 x^{2} y^{3}} \cdot 3 y}\end{array} \nonumber \]
Observe that the set of second monomial factors (\(2x\) and \(3y\)) contain no additional common factors.
You Try \(\PageIndex{4}\)
Find the greatest common factor of \(16xy^3\) and \(12x^4y^2\).
- Answer
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\(4 x y^{2}\)
Example \(\PageIndex{5}\)
Find the greatest common factor of \(12x^4\), \(18 x^3\), and \(30 x^2\).
Solution
To find the \(\mathrm{GCF}\) of \(12x^4\), \(18 x^3\), and \(30 x^2\), we note that:
- The greatest common factor of \(12\), \(18\), and \(30\) is \(6\).
- The monomials \(12x^4\), \(18 x^3\), and \(30 x^2\) have the variable \(x\) in common.
- The highest power of \(x\) in common is \(x^2\).
Thus, the greatest common factor is \(\mathrm{GCF}\left(12 x^{4}, 18 x^{3}, 30 x^{2}\right)=6 x^{2}\). Note what happens when we write each of the given monomials as a product of the greatest common factor and a second monomial:
\[\begin{array}{l}{12 x^{4}={\color {Red}6 x^{2}} \cdot 2 x^{2}} \\ {18 x^{3}={\color {Red}6 x^{2}} \cdot 3 x} \\ {30 x^{2}={\color {Red}6 x^{2}} \cdot 5}\end{array} \nonumber \]
Observe that the set of second monomial factors (\(2x^2\), \(3 x\), and \(5\)) contain no additional common factors.
You Try \(\PageIndex{5}\)
Find the greatest common factor of \(6y^3\), \(15 y^2\), and \(9 y^5\).
- Answer
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\(3 y^{2}\)
Factor Out the GCF
Earlier, we multiplied a monomial and polynomial by distributing the monomial times each term in the polynomial.
\[\begin{aligned} {\color {Red}2 x(}3 x^{2}+4 x-7{\color {Red})} &={\color {Red}2 x} \cdot 3 x^{2}+{\color {Red}2 x} \cdot 4 x-{\color {Red}2 x} \cdot 7 \\ &=6 x^{3}+8 x^{2}-14 x \end{aligned} \nonumber \]
In this section we reverse that multiplication process using the distributive property, \[a \cdot b+a\cdot c=a(b+c)\nonumber\]
We present you with the final product and ask you to bring back the original multiplication problem. In the case \(6 x^{3}+8 x^{2}-14 x\), the greatest common factor of \(6x^3\), \(8x^2\), and \(14x\) is \(2x\). We then use the distributive property to factor out \(2x\) from each term of the polynomial.
\[\begin{aligned} 6 x^{3}+8 x^{2}-14 x &={\color {Red}2 x} \cdot 3 x^{2}+{\color {Red}2 x} \cdot 4 x-{\color {Red}2 x} \cdot 7 \\ &={\color {Red}2 x(}3 x^{2}+4 x-7{\color {Red})} \end{aligned} \nonumber \]
Factoring
Factoring is the process of writing an expression as a product of polynomials.
Let’s look at a few examples that factor out the \(\mathrm{GCF}\).
Example \(\PageIndex{6}\)
Factor: \(6 x^{2}+10 x+14\)
Solution
The greatest common factor (\(\mathrm{GCF}\)) of \(6x^2\), \(10 x\) and \(14\) is \(2\). Factor out the \(\mathrm{GCF}\).
\[\begin{aligned} 6 x^{2}+10x+14 x &={\color {Red}2 } \cdot 3 x^{2}+{\color {Red}2 } \cdot 5 x+{\color {Red}2 } \cdot 7 \\ &={\color {Red}2 (}3 x^{2}+5 x+7{\color {Red})} \end{aligned} \nonumber \]
The factored form of \(6 x^{2}+10 x+14\) is \(2(3x^2+5x+7)\).
Check:
Check that you factored correctly by multiplying:
\[\begin{aligned} {\color {Red}2 (}3 x^{2}+5 x+7{\color {Red})} &={\color {Red}2 } \cdot 3 x^{2}+{\color {Red}2 } \cdot 5 x+{\color {Red}2 } \cdot 7 \\ &=6 x^{2}+10 x+14 \end{aligned} \nonumber \]
That’s the original polynomial, so we factored correctly.
You Try \(\PageIndex{6}\)
Factor: \(9 y^{2}-15 y+12\)
- Answer
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\(3\left(3 y^{2}-5 y+4\right)\)
Example \(\PageIndex{7}\)
Factor: \(12 y^{5}-32 y^{4}+8 y^{2}\)
Solution
The greatest common factor (\(\mathrm{GCF}\)) of \(12y^5\), \(32y^4\) and \(8y^2\) is \(4y^2\). Factor out the \(\mathrm{GCF}\).
\[\begin{aligned} 12 y^{5}-32 y^{4}+8 y^{2} &={\color {Red}4 y^{2}} \cdot 3 y^{3}-{\color {Red}4 y^{2}} \cdot 8 y^{2}+{\color {Red}4 y^{2}} \cdot 2 \\ &={\color {Red}4 y^{2} (}3 y^{3}-8 y^{2}+2{\color {Red})} \end{aligned} \nonumber \]
Check: Multiply. Distribute the monomial \(4y^2\).
\[\begin{aligned} {\color {Red}4 y^{2} (}3 y^{3}-8 y^{2}+2{\color {Red})} &={\color {Red}4 y^{2}} \cdot 3 y^{3}-{\color {Red}4 y^{2}} \cdot 8 y^{2}+{\color {Red}4 y^{2}} \cdot 2 \\ &=12 y^{5}-32 y^{4}+8 y^{2} \end{aligned} \nonumber \]
That’s the original polynomial. We have factored correctly.
You Try \(\PageIndex{7}\)
Factor: \(8 x^{6}+20 x^{4}-24 x^{3}\)
- Answer
-
\(4 x^{3}\left(2 x^{3}+5 x-6\right)\)
Example \(\PageIndex{8}\)
Factor: \(12 a^{3} b+24 a^{2} b^{2}+12 a b^{3}\)
Solution
The greatest common factor (\(\mathrm{GCF}\)) of \(12a^3b\), \(24 a^2b^2\) and \(12ab^3\) is \(12ab\). Factor out the \(\mathrm{GCF}\).
\[\begin{aligned} 12 a^{3} b+24 a^{2} b^{2}+12 a b^{3} &={\color {Red}12ab} \cdot a^{2}-{\color {Red}12ab} \cdot 2 a b+{\color {Red}12ab} \cdot b^{2} \\ &= {\color {Red}12ab (}a^{2}+2ab+b^{2}{\color {Red})} \end{aligned} \nonumber \]
Check: Multiply. Distribute the monomial \(12ab\).
\[\begin{aligned} {\color {Red}12ab (}a^{2}+2ab+b^{2}{\color {Red})} &={\color {Red}12ab} \cdot a^{2}-{\color {Red}12ab} \cdot 2 a b+{\color {Red}12ab} \cdot b^{2} \\ &=12 a^{3} b+24 a^{2} b^{2}+12 a b^{3} \end{aligned} \nonumber \]
That’s the original polynomial. We have factored correctly.
You Try \(\PageIndex{8}\)
Factor: \(15 s^{2} t^{4}+6 s^{3} t^{2}+9 s^{2} t^{2}\)
- Answer
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\(3 s^{2} t^{2}\left(5 t^{2}+2 s+3\right)\)
Remember that the distributive property allows us to pull the \(\mathrm{GCF}\) out in front of the expression or to pull it out in back. In symbols:
\({\color {Red}a}b+{\color {Red}a}c={\color {Red}a}(b+c) \quad\) or \(\quad b{\color {Red}a}+c{\color {Red}a}=(b+c){\color {Red}a}\)
We can also use the distributive property to factor out binomial factors.
Example \(\PageIndex{9}\)
Factor: \(2 x(3 x+2)+5(3 x+2)\)
Solution
In this case, the greatest common factor (\(\mathrm{GCF}\)) is \(3x + 2\).
\[\begin{aligned} 2x(3 x+2)+5(3 x+2) &=2x \cdot {\color {Red}(3 x+2)}+5 \cdot{\color {Red}(3 x+2)} \\ &=(2 x+5){\color {Red}(3 x+2)} \end{aligned} \nonumber \]
Because of the commutative property of multiplication, it is equally valid to pull the \(\mathrm{GCF}\) out in front.
\[\begin{aligned} 2x(3 x+2)+5(3 x+2) &=2x \cdot {\color {Red}(3 x+2)}+5 \cdot{\color {Red}(3 x+2)} \\ &={\color {Red}(3 x+2)} (2 x+5)\end{aligned}\]
Note that the order of factors differs from the first solution, but because of the commutative property of multiplication, the order does not matter. The answers are the same.
You Try \(\PageIndex{9}\)
Factor: \(3 x^{2}(4 x-7)+8(4 x-7)\)
- Answer
-
\(\left(3 x^{2}+8\right)(4 x-7)\)
Example \(\PageIndex{10}\)
Factor: \(15 a(a+b)-12(a+b)\)
Solution
In this case, the greatest common factor (\(\mathrm{GCF}\)) is \(3(a + b)\).
\[\begin{aligned} 15 a(a+b)-12(a+b) &={\color {Red}3(a+b)} \cdot 5 a-{\color {Red}3(a+b)} \cdot 4 \\ &={\color {Red}3(a+b)}(5 a-4) \end{aligned} \nonumber \]
Alternate solution:
It is possible that you might not notice that \(15\) and \(12\) are divisible by \(3\), factoring out only the common factor \(a + b\).
\[\begin{aligned} 15 a(a+b)-12(a+b) &=15a\cdot {\color {Red}(a+b)} - 12\cdot {\color {Red}(a+b)} \\ &=(15a-12) {\color {Red}(a+b)}\end{aligned} \nonumber \]
However, you now need to notice that you can continue to factor by factoring out \(3\) from both \(15a\) and \(12\).
\(=3(5 a-4)(a+b)\)
Note that the order of factors differs from the first solution, but because of the commutative property of multiplication, the order does not matter. The answers are the same.
You Try \(\PageIndex{10}\)
Factor: \(24 m(m-2 n)+20(m-2 n)\)
- Answer
-
\(4(6 m+5)(m-2 n)\)
Factoring by Grouping
The last factoring skill in this section involves four-term expressions. The technique for factoring a four-term expression is called factoring by grouping .
Example \(\PageIndex{11}\)
Factor by grouping: \(x^{2}+8 x+3 x+24\)
Solution
We “group” the first and second terms, noting that we can factor an \(x\) out of both of these terms. Then we “group” the third and fourth terms, noting that we can factor \(3\) out of both of these terms.
\[ \underbrace{x^2+8x} +\underbrace{3 x+24}=x(x+8)+3(x+8) \]
Now we can factor \(x + 8\) out of both of these terms.
\((x+3){\color {Red}(x+8)}\)
You Try \(\PageIndex{11}\)
Factor by grouping: \(x^{2}-6 x+2 x-12\)
- Answer
-
\((x+2)(x-6)\)
Let’s try a grouping that contains some negative signs.
Example \(\PageIndex{12}\)
Factor by grouping: \(x^{2}+4 x-7 x-28\)
Solution
We “group” the first and second terms, noting that we can factor \(x\) out of both of these terms. Then we “group” the third and fourth terms, then try to factor a \(7\) out of both these terms.
\[ \underbrace{x^2+4x} -\underbrace{7x-28}=x{\color{Red} (x+4)}+7{\color{Red} (-x-4)} \]
This does not lead to a common factor. Let’s try again, this time factoring a \(−7\) out of the third and fourth terms.
\[ \underbrace{x^2+4x} -\underbrace{7x-28}=x{\color{Red} (x+4)}-7{\color{Red} (x+4)} \]
That worked! We now factor out a common factor \(x + 4\).
\((x-7){\color {Red}(x+4)}\)
It is important to be careful with negatives when factoring by grouping. In the previous example, it would be incorrect to write \(x^{2}+4 x-7 x-28\) as \((x^2+4x)-(7x-28)\) because if you distribute the negative through, the last term is +28 which is not the original polynomial.
You Try \(\PageIndex{12}\)
Factor by grouping: \(x^{2}-5 x-4 x+20\)
- Answer
-
\((x-4)(x-5)\)
Let’s increase the size of the numbers a bit.
Example \(\PageIndex{13}\)
Factor by grouping: \(6 x^{2}-8 x+9 x-12\)
Solution
Note that we can factor \(2x\) out of the first two terms and \(3\) out of the second two terms.
\[ \underbrace{x^2+4x} -\underbrace{9x-12}=2x{\color{Red} (3 x-4)}-7{\color{Red} (3 x-4)} \]
Now we have a common factor \(3x−4\) which we can factor out.
\((2x+3){\color {Red}(3x-4)}\)
You Try \(\PageIndex{13}\)
Factor by grouping: \(15 x^{2}+9 x+10 x+6\)
- Answer
-
\((3 x+2)(5 x+3)\)
As the numbers get larger and larger, you need to factor out the (\(\mathrm{GCF}\)) from each grouping. If not, you won’t get a common factor to finish the factoring.
Example \(\PageIndex{14}\)
Factor by grouping: \(24 x^{2}-32 x-45 x+60\)
Solution
Suppose that we factor \(8x\) out of the first two terms and \(−5\) out of the second two terms.
\[ \underbrace{24x^2-32x} -\underbrace{45x+60}=8x{\color{Red} (3x-4)}-5{\color{Red} (9x-12)} \]
That did not work, as we don’t have a common factor to complete the factoring process. However, note that we can still factor out a \(3\) from \(9x−12\). As we’ve already factored out a \(5\), and now we see can factor out an additional \(3\), this means that we should have factored out \(3\) times \(5\), or \(15\), to begin with. Let’s start again, only this time we’ll factor \(15\) out of the second two terms.
\[ \underbrace{24x^2-32x} -\underbrace{45x+60}=8x{\color{Red} (3x-4)}-15{\color{Red} (3x-4)} \]
Beautiful! We can now factor out \(3x−4\).
\((8x-15){\color {Red}(3x-4)}\)
You Try \(\PageIndex{14}\)
Factor by grouping: \(36 x^{2}-84 x+15 x-35\)
- Answer
-
\((12 x+5)(3 x-7)\)


10.5: Types of Muscle Fibers
Learning Objectives
- Describe the types of skeletal muscle fibers
- Explain fast and slow muscle fibers
Two criteria to consider when classifying the types of muscle fibers are how fast some fibers contract relative to others, and how fibers produce ATP. Using these criteria, there are three main types of skeletal muscle fibers. Slow oxidative (SO) fibers contract relatively slowly and use aerobic respiration (oxygen and glucose) to produce ATP. Fast oxidative (FO) fibers have fast contractions and primarily use aerobic respiration, but because they may switch to anaerobic respiration (glycolysis), can fatigue more quickly than SO fibers. Lastly, fast glycolytic (FG) fibers have fast contractions and primarily use anaerobic glycolysis. The FG fibers fatigue more quickly than the others. Most skeletal muscles in a human contain(s) all three types, although in varying proportions.
The speed of contraction is dependent on how quickly myosin’s ATPase hydrolyzes ATP to produce cross-bridge action. Fast fibers hydrolyze ATP approximately twice as quickly as slow fibers, resulting in much quicker cross-bridge cycling (which pulls the thin filaments toward the center of the sarcomeres at a faster rate). The primary metabolic pathway used by a muscle fiber determines whether the fiber is classified as oxidative or glycolytic. If a fiber primarily produces ATP through aerobic pathways it is oxidative. More ATP can be produced during each metabolic cycle, making the fiber more resistant to fatigue. Glycolytic fibers primarily create ATP through anaerobic glycolysis, which produces less ATP per cycle. As a result, glycolytic fibers fatigue at a quicker rate.
The oxidative fibers contain many more mitochondria than the glycolytic fibers, because aerobic metabolism, which uses oxygen (O 2 ) in the metabolic pathway, occurs in the mitochondria. The SO fibers possess a large number of mitochondria and are capable of contracting for longer periods because of the large amount of ATP they can produce, but they have a relatively small diameter and do not produce a large amount of tension. SO fibers are extensively supplied with blood capillaries to supply O 2 from the red blood cells in the bloodstream. The SO fibers also possess myoglobin, an O 2 -carrying molecule similar to O 2 -carrying hemoglobin in the red blood cells. The myoglobin stores some of the needed O 2 within the fibers themselves (and gives SO fibers their red color). All of these features allow SO fibers to produce large quantities of ATP, which can sustain muscle activity without fatiguing for long periods of time.
The fact that SO fibers can function for long periods without fatiguing makes them useful in maintaining posture, producing isometric contractions, stabilizing bones and joints, and making small movements that happen often but do not require large amounts of energy. They do not produce high tension, and thus they are not used for powerful, fast movements that require high amounts of energy and rapid cross-bridge cycling.
FO fibers are sometimes called intermediate fibers because they possess characteristics that are intermediate between fast fibers and slow fibers. They produce ATP relatively quickly, more quickly than SO fibers, and thus can produce relatively high amounts of tension. They are oxidative because they produce ATP aerobically, possess high amounts of mitochondria, and do not fatigue quickly. However, FO fibers do not possess significant myoglobin, giving them a lighter color than the red SO fibers. FO fibers are used primarily for movements, such as walking, that require more energy than postural control but less energy than an explosive movement, such as sprinting. FO fibers are useful for this type of movement because they produce more tension than SO fibers but they are more fatigue-resistant than FG fibers.
FG fibers primarily use anaerobic glycolysis as their ATP source. They have a large diameter and possess high amounts of glycogen, which is used in glycolysis to generate ATP quickly to produce high levels of tension. Because they do not primarily use aerobic metabolism, they do not possess substantial numbers of mitochondria or significant amounts of myoglobin and therefore have a white color. FG fibers are used to produce rapid, forceful contractions to make quick, powerful movements. These fibers fatigue quickly, permitting them to only be used for short periods. Most muscles possess a mixture of each fiber type. The predominant fiber type in a muscle is determined by the primary function of the muscle.
Chapter Review
ATP provides the energy for muscle contraction. The three mechanisms for ATP regeneration are creatine phosphate, anaerobic glycolysis, and aerobic metabolism. Creatine phosphate provides about the first 15 seconds of ATP at the beginning of muscle contraction. Anaerobic glycolysis produces small amounts of ATP in the absence of oxygen for a short period. Aerobic metabolism utilizes oxygen to produce much more ATP, allowing a muscle to work for longer periods. Muscle fatigue, which has many contributing factors, occurs when muscle can no longer contract. An oxygen debt is created as a result of muscle use. The three types of muscle fiber are slow oxidative (SO), fast oxidative (FO) and fast glycolytic (FG). SO fibers use aerobic metabolism to produce low power contractions over long periods and are slow to fatigue. FO fibers use aerobic metabolism to produce ATP but produce higher tension contractions than SO fibers. FG fibers use anaerobic metabolism to produce powerful, high-tension contractions but fatigue quickly.
Review Questions
Q. Muscle fatigue is caused by ________.
A. buildup of ATP and lactic acid levels
B. exhaustion of energy reserves and buildup of lactic acid levels
C. buildup of ATP and pyruvic acid levels
D. exhaustion of energy reserves and buildup of pyruvic acid levels
Answer: B
Q. A sprinter would experience muscle fatigue sooner than a marathon runner due to ________.
A. anaerobic metabolism in the muscles of the sprinter
B. anaerobic metabolism in the muscles of the marathon runner
C. aerobic metabolism in the muscles of the sprinter
D. glycolysis in the muscles of the marathon runner
Answer: A
Q. What aspect of creatine phosphate allows it to supply energy to muscles?
A. ATPase activity
B. phosphate bonds
C. carbon bonds
D. hydrogen bonds
Answer: B
Q. Drug X blocks ATP regeneration from ADP and phosphate. How will muscle cells respond to this drug?
A. by absorbing ATP from the bloodstream
B. by using ADP as an energy source
C. by using glycogen as an energy source
D. none of the above
Answer: D
Critical Thinking Questions
Q. Why do muscle cells use creatine phosphate instead of glycolysis to supply ATP for the first few seconds of muscle contraction?
A. Creatine phosphate is used because creatine phosphate and ADP are converted very quickly into ATP by creatine kinase. Glycolysis cannot generate ATP as quickly as creatine phosphate.
Q. Is aerobic respiration more or less efficient than glycolysis? Explain your answer.
A. Aerobic respiration is much more efficient than anaerobic glycolysis, yielding 36 ATP per molecule of glucose, as opposed to two ATP produced by glycolysis.
Glossary
- fast glycolytic (FG)
- muscle fiber that primarily uses anaerobic glycolysis
- fast oxidative (FO)
- intermediate muscle fiber that is between slow oxidative and fast glycolytic fibers
- slow oxidative (SO)
- muscle fiber that primarily uses aerobic respiration


<IP_ADDRESS>: The Purpose of Research Questions
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Both professional researchers and successful student researchers develop research questions. That’s because research questions are more than handy tools; they are essential to the research process.
By defining exactly what the researcher is trying to find out, these questions influence most of the rest of the steps taken to conduct the research. That’s true even if the research is not for academic purposes but for other areas of our lives.
For instance, if you’re seeking information about a health problem in order to learn whether you have anything to worry about, research questions will make it possible for you to more effectively decide whether to seek medical help–and how quickly.
Or, if you’re researching a potential employer, having developed and used research questions will mean you’re able to more confidently decide whether to apply for an internship or job there.
The confidence you’ll have when making such decisions will come from knowing that the information they’re based on was gathered by conscious thought rather than serendipity and whim.


5.7: Ribosomes
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Ribosomes are composed of rRNA and protein and consist of 2 subunits. In eukaryotic cells, the subunits have densities of 60S and 40S ("S" refers to a unit of density called the Svedberg unit) and are composed of longer rRNA molecules and more proteins than the 50S and 30S subunits found in prokaryotic ribosomes. When the two ribosomal subunits join together during translation, they form a complete ribosome having a density of 80S.
The ribosomes are both attached to the endoplasmic reticulum and free in the cytoplasm. They serve as a workbench for protein synthesis, that is, they receive and translate genetic instructions for the formation of specific proteins or polypeptides.


6: Exponential and Logarithmic Functions
In this chapter, we will explore exponential functions, which can be used for, among other things, modeling growth patterns such as those found in bacteria. We will also investigate logarithmic functions, which are closely related to exponential functions. Both types of functions have numerous real-world applications when it comes to modeling and interpreting data.
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- 6.0: Prelude to Exponential and Logarithmic Functions
- Focus in on a square centimeter of your skin. Look closer. Closer still. If you could look closely enough, you would see hundreds of thousands of microscopic organisms.  In fact, the bacterial cells in your body at any given moment outnumber your own cells. But that is no reason to feel bad about yourself. While some bacteria can cause illness, many are healthy and even essential to the body.
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- 6.1: Exponential Functions
- When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions, which model this kind of rapid growth.
-
- 6.2: Graphs of Exponential Functions
- As we discussed in the previous section, exponential functions are used for many real-world applications such as finance, forensics, computer science, and most of the life sciences. Working with an equation that describes a real-world situation gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about things by seeing their pictorial representations, and that is exactly why graphing exponential equations is a powerful tool.
-
- 6.6: Exponential and Logarithmic Equations
- Uncontrolled population growth can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.
-
- 6.8: Fitting Exponential Models to Data
- We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key features and definitions of its associated function are included for review.
Thumbnail: The functions \(y=e^x\) and \(y=\ln(x)\) are inverses of each other, so their graphs are symmetric about the line \(y=x\). (CC BY-SA; OpenStax).


14.3: Chapter 14 Homework Problems
Exercise \(\PageIndex{1}\)
A flywheel with a diameter of 2 ft and a weight of 60 lbs is rotating at a rate of 600 rpm. A brake applies a friction force to the outer rim of the flywheel, bringing it to a stop in 1.5 seconds. Based on this information, what was the average friction force applied by the brake over this time?
- Solution
-
\(F_{brake} = 39.01 \ lbs\)
Exercise \(\PageIndex{2}\)
A ring-shaped space station can be approximated as a thin ring 60 meters in diameter with a mass of 500,000 kg. Centrifugal acceleration of the spinning station will be used to simulate gravity.
- To simulate the acceleration of Earth (9.81 m/s²), how fast will the station need to be spinning?
- If two thrusters each capable of exerting 10 kN of force will be used to get the station up to this speed, how long will we need to run the thrusters?
- Solution
-
\(\omega_f = 0.571 \ \frac{rad}{s}\)
\(t_{thrust} = 428.25 \ s\)


<IP_ADDRESS>: Strange Meeting
Strange Meeting
Strange Meeting License: Public DomainIt seemed that out of the battle I escaped
Down some profound dull tunnel, long since scooped
Through granites which Titanic wars had groined.
Yet also there encumbered sleepers groaned,
Too fast in thought or death to be bestirred.
Then, as I probed them, one sprang up, and stared
With piteous recognition in fixed eyes,
Lifting distressful hands as if to bless.
And by his smile, I knew that sullen hall;
With a thousand fears that vision's face was grained;
Yet no blood reached there from the upper ground,
And no guns thumped, or down the flues made moan.
"Strange, friend," I said, "Here is no cause to mourn."
"None," said the other, "Save the undone years,
The hopelessness. Whatever hope is yours,
Was my life also; I went hunting wild
After the wildest beauty in the world,
Which lies not calm in eyes, or braided hair,
But mocks the steady running of the hour,
And if it grieves, grieves richlier than here.
For by my glee might many men have laughed,
And of my weeping something has been left,
Which must die now. I mean the truth untold,
The pity of war, the pity war distilled.
Now men will go content with what we spoiled.
Or, discontent, boil bloody, and be spilled.
They will be swift with swiftness of the tigress,
None will break ranks, though nations trek from progress.
Courage was mine, and I had mystery;
Wisdom was mine, and I had mastery;
To miss the march of this retreating world
Into vain citadels that are not walled.
Then, when much blood had clogged their chariot-wheels
I would go up and wash them from sweet wells,
Even with truths that lie too deep for taint.
I would have poured my spirit without stint
But not through wounds; not on the cess of war.
Foreheads of men have bled where no wounds were.
I am the enemy you killed, my friend.
I knew you in this dark; for so you frowned
Yesterday through me as you jabbed and killed.
I parried; but my hands were loath and cold.
Let us sleep now . . ."


2.10: Factoring ax² + bx + c when a≠1
In this section we continue to factor trinomials of the form \(ax^2 + bx + c\). In the last section, all of our examples had \(a = 1\), and we were able to “Drop in place” our circled integer pair. However, in this section, \(a \neq 1\), and we’ll soon see that we will not be able to use the “Drop in place” technique. However, readers will be pleased to learn that the \(ac\)-method will still apply.
Factor: \(2x^2 −7x−15\).
Solution
We proceed as follows:
- Compare \(2x^2 −7x−15\) with \(ax^2 + bx + c\) and identify \(a = 2\), \(b = −7\), and \(c = −15\). Note that the leading coefficient is \(a = 2\), so this case is different from all of the cases discussed in in the previous section.
- Calculate \(ac\). Note that \(ac = (2)(−15)\), so \(ac = −30\).
- List all integer pairs whose product is \(ac =−30\). \[\begin{array}{ll}{1,-30} & {-1,30} \\ {2,-15} & {-2,15} \\ {3,-10} & {-3,10} \\ {5,-6} & {-5,6}\end{array} \nonumber \]
- Identify the ordered pair whose sum is \(b =−7\).\[\begin{array}{cc}{1,-30} & {-1,30} \\ {2,-15} & {-2,15} \\ {\color {Red}3,-10} & {-3,10} \\ {5,-6} & {-5,6}\end{array} \nonumber \]
- Note that if we “drop in place” the ordered pair whose sum is -7, \((x+3)(x-10) \neq 2x^2-7x-15\). Right off the bat, the product of the terms in the “First” position does not equal \(2x^2\). Instead, we break up the middle term of \(2x^2 −7x−15\) into a sum of like terms using our pair of integers \(3\) and \(−10\). \[2x^2{\color {Red}-7x}-15 = 2x^2{\color {Red}+3x-10x}-15 \nonumber \]Now we factor by grouping. Factor \(x\) out of the first two terms and \(−5\) out of the second two terms.\[= x(2x + 3)−5(2x + 3) \nonumber \]Now we can factor out (2x + 3).\[=(x−5)(2x + 3) \nonumber \]
-
Use the FOIL shortcut to mentally check your answer. To multiply \((x−5)(2x + 3)\), use these steps:
- Multiply the terms in the “First” positions: \(2x^2\).
- Multiply the terms in the “Outer” and “Inner” positions and combine the results mentally: \(3x−10x = −7x\).
- Multiply the terms in the “Last” positions: \(−15\). That is:\[(x-5)(2x+3)=\begin{array}{ccccccc} \color {Red}F & & \color {Red}O & & \color {Red}I & & \color {Red}L \\ 2x^2&+&3x&-&10x&-&15 \end{array} \nonumber \]
Combining like terms, \((x − 5)(2x + 3) = 2x^2 − 7x − 15\), which is the original trinomial, so our solution checks. Note that if you combine the “Outer” and “Inner” products mentally, the check goes even faster.
Factor: \(3v^2 + 13v + 14\)
- Answer
-
\((v + 2)(3v + 7)\)
Speeding Things Up a Bit
Some readers might already be asking “Do I really have to list all of those ordered pairs if I already see the pair I need?” The answer is “No!” If you see the pair you need, use it to break up the middle term of the trinomial as a sum of like terms.
Factor: \(3k^2 −7k−6\).
Solution
Compare \(3k^2−7k−6\) with \(ak^2+bk+c\) and note that \(a = 3\), \(b =−7\), and \(c = −6\). Calculate \(ac = (3)(−6)\), so \(ac = −18\). Now can you think of an integer pair whose product is \(ac = −18\) and whose sum is \(b = −7\)? For some, the pair just pops into their head: \(2\) and \(−9\). Break up the middle term into a sum of like terms using the pair \(2\) and \(−9\).
\[\begin{align*} 3k^2{\color {Red}-7k}-6 &= 3k^2{\color {Red}+2k-9k}-6 \quad \color {Red} -7k=2k-9k\\ &= k(3k+2)-3(3k+2) \quad \color {Red} \text {Factor by grouping.}\\ &= (k-3)(3k+2) \quad \color {Red} \text {Factor out} (3k+2). \end{align*} \nonumber \]
Use the FOIL shortcut to check your answer.
\[(k-3)(3k+2)=\begin{array}{ccccccc} \color {Red}F & & \color {Red}O & & \color {Red}I & & \color {Red}L \\ 3k^2&+&2x&-&9x&-&6 \end{array} \nonumber \]
Combining like terms, \((k−3)(3k+2) = 3k^2−7k−6\), the original trinomial. Our solution checks.
Factor: \(2p^4 −9p^2 + 10\)
- Answer
-
\((p^2−2)(2p^2−5)\)
On the other hand, some readers might be saying “Well, the needed ordered pair is not popping into my head. Do I have a way of cutting down the work?” The answer is “Yes!” As you are listing the ordered pairs whose product equals \(ac\), be mindful that you need the ordered pair whose sum is \(b\). If you stumble across the needed pair, stop the listing process and “drop” your ordered pair in place.
Factor: \(-3z^2 +33z - 54\).
Solution
Since \(a<0\) we begin by factoring out a \(-1\). \[-3z^2 +33z - 54=-(3z^2-33z+54) \nonumber \]
Next, compare \(3z^2 −33z + 54\) with \(ax^2 + bx + c\) and note that \(a = 3\), \(b = −33\), and \(c = 54\). Calculate \(ac = (3)(54)\), so \(ac = 162\). Ouch! That’s a big number! However, start listing the integer pairs whose product is \(ac = 162\), but be mindful that you need an integer pair whose sum is \(b = −33\).
\[\begin{array}{ll}{1,162} \\ {2,81} \\ {3,54} \\ {6,27} & {\color {Red}-6,-27}\end{array} \nonumber\]
As soon as we wrote down the pair \(6\) and \(27\), our mind said “the sum of \(6\) and \(7\) is \(33\).” However, we need the sum to equal \(b = −33\), so we boxed \(−6\) and \(−27\) instead.  Now we realize we can factor \(3\) out of each term in the first factor!
\[= 3(z−9)(z−2) \nonumber \]
We missed taking out the \(\mathrm {GCF}\) ! Let’s try again, only this time let’s do what we are always supposed to do in the first step: Factor out the \(\mathrm {GCF}\).
\[3z^2 −33z + 54 = 3(z^2 −11z + 18) \nonumber \]
Comparing \(z^2 − 11z + 18\) with \(ax^2 + bx + c\), we see that \(a = 1\), \(b = −11\), and \(c = 18\). We need an integer pair whose product is \(ac = 18\) and whose sum is \(b = −11\). Note that these numbers are considerably smaller than the numbers we had to deal with when we forgot to first factor out the \(\mathrm {GCF}\). Because the numbers are smaller, the integer pair \(−9\) and \(−2\) easily comes to mind. Furthermore, because \(a = 1\), we can factor \(z^2 − 11z + 18\) by simply dropping the integer pair \(−9\) and \(−2\) in place.
\[3(z^2 −11z + 18) = 3(z−9)(z−2) \nonumber \]
It is much nicer to factor out a GFC first!
Factor: \(5h^2 −35h−40\)
- Answer
-
\(5(h−8)(h + 1)\)
In Example \(\PageIndex{3}\), we saw how much more difficult we made the problem by forgetting to first factor out the greatest common factor (\(\mathrm {GCF}\)). Let’s try not to make that mistake again.
The first step in factoring any polynomial is to factor out the greatest common factor, if there is one.
Factor: \(30x^3 −21x^2 −18x\).
Solution
Note that the \(\mathrm {GCF}\) of \(30x^3\), \(21x^2\), and \(18x\) is \(3x\). Factor out this \(\mathrm {GCF}\).
\[\begin{align*} 30x^3-21x^2-18x &= {\color{Red}3x}\cdot 10x^2-{\color{Red}3x}\cdot 7x -{\color{Red}3x}\cdot 6\\ &= {\color{Red}3x}(10x^2-7x-6) \end{align*} \nonumber \]
Next, compare \(10x^2 −7x−6\) with \(ax^2 + bx + c\) and note that \(a = 10\), \(b =−7\), and \(c = −6\). Start listing the integer pairs whose product is \(ac = −60\), but be mindful that you need an integer pair whose sum is \(b = −7\).
\[\begin{array}{ll}{1,-60} \\ {2,-30} \\ {3,-20} \\ {4,-15} \\ {\color {Red}5,-12}\end{array} \nonumber\]
Break up the middle term into a sum of like terms using our circled pair.
\[\begin{align*} 3x(10x^2-7x-6) &= 3x(10x^2+5x-12x-18) \quad \color {Red} -7x=5x-12x\\ &= 3x[5x(2x+1)-6(2x+1)] \quad \color {Red} \text {Factor by grouping.}\\ &= 3x(5x-6)(2x+1) \quad \color {Red} \text {Factor out} (2x+1). \end{align*} \nonumber\]
Hence, \(30x^3 −21x^2 −18x =3x(5x−6)(2x + 1)\).
Check: First, use the FOIL shortcut to multiply the two binomial factors, then distribute the monomial factor.
\[\begin{align*} 3x(5x-6)(2x + 1) &= 3x(10x^2-7x-6) \quad \color {Red} \text {Apply the FOIL shortcut}\\ &= 3x^3-21x^2-18x \quad \color {Red} \text {Distribute the } 3x. \end{align*} \nonumber\]
Because this is the original polynomial, the solution checks.
Factor: \(12m^4 +2m^3 −30m^2\)
- Answer
-
\(2m^2(3m+5)(2m−3)\)
Nonlinear Equations Revisited
Let’s use the factoring technique of this chapter to solve some nonlinear equations.
Solve the equation \(2x^2 = 13 x−20\) both algebraically and graphically, then compare your answers.
Solution
Because there is a power of x larger than one, the equation is nonlinear. Make one side equal to zero.
\[\begin{align*} 2x^2 &= 13x-20 \quad \color {Red} \text {Original equation}\\2x^2-13x+20 &= 0 \quad \color {Red} \text {Make one side zero.} \end{align*} \nonumber\]
Compare \(2x^2 −13x + 20\) with \(ax^2 + bx + c\) and note that \(a = 2\), \(b = −13\) and \(c = 20\). We need an integer pair whose product is \(ac = 40\) and whose sum is \(b = −13\). The integer pair \(−5\) and \(−8\) comes to mind. Write the middle term as a sum of like terms using this pair.
\[\begin{align*} 2x^2-5x-8x+20 &= 0 \quad \color {Red} -13x=-5x-8x\\x(2x-5)(2x-5) &= 0 \quad \color {Red} \text {Factor by grouping.}\\(x-4)(2x-5) &= 0 \quad \color {Red} \text {Factor out } 2x-5 \end{align*} \nonumber\]
We have a product that equals zero. Use the zero product property to complete the solution.
\[\begin{align*} x-4 &= 0 \\x & = 4 \end{align*} \nonumber\]
or
\[\begin{align*} 2x-5 &= 0 \\2x & = 5\\x & = \dfrac{5}{2} \end{align*} \nonumber\]
Thus, the solutions of \(2x^2 = 13x−20\) are \(x = 4\) and \(x =5 /2\).
Graphical solution:
Input each side of the equation \(2x^2 = 13x−20\) into the Y= menu of your graphing calculator, \(y =2x^2\) in \(\mathbb{Y1}\), \(y = 13 x−20\) in \(\mathbb{Y2}\)
(see Figure \(\PageIndex{1}\)). Select 6:ZStandard from the ZOOM menu to produce the image on the left in Figure \(\PageIndex{1}\)). However, even after adjusting the WINDOW parameters (\(\mathbb{Xmin} =−10\), \(\mathbb{Xmax} = 10\), \(\mathbb{Ymin} = −10\), and \(\mathbb{Ymax} = 60\)), the image resulting from pushing the GRAPH button (see the image on the right in Figure \(\PageIndex{1}\)) does not clearly show the two points of intersection.
Let’s switch our strategy and work with the equation \(2x2 −13x + 20 = 0 \) instead. Load \(y =2x^2 − 13x + 20\) into \(\mathbb{Y1}\) in the Y= menu , then select 6:ZStandard from the ZOOM menu to produce the image at the right in Figure \(\PageIndex{2}\).
To find the solutions of \(2x^2−13x+20 = 0\), we must identify the \(x\)-intercepts of the graph in Figure \(\PageIndex{1}\). Select 2:zero from the CALC menu, then move the left- and right=arrows to move the cursor to the left of the first \(x\)-intercept. Press ENTER to mark the “Left bound,” then move the cursor to the right of the \(x\)-intercept and press ENTER to mark the “Right bound.” Finally, press ENTER to use the current position of the cursor for your “Guess.” The result is shown in the image on the left in Figure \(\PageIndex{3}\). Repeat the process to find the rightmost \(x\)-intercept. The result is shown in the image on the right in Figure \(\PageIndex{3}\).
Note how the graphical solutions of \(2x^2−13x+20=0\), namely \(x =2.5\) and \(x = 4\), match the solutions \(x =5 /2\) and \(x = 4\) found using the algebraic method. This is solid evidence that both methods of solution are correct. However, it doesn’t hurt to check the final answers in the original equation, substituting \(5/2\) for \(x\) and \(4\) for \(x\).
\[\begin{align*}
2x^2 &= 13x-20 \\
2\left (\dfrac{5}{2} \right )^2 & = 13\left (\dfrac{5}{2} \right ) -20\\
2\left (\dfrac{25}{4} \right ) & = 13\left (\dfrac{5}{2} \right ) -20\\
\dfrac{25}{2} & = \dfrac{65}{2} - \dfrac{40}{2}
\end{align*} \nonumber \]
and
\[\begin{align*} 2x^2 &= 13x-20 \\ 2(4)^2 & = 13(4) -20\\ 2(16) & = 13(4) -20\\ 32 & = 52 - 20 \end{align*} \nonumber\]
Because the last two statements are true statements, the solutions \(x =5 /2\) and \(x = 4\) check in the original equation \(2x^2 = 13x−20\).
Solve the equation \(5x^2 = 12x + 9\) both algebraically and graphically, then compare your answers.
- Answer
-
\(-3/5\), \(3\)
Solve the equation \(2x^3 + x^2 = 28x\) both algebraically and graphically, then compare your answers.
Solution
Because there is a power of x larger than one, the equation is nonlinear. Make one side equal to zero.
\[\begin{align*} 2x^3+x^2 &= 28x \quad \color {Red} \text {Original equation}\\2x^3+x^2-28x &= 0 \quad \color {Red} \text {Make one side zero.} \end{align*} \nonumber\]
Note that the \(\mathrm {GCF}\) of \(2x^3\), \(x^2\), and \(28x\) is \(x\). Factor out \(x\).
\[x(2x^2 + x−28) = 0 \quad \color {Red} \text {Factor out the GCF} \nonumber \]
Compare \(2x^2+x−28\) with \(ax^2+bx+c\) and note that \(a = 2\), \(b = 1\) and \(c = −28\). We need an integer pair whose product is \(ac = −56\) and whose sum is \(b = 1\). The integer pair \(−7\) and \(8\) comes to mind. Write the middle term as a sum of like terms using this pair.
\[\begin{align*} x(2x^2-7x+8x-28) &= 0 \quad \color {Red} x=-7x+8x\\x[x(2x-7)+4(2x-7)] &= 0 \quad \color {Red} \text {Factor by grouping.}\\x(x+4)(2x-7) &= 0 \quad \color {Red} \text {Factor out } 2x-7. \end{align*} \nonumber\]
We have a product of three factors that equals zero. By the zero product property, at least one of the factors must equal zero.
\[x=0 \nonumber \]
or
\[\begin{align*} x+4 &= 0 \\x & = -4 \end{align*} \nonumber\]
or
\[\begin{align*} 2x-7 & = 0 \\ 2x & = 7\\ x & = \dfrac{7}{2} \end{align*} \nonumber\]
Thus, the solutions of \(2x^3 + x^2 = 28x\) are \(x = 0\), \(x =−4\), and \(x =7 /2\).
Graphical solution:
Rather than working with \(2x^3+x^2 = 28x\), graphing each side separately and finding where the graphs intersect, we will work instead with \(2x^3 + x^2 − 28x = 0\), locating where the graph of \(y =2x^3 + x^2 − 28x\) crosses the \(x\)-axis. Load \(y =2x^3 + x^2 −28x\) into \(\mathbb{Y1}\) in the Y= menu , then select 6:ZStandard from the ZOOM menu to produce the image at the right in Figure \(\PageIndex{4}\).
In the image at the right in Figure \(\PageIndex{4}\), we saw the graph rise from the bottom of the screen, leave the top of the screen, return and leave via the bottom of the screen, and then finally return and leave via the top of the screen. Clearly, there are at least two turning points to the graph that are not visible in the current viewing window. Set the WINDOW settings as shown in the image on the left in Figure \(\PageIndex{5}\), then push the GRAPH button to produce the image on the right in Figure \(\PageIndex{5}\). Note that this window now shows the \(x\)-intercepts as well as the turning points of the graph of the polynomial.
To find the solutions of \(2x^3 + x2 −28x = 0\), we must identify the \(x\)-intercepts of the graph in Figure \(\PageIndex{5}\). Select 2:zero from the CALC menu, then use the left- and right-arrow keys to move the cursor to the left of the first \(x\)-intercept. Press ENTER to mark the “Left bound,” then move the cursor to the right of the \(x\)-intercept and press ENTER to mark the “Right bound.” Finally, press ENTER to use the current position of the cursor for your “Guess.” The result is shown in the first image on the left in Figure \(\PageIndex{6}\). Repeat the process to find the remaining \(x\)-intercepts. The results are shown in the next two images in Figure \(\PageIndex{6}\).
Note how the graphical solutions of \(2x^3+x^2−28x = 0\), namely \(x =−4\), \(x = 0\), and \(x =3.5\), match the solutions \(x = −4\), \(x = 0\), and \(x =7 /2\) found using the algebraic method.
Solve the equation \(4x^3 = −x^2 + 14x\) both algebraically and graphically, then compare your answers.
- Answer
-
\(−2\), \(0\), \(7/4\)


5.1: Carnot Cycle
The main contribution of Carnot to thermodynamics is his abstraction of the steam engine’s essential features into a more general and idealized heat engine. The definition of Carnot’s idealized cycle is as follows:
A Carnot cycle is an idealized process composed of two isothermal and two adiabatic transformations. Each transformation is either an expansion or a compression of an ideal gas. All transformations are assumed to be reversible, and no energy is lost to mechanical friction.
A Carnot cycle connects two “heat reservoirs” at temperatures \(T_h\) (hot) and \(T_l\) (low), respectively. The reservoirs have a large thermal capacity so that their temperatures are unaffected by the cycle. The system is composed exclusively by the ideal gas, which is the only substance that changes temperature throughout the cycle. If we report the four transformations of a Carnot cycle on a \(PV\) diagram, we obtain the following plot:
Stage 1: isothermal expansion \(A \rightarrow B\)
At this stage heat is released from the hot reservoir and is absorbed by the ideal gas particles within the system. Thus, the temperature of the system rises. The high temperature causes the gas particles to expand; pushing the piston upwards and doing work on the surroundings.
Starting the analysis of the cycle from point \(A\) in Figure \(\PageIndex{1}\),\(^1\) the first transformation we encounter is an isothermal expansion at \(T_h\). Since the transformation is isothermal:
\[ \Delta U_1 = \overbrace{W_1}^{<0} + \overbrace{Q_1}^{>0} = 0 \Rightarrow Q_1 = -W_1, \nonumber \]
and heat and work can be calculated for this stage using Equation 2.4.14 :
\[\begin{equation} \begin{aligned} Q_1 & = \left| Q_h \right| = nRT_h \overbrace{\ln \dfrac{V_B}{V_A}}^{>0 \text{ since } V_B>V_A} > 0, \\ W_1 & = -Q_1 = - nRT_h \ln \dfrac{V_B}{V_A} < 0, \end{aligned} \end{equation} \nonumber \]
where we denoted \(\left| Q_h \right|\) the absolute value of the heat that gets into the system from the hot reservoir.
Stage 2: adiabatic expansion \(B \rightarrow C\)
At this stage expansion continues, however there is no heat exchange between system and surroundings. Thus, the system is undergoing adiabatic expansion. The expansion allows the ideal gas particles to cool, decreasing the temperature of the system.
The second transformation is an adiabatic expansion between \(T_h\) and \(T_l\). Since we are at adiabatic conditions:
\[ Q_2 = 0 \Rightarrow \Delta U_2 = W_2, \nonumber \]
and the negative energy (expansion work) can be calculated using:
\[ \Delta U_2 = W_2 = n \underbrace{\int_{T_h}^{T_l} C_V dT}_{<0 \text{ since } T_\mathrm{l}<T_\mathrm{h}} < 0. \nonumber \]
Stage 3: isothermal compression \(C \rightarrow D\)
At this stage the surroundings do work on the system which causes heat to be released (qc). The temperature within the system remains the same. Thus, isothermal compression occurs.
The third transformation is an isothermal compression at \(T_l\). The formulas are the same as those used for stage 1, but they will results in heat and work with reversed signs (since this is a compression):
\[ \Delta U_3 = \overbrace{W_3}^{>0} + \overbrace{Q_3}^{<0} = 0 \Rightarrow Q_3 = -W_3, \nonumber \]
and:
\[\begin{equation} \begin{aligned} Q_3 & = \left| Q_l \right| = nRT_l \overbrace{\ln \dfrac{V_D}{V_C}}^{<0 \text{ since } V_D<V_C} < 0 , \\ W_3 & = -Q_3 = - nRT_l \ln \dfrac{V_D}{V_C} > 0, \end{aligned} \end{equation} \nonumber \]
where \(\left| Q_l \right|\) is the absolute value of the heat that gets out of the system to the cold reservoir (\(\left| Q_l \right|\) being the heat entering the system).
Stage 4: adiabatic compression \(D \rightarrow A\)
No heat exchange occurs at this stage, however, the surroundings continue to do work on the system. Adiabatic compression occurs which raises the temperature of the system as well as the location of the piston back to its original state (prior to stage one).
The fourth and final transformation is an adiabatic comprssion that restores the system to point \(A\), bringing it from \(T_l\) to \(T_h\). Similarly to stage 3:
\[ Q_4 = 0 \Rightarrow \Delta U_4 = W_4, \nonumber \]
Since we are at adiabatic conditions. The energy associated with this process is now positive (compression work), and can be calculated using:
\[ \Delta U_4 = W_4 = n \underbrace{\int_{T_l}^{T_h} C_V dT}_{>0 \text{ since } T_\mathrm{h}>T_\mathrm{l}} > 0. \nonumber \]
Notice how \(\Delta U_4 = - \Delta U_2\) because \(\int_x^y=-\int_y^x\).
- The stages of a Carnot depicted at the beginning of each of this section and the following three ones are genetaken from Wikipedia , and have been generated and distributed by Author BlyumJ under CC-BY-SA license.


6.6: Proving the contrapositive
Recall. Modus tollens : \(P \rightarrow Q \Leftrightarrow \neg Q \rightarrow \neg P\text{.}\)
To prove \(P \Rightarrow Q\text{,}\) you can instead prove \(\neg Q \Rightarrow \neg P\text{.}\)
In Worked Example 6.3.1 , we proved that the square of an even number is also even. Therefore, this also constitutes a proof of the contrapositive statement: if the square of a number is odd, then that number is also odd.
Prove that every prime number larger than \(2\) is odd.
Solution
We want to prove the following universally quantified conditional (“for all \(p\)” omitted, domain is positive integers).
| conditional | if (\(p\) is prime and \(p>2\)) then \(p\) is odd. |
| contrapositive | if \(p\) is not odd, then not (\(p\) is prime and \(p>2\)) |
| DeMorgan Subsitution | if \(p\) is not odd, then (\(p\) is not prime or \(p \le 2\)) |
These are all equivalent.
Let's prove the last statement: as in the procedure for proving conditionals with a disjunction, start by assuming that \(p\) is not odd and \(p \gt 2\text{.}\) We must then show that \(p\) is not prime. Since \(p\) is not odd, it is divisible by \(2\text{.}\) But since \(p \gt 2\text{,}\) \(p\) is divisible by a number other than \(1\) and \(p\) itself. Therefore, \(p\) is not prime.
Check your understanding. Attempt Exercise 6.12.8 .


11.8.3: Sanitary Landfills and Combustion
Much of the trash we produce ends up in landfills, which are essentially encapsulated and buried under layers of dirt. In order to protect groundwater and surface water from potentially toxic leachate - liquids that accrue from the decomposition of waste - the sanitary landfill needs to be lined with plastic and have adequate drainage. The leachate may be collected and treated before it too is stored. Additionally, decomposition of waste produces gasses and heat - which can cause emissions or even catch fire if not properly vented. Sometimes the methane produced by a landfill can be collected, treated, and used for energy. After waste is dumped into the landfill, it needs to be covered with soil in order to prevent pests (rodents, birds, or insects) from getting into it or the wind from blowing it out of the landfill (Frumkin, 2016). It may seem obvious that landfills are limited by capacity and location: no one wants to live near a landfill, and what should be done once the landfill is full?
Combustion of waste is another option. Although open incineration of trash was often practiced in backyards up until the mid 20th century, this practice creates high levels of pollution - particulate matter and GHGs - that pose a risk to the community and planet. The Clean Air Act of 1970 essentially put an end to this practice in the U.S., although it is still common in other developing countries. Incinerators in the U.S. can be used to produce energy and get rid of waste, but they must have expensive technology to reduce air pollution. The resulting ash can also become problematic to dispose of (Frumkin, 2016, Seabert et al., 2021).


1.1: Microsoft Word Window and Menu Environment
Microsoft Word is a popular word processing program that is part of the Microsoft Office Suite. Many businesses use it to create, edit, format, and organize documents.
How to Open a Microsoft Word Document
Watch this introductory video to learn how to open a new document, insert text, name your document, use spell check, center your text, change your font, and use the basic navigation tools in Microsoft Word.
Microsoft Word Layout Basics
Watch this video to explore the Microsoft Word interface, such as the title bar, the task bar, ribbon display options, and the quick access toolbar. The presenter also shows how to maximize and minmize the windows in your screen space.
Microsoft Word's Quick Access Toolbar
Watch this video to learn how to use Microsoft Word's Quick Access toolbar to easily perform common commands.


Skip to main content
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12.1: Phase Stability
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We have already encountered the gas, liquid, and solid phases and already discussed some of their properties. These terms are intuitive since these are the three most common states of matter.
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12.2: Gibbs Phase Rule
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In chapter 1, we have already seen that the number of independent variables required to describe an ideal gas is two. This number was derived by counting the total number of variables (3:P,V¯,T), and reduce it by one because the ideal gas law constrains the value of one of them, once the other two are fixed.
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12.3: PT Phase Diagrams
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Let’s now discuss the pressure–temperature diagram of a typical substance, as reported in Figure 12.3.1. Each of the lines reported in the diagram represents an equilibrium between two phases, and therefore it represents a condition that reduces the number of degrees of freedom to one.
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12.4: The Clausius-Clapeyron Equation
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Let’s now take a closer look at the equilibrium between a condensed phase and the gas phase.


3.5: Electric Potential
By the end of this section, you will be able to:
- Define electric potential, voltage, and potential difference.
- Calculate electric potential and potential difference from potential energy.
- Define the electron-volt.
- Describe systems in which the electron-volt is a useful unit.
- Apply conservation of energy to systems with electric charges.
Recall that earlier we defined electric field to be a quantity independent of the test charge in a given system, which would nonetheless allow us to calculate the force that would result on an arbitrary test charge. (The default assumption in the absence of other information is that the test charge is positive.) We briefly defined a field for gravity, but gravity is always attractive, whereas the electric force can be either attractive or repulsive. Therefore, although potential energy is perfectly adequate in a gravitational system, it is convenient to define a quantity that allows us to calculate the work on a charge independent of the magnitude of the charge. Calculating the work directly may be difficult, since \(W = \vec{F} \cdot \vec{d}\) and the direction and magnitude of \(\vec{F}\) can be complex for multiple charges, for odd-shaped objects, and along arbitrary paths. But we do know that because \(\vec{F}\), the work, and hence \(\Delta U\) is proportional to the test charge \(q\). To have a physical quantity that is independent of test charge, we define electric potential \(V\) (or simply potential, since electric is understood) to be the potential energy per unit charge:
The electric potential energy per unit charge is
\[V = \dfrac{U}{q}. \label{eq-1}\]
Since the potential energy \(U\) is proportional to the test charge \(q\), the dependence on \(q\) cancels. Thus, \(V\) does not depend on \(q\). The change in potential energy \(\Delta U\) is crucial, so we are concerned with the difference in potential or potential difference \(\Delta V\) between two points, where
\[ \Delta V = V_B - V_A = \dfrac{\Delta U}{q} \]
The electric potential difference between points \(A\) and \(B\), \(V_B - V_A\) is defined to be the change in potential energy of a charge \(q\) moved from \(A\) to \(B\), divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta .
\[1 \, \mathrm{V} = 1 \, \mathrm{J/C} \label{eq0}\]
The familiar term voltage is the common name for electric potential difference. Keep in mind that whenever a voltage is quoted, it is understood to be the potential difference between two points. For example, every battery has two terminals, and its voltage is the potential difference between them. More fundamentally, the point you choose to be zero volts is arbitrary. This is analogous to the fact that gravitational potential energy has an arbitrary zero, such as sea level or perhaps a lecture hall floor. It is worthwhile to emphasize the distinction between potential difference and electrical potential energy.
The relationship between potential difference (or voltage) and electrical potential energy is given by
\[\Delta V = \dfrac{\Delta U}{q} \label{eq1}\]
or
\[ \Delta U = q \Delta V. \label{eq2}\]
Voltage is not the same as energy. Voltage is the energy per unit charge. Thus, a motorcycle battery and a car battery can both have the same voltage (more precisely, the same potential difference between battery terminals), yet one stores much more energy than the other because \(\Delta U = q\Delta V\). The car battery can move more charge than the motorcycle battery, although both are 12-V batteries.
You have a 12.0-V motorcycle battery that can move 5000 C of charge, and a 12.0-V car battery that can move 60,000 C of charge. How much energy does each deliver? (Assume that the numerical value of each charge is accurate to three significant figures.)
Strategy
To say we have a 12.0-V battery means that its terminals have a 12.0-V potential difference. When such a battery moves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to \(\Delta U = q\Delta V\). To find the energy output, we multiply the charge moved by the potential difference.
Solution
For the motorcycle battery, \(q = 5000 \, \mathrm{C}\) and \(\Delta V = 12.0 \, \mathrm{V}\). The total energy delivered by the motorcycle battery is
\[\Delta U_{cycle} = (5000 \, \mathrm{C})(12.0 \, \mathrm{V}) = (5000 \, \mathrm{C})(12.0 \, \mathrm{J/C}) = 6.00 \times 10^4 \, \mathrm{J}. \nonumber\]
Similarly, for the car battery, \(q = 60,000 \, \mathrm{C}\) and
\[\Delta U_{car} = (60,000 \, \mathrm{C})(12.0 \, \mathrm{V}) = 7.20 \times 10^5 \, \mathrm{J}. \nonumber\]
Significance
Voltage and energy are related, but they are not the same thing. The voltages of the batteries are identical, but the energy supplied by each is quite different. A car battery has a much larger engine to start than a motorcycle. Note also that as a battery is discharged, some of its energy is used internally and its terminal voltage drops, such as when headlights dim because of a depleted car battery. The energy supplied by the battery is still calculated as in this example, but not all of the energy is available for external use.
How much energy does a 1.5-V AAA battery have that can move 100 C?
- Answer
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\(\Delta U = q\Delta V = (100 \, \mathrm{C})(1.5 \, \mathrm{V}) = 150 \, \mathrm{J}\)
Note that the energies calculated in the previous example are absolute values. The change in potential energy for the battery is negative, since it loses energy. These batteries, like many electrical systems, actually move negative charge—electrons in particular. The batteries repel electrons from their negative terminals (\(A\)) through whatever circuitry is involved and attract them to their positive terminals (\(B\)), as shown in Figure \(\PageIndex{1}\). The change in potential is \(\Delta V = V_B - V_A = +12 \, V\) and the charge \(q\) is negative, so that \(\Delta U = q \Delta V\) is negative, meaning the potential energy of the battery has decreased when \(q\) has moved from \(A\) to \(B\).
When a 12.0-V car battery powers a single 30.0-W headlight, how many electrons pass through it each second?
Strategy
To find the number of electrons, we must first find the charge that moves in 1.00 s. The charge moved is related to voltage and energy through the equations \(\Delta U = q \Delta V\). A 30.0-W lamp uses 30.0 joules per second. Since the battery loses energy, we have \(\Delta U = - 30 \, \mathrm{J}\) and, since the electrons are going from the negative terminal to the positive, we see that \(\Delta V = +12.0 \, \mathrm{V}\).
Solution
To find the charge \(q\) moved, we solve the equation \(\Delta U = q\Delta V\):
\[q = \dfrac{\Delta U}{\Delta V}.\]
Entering the values for \(\Delta U\) and \(\Delta V\), we get
\[q = \dfrac{-30.0 \, \mathrm{J}}{+12.0 \, \mathrm{V}} = \dfrac{-30.0 \, \mathrm{J}}{+12.0 \, \mathrm{J/C}} = -2.50 \, \mathrm{C}.\]
The number of electrons \(n_e\) is the total charge divided by the charge per electron. That is,
\[n_e = \dfrac{-2.50 \, \mathrm{C}}{-1.60 \times 10^{-19} \mathrm{C}/e^-} = 1.56 \times 10^{19} \,\mathrm{electrons}.\]
Significance
This is a very large number. It is no wonder that we do not ordinarily observe individual electrons with so many being present in ordinary systems. In fact, electricity had been in use for many decades before it was determined that the moving charges in many circumstances were negative. Positive charge moving in the opposite direction of negative charge often produces identical effects; this makes it difficult to determine which is moving or whether both are moving.
How many electrons would go through a 24.0-W lamp?
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\(-2.00 \, \mathrm{C}, \, n_e = 1.25 \times 10^{19} \,\mathrm{electrons}\)
The Electron-Volt
The energy per electron is very small in macroscopic situations like that in the previous example—a tiny fraction of a joule. But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. For example, even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue. The particle may do its damage by direct collision, or it may create harmful X-rays, which can also inflict damage. It is useful to have an energy unit related to submicroscopic effects.
Figure \(\PageIndex{2}\) shows a situation related to the definition of such an energy unit. An electron is accelerated between two charged metal plates, as it might be in an old-model television tube or oscilloscope. The electron gains kinetic energy that is later converted into another form—light in the television tube, for example. (Note that in terms of energy, “downhill” for the electron is “uphill” for a positive charge.) Since energy is related to voltage by \(\Delta U = q\Delta V\), we can think of the joule as a coulomb-volt.
On the submicroscopic scale, it is more convenient to define an energy unit called the electron-volt (eV), which is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,
\[1 \, \mathrm{eV} = (1.60 \times 10^{-19} \mathrm{C})(1 \, \mathrm{V}) = (1.60 \times 10^{-19} \mathrm{C})(1 \, \mathrm{J/C}) = 1.60 \times 10^{-19} \, \mathrm{J}.\]
An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. It follows that an electron accelerated through 50 V gains 50 eV. A potential difference of 100,000 V (100 kV) gives an electron an energy of 100,000 eV (100 keV), and so on. Similarly, an ion with a double positive charge accelerated through 100 V gains 200 eV of energy. These simple relationships between accelerating voltage and particle charges make the electron-volt a simple and convenient energy unit in such circumstances.
The electron-volt is commonly employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies are among the quantities often expressed in electron-volts. For example, about 5 eV of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of 30 kV, it acquires an energy of 30 keV (30,000 eV) and can break up as many as 6000 of these molecules \((30,000 \,\mathrm{eV} \div 5 \, \mathrm{eV \, per \, molecule} = 6000 \, \mathrm{molecules})\). Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and can thus produce significant biological damage.
Conservation of Energy
The total energy of a system is conserved if there is no net addition (or subtraction) due to work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.
Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, \(K + U = \mathrm{constant}\). A loss of \(U\) for a charged particle becomes an increase in its \(K\). Conservation of energy is stated in equation form as
\[K + U = \mathrm{constant} \] or \[K_i + U_i = K_f + U_f\]
where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.
Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value is accurate to three significant figures.)
Strategy
We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (we will check on this assumption later), all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms of energy to be
\(K_i = 0\), \(K_f = \frac{1}{2}mv^2\), \(U_i = qV\), \(U_f = 0\).
Solution
Conservation of energy states that
\[K_i + U_i = K_f + U_f.\]
Entering the forms identified above, we obtain
\[qV = \dfrac{mv^2}{2}.\]
We solve this for \(v\):
\[v = \sqrt{\dfrac{2qV}{m}}.\]
Entering values for \(q\), \(V\), and \(m\) gives
\[v = \sqrt{\dfrac{2(-1.60 \times 10^{-19}\mathrm{C})(-100 \, \mathrm{J/C})}{9.11 \times 10^{-31} \mathrm{kg}}} = 5.93 \times 10^6 \, \mathrm{m/s}.\]
Significance
Note that both the charge and the initial voltage are negative, as in Figure \(\PageIndex{2}\). From the discussion of electric charge and electric field, we know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. These higher voltages produce electron speeds so great that effects from special relativity must be taken into account and will be discussed elsewhere. That is why we consider a low voltage (accurately) in this example.
How would this example change with a positron? A positron is identical to an electron except the charge is positive.
- Answer
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It would be going in the opposite direction, with no effect on the calculations as presented.
Before presenting problems involving electrostatics, we suggest a problem-solving strategy to follow for this topic.
- Examine the situation to determine if static electricity is involved; this may concern separated stationary charges, the forces among them, and the electric fields they create.
- Identify the system of interest. This includes noting the number, locations, and types of charges involved.
- Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. Determine whether the electric force is to be considered directly—if so, it may be useful to draw a free-body diagram, using electric field lines.
- Make a list of what is given or can be inferred from the problem as stated (identify the knowns). For example, it is important to distinguish the electric force \(\vec{F}\) from the electric field \(\vec{E}\) or the electric potential energy \(U\) from the electric potential \(V\).
- Solve the appropriate equation for the quantity to be determined (the unknown) or draw the field lines as requested.
- Examine the answer to see if it is reasonable: Does it make sense? Are units correct and the numbers involved reasonable?


45.3: Environmental Limits to Population Growth
Skills to Develop
- Explain the characteristics of and differences between exponential and logistic growth patterns
- Give examples of exponential and logistic growth in natural populations
- Describe how natural selection and environmental adaptation led to the evolution of particular life history patterns
Although life histories describe the way many characteristics of a population (such as their age structure) change over time in a general way, population ecologists make use of a variety of methods to model population dynamics mathematically. These more precise models can then be used to accurately describe changes occurring in a population and better predict future changes. Certain models that have been accepted for decades are now being modified or even abandoned due to their lack of predictive ability, and scholars strive to create effective new models.
Exponential Growth
Charles Darwin, in his theory of natural selection, was greatly influenced by the English clergyman Thomas Malthus. Malthus published a book in 1798 stating that populations with unlimited natural resources grow very rapidly, and then population growth decreases as resources become depleted. This accelerating pattern of increasing population size is called exponential growth .
The best example of exponential growth is seen in bacteria. Bacteria are prokaryotes that reproduce by prokaryotic fission. This division takes about an hour for many bacterial species. If 1000 bacteria are placed in a large flask with an unlimited supply of nutrients (so the nutrients will not become depleted), after an hour, there is one round of division and each organism divides, resulting in 2000 organisms—an increase of 1000. In another hour, each of the 2000 organisms will double, producing 4000, an increase of 2000 organisms. After the third hour, there should be 8000 bacteria in the flask, an increase of 4000 organisms. The important concept of exponential growth is that the population growth rate —the number of organisms added in each reproductive generation—is accelerating; that is, it is increasing at a greater and greater rate. After 1 day and 24 of these cycles, the population would have increased from 1000 to more than 16 billion. When the population size, N , is plotted over time, a J-shaped growth curve is produced (Figure \(\PageIndex{1}\)).
The bacteria example is not representative of the real world where resources are limited. Furthermore, some bacteria will die during the experiment and thus not reproduce, lowering the growth rate. Therefore, when calculating the growth rate of a population, the death rate ( D ) (number organisms that die during a particular time interval) is subtracted from the birth rate ( B ) (number organisms that are born during that interval). This is shown in the following formula:
The birth rate is usually expressed on a per capita (for each individual) basis. Thus, B (birth rate) = bN (the per capita birth rate “ b ” multiplied by the number of individuals “ N ”) and D (death rate) = dN (the per capita death rate “d” multiplied by the number of individuals “ N ”). Additionally, ecologists are interested in the population at a particular point in time, an infinitely small time interval. For this reason, the terminology of differential calculus is used to obtain the “instantaneous” growth rate, replacing the change in number and time with an instant-specific measurement of number and time.
Notice that the “ d ” associated with the first term refers to the derivative (as the term is used in calculus) and is different from the death rate, also called “\(d\).” The difference between birth and death rates is further simplified by substituting the term “ r ” (intrinsic rate of increase) for the relationship between birth and death rates:
The value “\(r\)” can be positive, meaning the population is increasing in size; or negative, meaning the population is decreasing in size; or zero, where the population’s size is unchanging, a condition known as zero population growth . A further refinement of the formula recognizes that different species have inherent differences in their intrinsic rate of increase (often thought of as the potential for reproduction), even under ideal conditions. Obviously, a bacterium can reproduce more rapidly and have a higher intrinsic rate of growth than a human. The maximal growth rate for a species is its biotic potential , or \(r_{max}\), thus changing the equation to:
Logistic Growth
Exponential growth is possible only when infinite natural resources are available; this is not the case in the real world. Charles Darwin recognized this fact in his description of the “struggle for existence,” which states that individuals will compete (with members of their own or other species) for limited resources. The successful ones will survive to pass on their own characteristics and traits (which we know now are transferred by genes) to the next generation at a greater rate (natural selection). To model the reality of limited resources, population ecologists developed the logistic growth model.
Carrying Capacity and the Logistic Model
In the real world, with its limited resources, exponential growth cannot continue indefinitely. Exponential growth may occur in environments where there are few individuals and plentiful resources, but when the number of individuals gets large enough, resources will be depleted, slowing the growth rate. Eventually, the growth rate will plateau or level off (Figure \(\PageIndex{1}\)). This population size, which represents the maximum population size that a particular environment can support, is called the carrying capacity , or \(K\).
The formula we use to calculate logistic growth adds the carrying capacity as a moderating force in the growth rate. The expression “\(K – N\)” is indicative of how many individuals may be added to a population at a given stage, and “\(K – N\)” divided by “\(K\)” is the fraction of the carrying capacity available for further growth. Thus, the exponential growth model is restricted by this factor to generate the logistic growth equation:
&= r_\text{max} \frac{d N} {d T} \nonumber \\[5pt]
&= r_\text{max}N \frac{(K-N)} {K}\nonumber
\end{align} \nonumber\]
Notice that when \(N\) is very small, \((K-N)/K\) becomes close to \(K/K\) or \(1\), and the right side of the equation reduces to \(r_{max}N\), which means the population is growing exponentially and is not influenced by carrying capacity. On the other hand, when \(N\) is large, \((K-N)/K\) come close to zero, which means that population growth will be slowed greatly or even stopped. Thus, population growth is greatly slowed in large populations by the carrying capacity \(K\). This model also allows for the population of a negative population growth, or a population decline. This occurs when the number of individuals in the population exceeds the carrying capacity (because the value of \((K-N)/K\) is negative).
A graph of this equation yields an S-shaped curve (Figure \(\PageIndex{1}\)), and it is a more realistic model of population growth than exponential growth. There are three different sections to an S-shaped curve. Initially, growth is exponential because there are few individuals and ample resources available. Then, as resources begin to become limited, the growth rate decreases. Finally, growth levels off at the carrying capacity of the environment, with little change in population size over time.
Role of Intraspecific Competition
The logistic model assumes that every individual within a population will have equal access to resources and, thus, an equal chance for survival. For plants, the amount of water, sunlight, nutrients, and the space to grow are the important resources, whereas in animals, important resources include food, water, shelter, nesting space, and mates.
In the real world, phenotypic variation among individuals within a population means that some individuals will be better adapted to their environment than others. The resulting competition between population members of the same species for resources is termed intraspecific competition (intra- = “within”; -specific = “species”). Intraspecific competition for resources may not affect populations that are well below their carrying capacity—resources are plentiful and all individuals can obtain what they need. However, as population size increases, this competition intensifies. In addition, the accumulation of waste products can reduce an environment’s carrying capacity.
Examples of Logistic Growth
Yeast, a microscopic fungus used to make bread and alcoholic beverages, exhibits the classical S-shaped curve when grown in a test tube (Figure \(\PageIndex{2}\)a). Its growth levels off as the population depletes the nutrients that are necessary for its growth. In the real world, however, there are variations to this idealized curve. Examples in wild populations include sheep and harbor seals (Figure \(\PageIndex{2}\)b). In both examples, the population size exceeds the carrying capacity for short periods of time and then falls below the carrying capacity afterwards. This fluctuation in population size continues to occur as the population oscillates around its carrying capacity. Still, even with this oscillation, the logistic model is confirmed.
Art Connection
If the major food source of the seals declines due to pollution or overfishing, which of the following would likely occur?
- The carrying capacity of seals would decrease, as would the seal population.
- The carrying capacity of seals would decrease, but the seal population would remain the same.
- The number of seal deaths would increase but the number of births would also increase, so the population size would remain the same.
- The carrying capacity of seals would remain the same, but the population of seals would decrease.
Summary
Populations with unlimited resources grow exponentially, with an accelerating growth rate. When resources become limiting, populations follow a logistic growth curve. The population of a species will level off at the carrying capacity of its environment.
Art Connections
Figure \(\PageIndex{2}\) b If the major food source of the seals declines due to pollution or overfishing, which of the following would likely occur?
- The carrying capacity of seals would decrease, as would the seal population.
- The carrying capacity of seals would decrease, but the seal population would remain the same.
- The number of seal deaths would increase but the number of births would also increase, so the population size would remain the same.
- The carrying capacity of seals would remain the same, but the population of seals would decrease.
- Answer
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A
Glossary
- biotic potential (r max )
- maximal potential growth rate of a species
- birth rate ( B )
- number of births within a population at a specific point in time
- carrying capacity ( K )
- number of individuals of a species that can be supported by the limited resources of a habitat
- death rate ( D )
- number of deaths within a population at a specific point in time
- exponential growth
- accelerating growth pattern seen in species under conditions where resources are not limiting
- intraspecific competition
- competition between members of the same species
- J-shaped growth curve
- shape of an exponential growth curve
- logistic growth
- leveling off of exponential growth due to limiting resources
- population growth rate
- number of organisms added in each reproductive generation
- S-shaped growth curve
- shape of a logistic growth curve
- zero population growth
- steady population size where birth rates and death rates are equal