4.1: Exponential Functions
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Learning Objectives
- Evaluate exponential functions.
- Find the equation of an exponential function.
- Use compound interest formulas.
- Evaluate exponential functions with base \(e\).
India is the second most populous country in the world with a population of about \(1.25\) billion people in 2013. The population is growing at a rate of about \(.2\%\) each year. If this rate continues, the population of India will exceed China’s population by the year 2031. 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.
Identifying Exponential Functions
When exploring linear growth, we observed a constant rate of change—a constant number by which the output increased for each unit increase in input. For example, in the equation \(f(x)=3x+4\), the slope tells us the output increases by \(3\) each time the input increases by \(1\). The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people.
Defining an Exponential Function
A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, \(2.5\%\) of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up \(10\%\) of the U.S. population in 2015, \(40\%\) in 2019, and \(80\%\) in 2050.
What exactly does it mean to grow exponentially ? What does the word double have in common with percent increase ? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media.
- Percent change refers to a change based on a percent of the original amount.
- Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time.
- Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time.
For us to gain a clear understanding of exponential growth , let us contrast exponential growth with linear growth . We will construct two functions. The first function is exponential. We will start with an input of \(0\), and increase each input by \(1\). We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of \(0\), and increase each input by \(1\). We will add \(2\) to the corresponding consecutive outputs (Table \(\PageIndex{1}\)).
From Table \(\PageIndex{1}\) we can infer that for these two functions, exponential growth dwarfs linear growth.
- Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain.
- Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain.
| \(x\) | \(f(x)=2^x\) | \(g(x)=2x\) |
|---|---|---|
| 0 | 1 | 0 |
| 1 | 2 | 2 |
| 2 | 4 | 4 |
| 3 | 8 | 6 |
| 4 | 16 | 8 |
| 5 | 32 | 10 |
| 6 | 64 | 12 |
Apparently, the difference between “the same percentage” and “the same amount” is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding \(2\) to the output whenever the input was increased by one.
The general form of the exponential function is \(f(x)=ab^x\), where \(a\) is any nonzero number, \(b\) is a positive real number not equal to \(1\).
- If \(b>1\),the function grows at a rate proportional to its size.
- If \(0<b<1\), the function decays at a rate proportional to its size.
Let’s look at the function \(f(x)=2^x\) from our example. We will create a table (Table \(\PageIndex{2}\)) to determine the corresponding outputs over an interval in the domain from \(−3\) to \(3\).
| \(x\) | \(−3\) | \(−2\) | \(−1\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|---|---|
| \(f(x)=2^x\) | \(2^{−3}=\dfrac{1}{8}\) | \(2^{−2}=\dfrac{1}{4}\) | \(2^{−1}=\dfrac{1}{2}\) | \(2^0=1\) | \(2^1=2\) | \(2^2=4\) | \(2^3=8\) |
Let us examine the graph of \(f\) by plotting the ordered pairs from Table \(\PageIndex{2}\) and then make a few observations \(\PageIndex{1}\).
Let’s define the behavior of the graph of the exponential function \(f(x)=2^x\) and highlight some its key characteristics.
- the domain is \((−\infty,\infty)\),
- the range is \((0,\infty)\),
- as \(x\rightarrow \infty\), \(f(x)\rightarrow \infty\),
- as \(x\rightarrow −\infty\), \(f(x)\rightarrow 0\),
- \(f(x)\) is always increasing,
- the graph of \(f(x)\) will never touch the x -axis because base two raised to any exponent never has the result of zero.
- \(y=0\) is the horizontal asymptote.
- the y -intercept is \(1\).
Definition: Exponential Functions
For any real number \(x\),an exponential function is a function with the form
\[f(x)=ab^x\]
where
- \(a\) is a non-zero real number called the initial value and
- \(b\) is any positive real number such that \(b≠1\).
- The domain of \(f\) is all real numbers.
- The range of \(f\) is all positive real numbers if \(a>0\).
- The range of \(f\) is all negative real numbers if \(a<0\).
- The y -intercept is \((0,a)\),and the horizontal asymptote is \(y=0\).
Example \(\PageIndex{1}\): Identifying Exponential Functions
Which of the following equations are not exponential functions?
- \(f(x)=4^{3(x−2)}\)
- \(g(x)=x^3\)
- \(h(x)=\left(\dfrac{1}{3}\right)^x\)
- \(j(x)=(−2)^x\)
Solution
By definition, an exponential function has a constant as a base and an independent variable as an exponent. Thus, \(g(x)=x^3\) does not represent an exponential function because the base is an independent variable. In fact, \(g(x)=x^3\) is a power function.
Recall that the base \(b\) of an exponential function is always a positive constant, and \(b≠1\). Thus, \(j(x)={(−2)}^x\) does not represent an exponential function because the base, \(−2\), is less than \(0\).
Exercise \(\PageIndex{1}\)
Which of the following equations represent exponential functions?
- \(f(x)=2x^2−3x+1\)
- \(g(x)={0.875}^x\)
- \(h(x)=1.75x+2\)
- \(j(x)={1095.6}^{−2x}\)
- Answer
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\(g(x)={0.875}^x\) and \(j(x)={1095.6}^{−2x}\) represent exponential functions.
Evaluating Exponential Functions
Recall that the base of an exponential function must be a positive real number other than \(1\).Why do we limit the base bb to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:
- Let \(b=−9\) and \(x=\dfrac{1}{2}\). Then \(f(x)=f\left(\dfrac{1}{2}\right)={(−9)}^{\dfrac{1}{2}}=\sqrt{−9}\),which is not a real number.
Why do we limit the base to positive values other than \(1\)? Because base \(1\) results in the constant function. Observe what happens if the base is \(1\):
- Let \(b=1\). Then \(f(x)=1^x=1\) for any value of \(x\).
To evaluate an exponential function with the form \(f(x)=b^x\),we simply substitute \(x\) with the given value, and calculate the resulting power. For example:
Let \(f(x)=2^x\). What is \(f(3)\)?
\[\begin{align*} f(x)&= 2^x\\ f(3)&= 2^3 \qquad \text{Substitute } x=3\\ &= 8 \qquad \text{Evaluate the power} \end{align*}\]
To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:
Let \(f(x)=30{(2)}^x\). What is \(f(3)\)?
\[\begin{align*} f(x)&= 30{(2)}^x\\ f(3)&= 30{(2)}^3 \qquad \text{Substitute } x=3\\ &= 30(8) \qquad \text{Simplify the power first}\\ &= 240 \qquad \text{Multiply} \end{align*}\]
Note that if the order of operations were not followed, the result would be incorrect:
\[f(3)=30{(2)}^3≠{60}^3=216,000 \nonumber\]
Example \(\PageIndex{2}\): Evaluating Exponential Functions
Let \(f(x)=5{(3)}^{x+1}\). Evaluate \(f(2)\) without using a calculator.
Solution
Follow the order of operations. Be sure to pay attention to the parentheses.
\[\begin{align*} f(x)&= 5{(3)}^{x+1}\\ f(2)&= 5{(3)}^{2+1} \qquad \text{Substitute } x=2\\ &= 5{(3)}^3 \qquad \text{Add the exponents}\\ &= 5(27) \qquad \text{Simplify the power}\\ &= 135 \qquad \text{Multiply} \end{align*}\]
Exercise \(\PageIndex{2}\)
Let \(f(x)=8{(1.2)}^{x−5}\). Evaluate \(f(3)\) using a calculator. Round to four decimal places.
- Answer
-
\(5.5556\)
Defining Exponential Growth
Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.
Definition: Exponential Growth
A function that models exponential growth grows by a rate proportional to the amount present. For any real number \(x\) and any positive real numbers \(a\) and \(b\) such that \(b≠1\),an exponential growth function has the form
\[f(x)=ab^x\]
where
- \(a\) is the initial or starting value of the function.
- \(b\) is the growth factor or growth multiplier per unit \(x\).
In more general terms, we have an exponential function , in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has \(100\) stores and expands by opening \(50\) new stores a year, so its growth can be represented by the function \(A(x)=100+50x\). Company B has \(100\) stores and expands by increasing the number of stores by \(50\%\) each year, so its growth can be represented by the function \(B(x)=100{(1+0.5)}^x\).
A few years of growth for these companies are illustrated in Table \(\PageIndex{3}\).
| Year, \(x\) | Stores, Company A | Stores, Company B |
|---|---|---|
| \(0\) | \(100+50(0)=100\) | \(100{(1+0.5)}^0=100\) |
| \(1\) | \(100+50(1)=150\) | \(100{(1+0.5)}^1=150\) |
| \(2\) | \(100+50(2)=200\) | \(100{(1+0.5)}^2=225\) |
| \(3\) | \(100+50(3)=250\) | \(100{(1+0.5)}^3=337.5\) |
| \(x\) | \(A(x)=100+50x\) | \(B(x)=100{(1+0.5)}^x\) |
The graphs comparing the number of stores for each company over a five-year period are shown in Figure \(\PageIndex{2}\). We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.
Notice that the domain for both functions is \([0,\infty)\),and the range for both functions is \([100,\infty)\). After year 1, Company B always has more stores than Company A.
Now we will turn our attention to the function representing the number of stores for Company \(B\), \(B(x)=100{(1+0.5)}^x\). In this exponential function, \(100\) represents the initial number of stores, \(0.50\) represents the growth rate, and \(1+0.5=1.5\) represents the growth factor. Generalizing further, we can write this function as \(B(x)=100{(1.5)}^x\),where \(100\) is the initial value, \(1.5\) is called the base , and \(x\) is called the exponent .
Example \(\PageIndex{3}\): Evaluating a Real-World Exponential Model
At the beginning of this section, we learned that the population of India was about \(1.25\) billion in the year 2013, with an annual growth rate of about \(1.2\%\). This situation is represented by the growth function \(P(t)=1.25{(1.012)}^t\), where \(t\) is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?
Solution
To estimate the population in 2031, we evaluate the models for \(t=18\), because 2031 is \(18\) years after 2013. Rounding to the nearest thousandth,
\[P(18)=1.25{(1.012)}^{18}≈1.549 \nonumber\]
There will be about \(1.549\) billion people in India in the year 2031.
Exercise \(\PageIndex{3}\)
The population of China was about \(1.39\) billion in the year 2013, with an annual growth rate of about \(0.6\%\). This situation is represented by the growth function \(P(t)=1.39{(1.006)}^t\), where \(t\) is the number of years since 2013.To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example \(\PageIndex{3}\)?
- Answer
-
About \(1.548\) billion people; by the year 2031, India’s population will exceed China’s by about \(0.001\) billion, or \(1\) million people.
Finding Equations of Exponential Functions
In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants \(a, a\) and \(b, b\),and evaluate the function.
How to: Given two data points, write an exponential model
- If one of the data points has the form \((0,a)\), then \(a\) is the initial value. Using \(a\), substitute the second point into the equation \(f(x)=a{(b)}^x\), and solve for \(b\).
- If neither of the data points have the form \((0,a)\), substitute both points into two equations with the form \(f(x)=a{(b)}^x\). Solve the resulting system of two equations in two unknowns to find \(a\) and \(b\).
- Using the \(a\) and \(b\) found in the steps above, write the exponential function in the form \(f(x)=a{(b)}^x\).
Example \(\PageIndex{4}\): Writing an Exponential Model When the Initial Value Is Known
In 2006, \(80\) deer were introduced into a wildlife refuge. By 2012, the population had grown to \(180\) deer. The population was growing exponentially. Write an algebraic function \(N(t)\) representing the population \((N)\) of deer over time \(t\).
Solution
We let our independent variable \(t\) be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, \(a=80\). We can now substitute the second point into the equation \(N(t)=80b^t\) to find \(b\):
\[\begin{align*} N(t)&= 80b^t\\ 180&= 80b^6 \qquad \text{Substitute using point } (6, 180)\\ \dfrac{9}{4}&= b^6 \qquad \text{Divide and write in lowest terms}\\ b&= {\left (\dfrac{9}{4} \right )}^{\tfrac{1}{6}} \qquad \text{Isolate b using properties of exponents}\\ b&\approx 1.1447 \qquad \text{Round to 4 decimal places} \end{align*}\]
Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.
The exponential model for the population of deer is \(N(t)=80{(1.1447)}^t\). (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)
We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in Figure \(\PageIndex{3}\) passes through the initial points given in the problem, \((0, 80)\) and \((6, 180)\). We can also see that the domain for the function is \([0,\infty)\),and the range for the function is \([80,\infty)\).
Exercise \(\PageIndex{4}\)
A wolf population is growing exponentially. In 2011, \(129\) wolves were counted. By 2013, the population had reached \(236\) wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population \(N\) of wolves over time \(t\).
- Answer
-
\((0,129)\) and \((2,236)\); \(N(t)=129{(1.3526)}^t\)
Example \(\PageIndex{5}\): Writing an Exponential Model When the Initial Value is Not Known
Find an exponential function that passes through the points \((−2,6)\) and \((2,1)\).
Solution
Because we don’t have the initial value, we substitute both points into an equation of the form \(f(x)=ab^x\), and then solve the system for \(a\) and \(b\).
- Substituting \((−2,6)\) gives \(6=ab^{−2}\)
- Substituting \((2,1)\) gives \(1=ab^2\)
Use the first equation to solve for \(a\) in terms of \(b\):
\[\begin{align*} 6&= ab^{-2}\\ \dfrac{6}{b^{-2}}&= a \qquad \text{Divide}\\ a&= 6b^2 \qquad \text{Use properties of exponents to rewrite the denominator} \end{align*}\]
Substitute a in the second equation, and solve for \(b\):
\[\begin{align*} 1&= ab^{2}\\ 1&= 6b^2 b^2\\ &= 6b^4 \qquad \text{Substitute a}\\ b&= \left (\dfrac{1}{6} \right )^{\tfrac{1}{4}} \qquad \text{Round 4 decimal places rewrite the denominator}\\ b&\approx 0.6389 \end{align*}\]
Use the value of \(b\) in the first equation to solve for the value of \(a\):
\[\begin{align*} a&= 6b^{2}\\ &\approx 6(0.6389)^2 \\ &\approx 2.4492 \end{align*}\]
Thus, the equation is \(f(x)=2.4492{(0.6389)}^x\).
We can graph our model to check our work. Notice that the graph in Figure \(\PageIndex{4}\) passes through the initial points given in the problem, \((−2, 6)\) and \((2, 1)\). The graph is an example of an exponential decay function.
Exercise \(\PageIndex{5}\)
Given the two points \((1,3)\) and \((2,4.5)\),find the equation of the exponential function that passes through these two points.
- Answer
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\(f(x)=2{(1.5)}^x\)
Q&A: Do two points always determine a unique exponential function?
Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in \(x\), which in many real world cases involves time.
How to: Given the graph of an exponential function, write its equation
- First, identify two points on the graph. Choose the \(y\)-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error.
- If one of the data points is the \(y\)-intercept \((0,a)\), then \(a\) is the initial value. Using \(a\), substitute the second point into the equation \(f(x)=a{(b)}^x\), and solve for \(b\)
- If neither of the data points have the form \((0,a)\), substitute both points into two equations with the form \(f(x)=a{(b)}^x\). Solve the resulting system of two equations in two unknowns to find \(a\) and \(b\).
- Write the exponential function, \(f(x)=a{(b)}^x\).
Example \(\PageIndex{6}\): Writing an Exponential Function Given Its Graph
Find an equation for the exponential function graphed in Figure \(\PageIndex{5}\).
Solution
We can choose the \(y\) -intercept of the graph, \((0,3)\), as our first point. This gives us the initial value, \(a=3\). Next, choose a point on the curve some distance away from \((0,3)\) that has integer coordinates. One such point is \((2,12)\).
\[\begin{align*} y&= ab^x \qquad \text{Write the general form of an exponential equation}\\ y&= 3b^x \qquad \text{Substitute the initial value } 3 \text{ for } a\\ 12&= 3b^2 \qquad \text{Substitute in 12 for } y \text{ and } 2 \text{ for } x\\ 4&= b^2 \qquad \text{Divide by }3\\ b&= \pm 2 \qquad \text{Take the square root} \end{align*}\]
Because we restrict ourselves to positive values of \(b\), we will use \(b=2\). Substitute \(a\) and \(b\) into the standard form to yield the equation \(f(x)=3{(2)}^x\).
Exercise \(\PageIndex{6}\)
Find an equation for the exponential function graphed in Figure \(\PageIndex{6}\).
- Answer
-
\(f(x)=\sqrt{2}{(\sqrt{2})}^x\). Answers may vary due to round-off error. The answer should be very close to \(1.4142{(1.4142)}^x\).
How to: Given two points on the curve of an exponential function, use a graphing calculator to find the equation
- Press [STAT].
- Clear any existing entries in columns L1 or L2.
- In L1 , enter the x -coordinates given.
- In L2 , enter the corresponding y -coordinates.
- Press [STAT] again. Cursor right to CALC , scroll down to ExpReg (Exponential Regression) , and press [ENTER].
- The screen displays the values of a and b in the exponential equation \(y=a⋅b^x\).
Example \(\PageIndex{7}\): Using a Graphing Calculator to Find an Exponential Function
Use a graphing calculator to find the exponential equation that includes the points \((2,24.8)\) and \((5,198.4)\).
Solution
Follow the guidelines above. First press [STAT] , [EDIT] , [1: Edit…], and clear the lists L1 and L2 . Next, in the L1 column, enter the \(x\)-coordinates, \(2\) and \(5\). Do the same in the L2 column for the \(y\)-coordinates, \(24.8\) and \(198.4\).
Now press [STAT] , [CALC] , [0: ExpReg] and press [ENTER] . The values \(a=6.2\) and \(b=2\) will be displayed. The exponential equation is \(y=6.2⋅2^x\).
Exercise \(\PageIndex{7}\)
Use a graphing calculator to find the exponential equation that includes the points \((3, 75.98)\) and \((6, 481.07)\).
- Answer
-
\(y≈12⋅{1.85}^x\)
Applying the Compound-Interest Formula
Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest . The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account.
The annual percentage rate (APR) of an account, also called the nominal rate , is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing.
We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time \(t\), principal \(P\), \(APR\) \(r\), and number of compounding periods in a year \(n\):
\[A(t)=P{\left (1+\dfrac{r}{n} \right )}^{nt} \nonumber\]
For example, observe Table \(\PageIndex{4}\), which shows the result of investing \($1,000\) at \(10\%\) for one year. Notice how the value of the account increases as the compounding frequency increases.
| Frequency | Value after \(1\) year |
|---|---|
| Annually | \($1100\) |
| Semiannually | \($1102.50\) |
| Quarterly | \($1103.81\) |
| Monthly | \($1104.71\) |
| Daily | \($1105.16\) |
Definition: Compound Interest
Compound interest can be calculated using the formula
\[A(t)=P{\left (1+\dfrac{r}{n} \right )}^{nt}\]
where
- \(A(t)\) is the account value,
- \(t\) is measured in years,
- \(P\) is the starting amount of the account, often called the principal, or more generally present value,
- \(r\) is the annual percentage rate (APR) expressed as a decimal, and
- \(n\) is the number of compounding periods in one year.
Example \(\PageIndex{8}\): Calculating Compound Interest
If we invest \($3,000\) in an investment account paying \(3\%\) interest compounded quarterly, how much will the account be worth in \(10\) years?
Solution
Because we are starting with \($3,000\), \(P=3000\). Our interest rate is \(3\%\), so \(r = 0.03\). Because we are compounding quarterly, we are compounding \(4\) times per year, so \(n=4\). We want to know the value of the account in \(10\) years, so we are looking for \(A(10)\),the value when \(t = 10\).
\[\begin{align*} A(t)&= P{\left (1+\dfrac{r}{n} \right )}^{nt} \qquad \text{Use the compound interest formula}\\ A(10)&= 3000{\left (1+\dfrac{0.03}{4} \right )}^{(4)\cdot (10)} \qquad \text{Substitute using given values}\\ &\approx \$4045.05 \qquad \text{Round to two decimal places} \end{align*}\]
The account will be worth about \($4,045.05\) in \(10\) years.
Exercise \(\PageIndex{8}\)
An initial investment of \($100,000\) at \(12\%\) interest is compounded weekly (use \(52\) weeks in a year). What will the investment be worth in \(30\) years?
- Answer
-
about \($3,644,675.88\)
Example \(\PageIndex{9}\): Using the Compound Interest Formula to Solve for the Principal
A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child’s future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to \($40,000\) over \(18\) years. She believes the account will earn \(6\%\) compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now?
Solution
The nominal interest rate is \(6\%\), so \(r=0.06\). Interest is compounded twice a year, so \(k=2\).
We want to find the initial investment, \(P\), needed so that the value of the account will be worth \($40,000\) in \(18\) years. Substitute the given values into the compound interest formula, and solve for \(P\).
\[\begin{align*} A(t)&= P{\left (1+\dfrac{r}{n} \right )}^{nt} \qquad \text{Use the compound interest formula}\\ 40,000&= P{\left (1+\dfrac{0.06}{2} \right )}^{2(18)} \qquad \text{Substitute using given values } A, r, n, t\\ 40,000&= P{(1.03)}^{36} \qquad \text{Simplify}\\ \dfrac{40,000}{ {(1.03)}^{36} }&= P \qquad \text{Isolate } P\\ P&\approx \$13,801 \qquad \text{Divide and round to the nearest dollar} \end{align*}\]
Lily will need to invest \($13,801\) to have \($40,000\) in \(18\) years.
Exercise \(\PageIndex{9}\)
Refer to Example \(\PageIndex{9}\). To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?
- Answer
-
\($13,693\)
Evaluating Functions with Base \(e\)
As we saw earlier, the amount earned on an account increases as the compounding frequency increases. Table \(\PageIndex{5}\) shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.
Examine the value of \($1\) invested at \(100\%\) interest for \(1\) year, compounded at various frequencies, listed in Table \(\PageIndex{5}\).
| Frequency | \(A(t)={\left (1+\dfrac{1}{n} \right )}^n\) | Value |
|---|---|---|
| Annually | \({\left (1+\dfrac{1}{1} \right )}^1\) | \($2\) |
| Semiannually | \({\left (1+\dfrac{1}{2} \right )}^2\) | \($2.25\) |
| Quarterly | \({\left (1+\dfrac{1}{4} \right )}^4\) | \($2.441406\) |
| Monthly | \({\left (1+\dfrac{1}{12} \right )}^{12}\) | \($2.613035\) |
| Daily | \({\left (1+\dfrac{1}{365} \right )}^{365}\) | \($2.714567\) |
| Hourly | \({\left (1+\dfrac{1}{8760} \right )}^{8760}\) | \($2.718127\) |
| Once per minute | \({\left (1+\dfrac{1}{525600} \right )}^{525600}\) | \($2.718279\) |
| Once per second | \({\left (1+\dfrac{1}{31536000} \right )}^{31536000}\) | \($2.718282\) |
These values appear to be approaching a limit as \(n\) increases without bound. In fact, as \(n\) gets larger and larger, the expression \({\left (1+\dfrac{1}{n} \right )}^n\) approaches a number used so frequently in mathematics that it has its own name: the letter \(e\). This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.
Definition: The Number e
The letter \(e\) represents the irrational number
\[{\left (1+\dfrac{1}{n} \right )}^n\]
as \(n\) increases without bound
The letter \(e\) is used as a base for many real-world exponential models. To work with base \(e\), we use the approximation, \(e≈2.718282\). The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties.
Example \(\PageIndex{10}\): Using a Calculator to Find Powers of \(e\)
Calculate \(e^{3.14}\). Round to five decimal places.
Solution
On a calculator, press the button labeled \([e^x]\). The window shows \([e {}^( ]\). Type \(3.14\) and then close parenthesis, \([)]\). Press [ENTER]. Rounding to \(5\) decimal places, \(e^{3.14}≈23.10387\). Caution: Many scientific calculators have an “Exp” button, which is used to enter numbers in scientific notation. It is not used to find powers of \(e\).
Exercise \(\PageIndex{10}\)
Use a calculator to find \(e^{−0.5}\). Round to five decimal places.
- Answer
-
\(e^{−0.5}≈0.60653\)
Investigating Continuous Growth
So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, \(e\) is used as the base for exponential functions. Exponential models that use \(e\) as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.
Definition: The Continuous Growth/decay Formula
For all real numbers \(t\),and all positive numbers \(a\) and \(r\), continuous growth or decay is represented by the formula
\[A(t)=ae^{rt}\]
where
- \(a\) is the initial value,
- \(r\) is the continuous growth rate per unit time,
- \(t\) is the elapsed time.
If \(r>0\), then the formula represents continuous growth. If \(r<0\), then the formula represents continuous decay.
For business applications, the continuous growth formula is called the continuous compounding formula and takes the form
\[A(t)=Pe^{rt}\]
where
- \(P\) is the principal or the initial invested,
- \(r\) is the growth or interest rate per unit time,
- \(t\) is the period or term of the investment.
How to: Given the initial value, rate of growth or decay, and time \(t\), solve a continuous growth or decay function
- Use the information in the problem to determine \(a\), the initial value of the function.
-
Use the information in the problem to determine the growth rate \(r\).
- If the problem refers to continuous growth, then \(r>0\).
- If the problem refers to continuous decay, then \(r<0\).
- Use the information in the problem to determine the time \(t\).
- Substitute the given information into the continuous growth formula and solve for \(A(t)\).
Example \(\PageIndex{11}\): Calculating Continuous Growth
A person invested \($1,000\) in an account earning a nominal \(10\%\) per year compounded continuously. How much was in the account at the end of one year?
Solution
Since the account is growing in value, this is a continuous compounding problem with growth rate \(r=0.10\). The initial investment was \($1,000\), so \(P=1000\). We use the continuous compounding formula to find the value after \(t=1\) year:
\[\begin{align*} A(t)&= Pe^{rt} \qquad \text{Use the continuous compounding formula}\\ &= 1000{(e)}^{0.1} \qquad \text{Substitute known values for } P, r, t\\ &\approx 1105.17 \qquad \text{Use a calculator to approximate} \end{align*}\]
The account is worth \($1,105.17\) after one year.
Exercise \(\PageIndex{11}\)
A person invests \($100,000\) at a nominal \(12\%\) interest per year compounded continuously. What will be the value of the investment in \(30\) years?
- Answer
-
\($3,659,823.44\)
Example \(\PageIndex{12}\): Calculating Continuous Decay
\(Radon-222\) decays at a continuous rate of \(17.3\%\) per day. How much will \(100 mg\) of \(Radon-222\) decay to in \(3\) days?
Solution
Since the substance is decaying, the rate, \(17.3\%\), is negative. So, \(r = −0.173\). The initial amount of \(Radon-222\) was \(100\) mg, so \(a=100\). We use the continuous decay formula to find the value after \(t=3\) days:
\[\begin{align*} A(t)&= ae^{rt} \qquad \text{Use the continuous growth formula}\\ &= 100e6{-0.173(3)} \qquad \text{Substitute known values for } a, r, t\\ &\approx 59.5115 \qquad \text{Use a calculator to approximate} \end{align*}\]
So \(59.5115\) mg of \(Radon-222\) will remain.
Exercise \(\PageIndex{12}\)
Using the data in Example \(\PageIndex{12}\), how much \(Radon-222\) will remain after one year?
- Answer
-
\(3.77E-26\) (This is calculator notation for the number written as \(3.77×10^{−26}\) in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)
Media
Access these online resources for additional instruction and practice with exponential functions.
- Exponential Growth Function
- Compound Interest
Key Equations
| definition of the exponential function | \(f(x)=b^x\), where \(b>0\), \(b≠1\) |
| definition of exponential growth | \(f(x)=ab^x\), where \(a>0\), \(b>0\), \(b≠1\) |
| compound interest formula |
\(A(t)=P{(1+\dfrac{r}{n})}^{nt}\), where \(A(t)\) is the account value at time \(t\) \(t\) is the number of years \(P\) is the initial investment, often called the principal \(r\) is the annual percentage rate (APR), or nominal rate \(n\) is the number of compounding periods in one year |
| continuous growth formula | \(A(t)=ae^{rt}\), where \(t\) is the number of unit time periods of growth \(a\) is the starting amount (in the continuous compounding formula a is replaced with \(P\), the principal) \(e\) is the mathematical constant, \(e≈2.718282\) |
Key Concepts
- An exponential function is defined as a function with a positive constant other than \(1\) raised to a variable exponent. See Example .
- A function is evaluated by solving at a specific value. See Example and Example .
- An exponential model can be found when the growth rate and initial value are known. See Example .
- An exponential model can be found when the two data points from the model are known. See Example .
- An exponential model can be found using two data points from the graph of the model. See Example .
- An exponential model can be found using two data points from the graph and a calculator. See Example .
- The value of an account at any time \(t\) can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known. See Example .
- The initial investment of an account can be found using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known. See Example .
- The number \(e\) is a mathematical constant often used as the base of real world exponential growth and decay models. Its decimal approximation is \(e≈2.718282\).
- Scientific and graphing calculators have the key \([ex]\) or \([exp(x)]\) for calculating powers of \(e\). See Example .
- Continuous growth or decay models are exponential models that use \(e\) as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known. See Example and Example .


4.3: Questions to Guide Your Reading
1. During China’s Neolithic Age, how did Longshan culture differ from Yangshao culture?
2. What kinds of evidence do we have to reconstruct the first dynasties in Chinese history–the Xia and Shang Dynasties? What can we learn from this evidence?
3. During the Western Zhou Dynasty, how did Zhou kings justify overthrowing the Shang Dynasty, and how did they govern newly-conquered lands?
4. During the Warring States Period, how did the nature of warfare change, and how did the demands of that changing warfare reshape Zhou feudal states?
5. Explain the major ideas of Confucius.
6. Explain the major ideas and practices of Daoism.
7. Why was the Qin Dynasty so important to the history of China? What did the First Emperor of Qin accomplish?
8. Describe prominent features of Han Dynasty society and governing.
9. How did Buddhism become a major religious tradition in China?
10. How was state development in Korea and Japan differently impacted by China?
11. What were the foundations of Tang Dynasty power in East Asia?
12. During the Song Dynasty, China was one of the most developed countries in the world. What features of that dynasty provide evidence supporting that judgment?
13. Describe the status of women during the Song Dynasty.
14. Explain the origins of the Yuan Dynasty and assess its impact on China.


3.8: Putting It Together- Segmentation and Targeting
Remember Chumber, your new employer from the beginning of this module?
Now that you’ve learned something about segmentation and targeting strategy, let’s return to the request your boss made for recommendations about whom Chumber ought to target and why.
Remember that Chumber’s product is an automated, fully online system for checking the references of job candidates. Chumber’s customers are other companies. After learning about market segmentation, you know that “all companies” is too broad to be a useful target market. Even on your first day of work, you can guess that marketing to every company you can find isn’t going to be a smart strategy.
Instead, you do a little research. It stands to reason that Chumber will be most valuable to companies that do a lot of hiring. A Google search for “employment by industry” brings up U.S. Bureau of Labor statistics data to help you identify which industries are expected to post the biggest gains in employment in the coming years.
| Industry Sector | Thousands of Jobs | Change | Percent Distribution | Compound Annual Rate of Change | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| 2008 | 2018 | 2028 | 2008–2018 | 2018–2028 | 2008 | 2018 | 2028 | 2008–2018 | 2018–2028 | |
| Total [1] [2] | 149,276.0 | 161,037.7 | 169,435.9 | 11,761.7 | 8,398.2 | 100.0 | 100.0 | 100.0 | 0.8 | 0.5 |
| Nonagriculture wage and salary [3] | 137,991.0 | 149,803.7 | 157,662.0 | 11,812.7 | 7,858.3 | 92.4 | 93.0 | 93.1 | 0.8 | 0.5 |
| Goods-producing, excluding agriculture | 21,277.9 | 20,661.3 | 20,872.7 | −616.6 | 211.4 | 14.3 | 12.8 | 12.3 | −0.3 | 0.1 |
| Mining | 709.9 | 683.3 | 727.9 | −26.6 | 44.6 | 0.5 | 0.4 | 0.4 | −0.4 | 0.6 |
| Construction | 7,162.5 | 7,289.3 | 8,096.8 | 126.8 | 807.5 | 4.8 | 4.5 | 4.8 | 0.2 | 1.1 |
| Manufacturing | 13,405.5 | 12,688.7 | 12,048.0 | −716.8 | −640.7 | 9.0 | 7.9 | 7.1 | −0.5 | −0.5 |
| Services-providing excluding special industries | 116,713.1 | 129,142.4 | 136,789.3 | 12,429.3 | 7,646.9 | 78.2 | 80.2 | 80.7 | 1.0 | 0.6 |
| Utilities | 558.8 | 554.6 | 537.2 | −4.2 | −17.4 | 0.4 | 0.3 | 0.3 | −0.1 | −0.3 |
| Wholesale trade | 5,875.0 | 5,852.5 | 5,754.0 | −22.5 | −98.5 | 3.9 | 3.6 | 3.4 | 0.0 | -0.2 |
| Retail trade | 15,289.1 | 15,833.1 | 15,679.4 | 544.0 | −153.7 | 10.2 | 9.8 | 9.3 | 0.4 | -0.1 |
| Transportation and warehousing | 4,513.6 | 5,419.1 | 5,741.4 | 905.5 | 322.3 | 3.0 | 3.4 | 3.4 | 1.8 | 0.6 |
| Information | 2,983.8 | 2,828.1 | 2,833.7 | −155.7 | 5.6 | 2.0 | 1.8 | 1.7 | −0.5 | 0.0 |
| Financial activities | 8,206.1 | 8,568.8 | 8,849.4 | 362.7 | 280.6 | 5.5 | 5.3 | 5.2 | 0.4 | 0.3 |
| Professional and business services | 17,792.3 | 20,999.5 | 22,661.9 | 3,207.2 | 1,662.4 | 11.9 | 13.0 | 13.4 | 1.7 | 0.8 |
| Educational services | 3,039.8 | 3,727.5 | 4,201.0 | 687.7 | 473.5 | 2.0 | 2.3 | 2.5 | 2.1 | 1.2 |
| Health care and social assistance | 16,188.6 | 19,939.3 | 23,335.4 | 3,750.7 | 3,396.1 | 10.8 | 12.4 | 13.8 | 2.1 | 1.6 |
| Leisure and hospitality | 13,436.2 | 16,348.5 | 17,904.9 | 2,912.3 | 1,556.4 | 9.0 | 10.2 | 10.6 | 2.0 | 0.9 |
| Other services | 6,320.5 | 6,622.4 | 6,716.7 | 301.9 | 94.3 | 4.2 | 4.1 | 4.0 | 0.5 | 0.1 |
| Federal government | 2,762.0 | 2,796.0 | 2,670.2 | 34.0 | −125.8 | 1.9 | 1.7 | 1.6 | 0.1 | -0.5 |
| State and local government | 19,747.3 | 19,653.0 | 19,904.0 | −94.3 | 251.0 | 13.2 | 12.2 | 11.7 | 0.0 | 0.1 |
| Agriculture, forestry, fishing, and hunting [4] | 2,071.4 | 2,310.0 | 2,320.6 | 238.6 | 10.6 | 1.4 | 1.4 | 1.4 | 1.1 | 0.0 |
| Agriculture wage and salary | 1,208.6 | 1,547.2 | 1,587.2 | 338.6 | 40.0 | 0.8 | 1.0 | 0.9 | 2.5 | 0.3 |
| Agriculture self-employed | 862.8 | 762.8 | 733.4 | −100.0 | −29.4 | 0.6 | 0.5 | 0.4 | −1.2 | −0.4 |
| Nonagriculture self-employed | 9,213.6 | 8,924.0 | 9,453.4 | −289.6 | 529.4 | 6.2 | 5.5 | 5.6 | −0.3 | 0.6 |
Segmenting by industry makes a lot of sense in this case because some industries clearly do more hiring than others. You decide that Chumber should focus on industries with the highest projected hiring increases in the next decade: health care; professional and business services; construction; leisure and hospitality; and retail. Companies in growth industries will definitely get the most value from Chumber.
Next you want to understand more about which decision makers in these companies will be the best targets for Chumber. Having just come through the hiring process, you know who is interested in reference checking: human resources professionals, job recruiters, and hiring managers. You email Ken, the Chumber HR person who handled your hiring process, to see if he can answer a few questions about how decisions are made in HR departments.
Ken is very helpful. Prior to Chumber, he worked in HR for a health care company and a consulting firm. He confirms that an HR manager or director of recruiting oversees the reference-checking process for new hires. This person would also be the primary decision maker for a product like Chumber.
Ken explains that the requirements for reference checking differ by industry. In health care, for instance, where people routinely handle life-and-death situations, reference checks are essential and thorough. Ken mentions a couple of features Chumber could add to fit the specific requirements of the health care industry. You take notes about product improvements that could be part of the marketing mix for this segment.
When you’re back at your desk, Ken sends you a list of Web sites, publications, and conferences where many HR recruiters go for professional information. This will be really useful when your boss wants to talk about promotion and place!
You invite Ken out for lunch to thank him for his valuable input.
You still have a lot to learn about Chumber and product marketing. But applying your knowledge about segmentation and targeting is giving you a good feel for how you might help the company succeed.
- Employment data for wage and salary workers are from the BLS Current Employment Statistics survey, which counts jobs, whereas self-employed and agriculture, forestry, fishing, and hunting are from the Current Population Survey (household survey), which counts workers. ↵
- ndividual sectors do not necessarily add to major sectors due to rounding. ↵
- Includes wage and salary data from the Current Employment Statistics survey, except private households, which is from the Current Populations Survey. Logging workers are excluded. ↵
- Includes agriculture, forestry, fishing, and hunting data from the Current Population Survey, except logging, which is from Current Employment Statistics survey. Government wage and salary workers are excluded. ↵
Contributors and Attributions
- Putting It Together: Segmentation and Targeting. Provided by : Lumen Learning. License : CC BY: Attribution
- Untitled. Authored by : Marius Ciocirlan. Provided by : Unsplash. Located at : unsplash.com/photos/vMV6r4VRhJ8. License : CC0: No Rights Reserved . License Terms : Unsplash License
- Employment by major industry sector. Provided by : Bureau of Labor Statistics. Located at : https://www.bls.gov/emp/tables/employment-by-major-industry-sector.htm . License : Public Domain: No Known Copyright


3.5: Determining Field from Potential
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Learning Objectives
By the end of this section, you will be able to:
- Explain how to calculate the electric field in a system from the given potential
- Calculate the electric field in a given direction from a given potential
- Calculate the electric field throughout space from a given potential
Recall that we were able, in certain systems, to calculate the potential by integrating over the electric field. As you may already suspect, this means that we may calculate the electric field by taking derivatives of the potential, although going from a scalar to a vector quantity introduces some interesting wrinkles. We frequently need \(\vec{E}\) to calculate the force in a system; since it is often simpler to calculate the potential directly, there are systems in which it is useful to calculate \(V\) and then derive \(\vec{E}\) from it.
In general, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because the force on a positive charge is in the direction of \(\vec{E}\) and also in the direction of lower potential \(V\). Furthermore, the magnitude of \(\vec{E}\) equals the rate of decrease of \(V\) with distance. The faster \(V\) decreases over distance, the greater the electric field. This gives us the following result.
Relationship between Voltage and Uniform Electric Field
In equation form, the relationship between voltage and uniform electric field is
\[E = - \dfrac{\Delta V}{\Delta s} \nonumber \]
where \(\Delta s\) is the distance over which the change in potential \(\Delta V\) takes place. The minus sign tells us that \(E\) points in the direction of decreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electric potential.
For continually changing potentials, \(\Delta V\) and \(\Delta s\) become infinitesimals, and we need differential calculus to determine the electric field. As shown in Figure \(\PageIndex{1}\), if we treat the distance \(\Delta s\) as very small so that the electric field is essentially constant over it, we find that
\[E_s = - \dfrac{dV}{ds}. \nonumber \]
Therefore, the electric field components in the Cartesian directions are given by
\[E_x = - \dfrac{\partial V}{\partial x}, \, E_y = - \dfrac{\partial V}{\partial y}, \, E_z = - \dfrac{\partial V}{\partial z}. \nonumber \]
This allows us to define the “grad” or “del” vector operator, which allows us to compute the gradient in one step. In Cartesian coordinates, it takes the form
\[\vec{\nabla} = \hat{i} \dfrac{\partial}{\partial x} + \hat{j} \dfrac{\partial}{\partial y} + \hat{k} \dfrac{\partial}{\partial z}. \nonumber \]
With this notation, we can calculate the electric field from the potential with
\[\vec{E} = - \vec{\nabla}V, \label{eq20} \]
a process we call calculating the gradient of the potential .
If we have a system with either cylindrical or spherical symmetry, we only need to use the del operator in the appropriate coordinates:
\[\begin{align} \vec{\nabla}_{cyl} &= \underbrace{\hat{r} \dfrac{\partial}{\partial r} + \hat{\varphi}\dfrac{1}{r} \dfrac{\partial}{\partial \varphi} + \hat{z} \dfrac{\partial}{\partial z}}_{\text{Cylindrical}} \label{cylindricalnabla} \\[4pt] \vec{\nabla}_{sph} &= \underbrace{ \hat{r} \dfrac{\partial}{\partial r} + \hat{\theta}\dfrac{1}{r} \dfrac{\partial}{\partial \theta} + \hat{\varphi} \dfrac{1}{r \sin\theta}\dfrac{\partial}{\partial \varphi}}_{\text{Spherical}} \label{spherenabla} \end{align} \]
Example \(\PageIndex{1}\): Electric Field of a Point Charge
Calculate the electric field of a point charge from the potential.
Strategy
The potential is known to be \(V = k\dfrac{q}{r}\), which has a spherical symmetry. Therefore, we use the spherical del operator (Equation \ref{spherenabla}) into Equation \ref{eq20}:
\[\vec{E} = - \vec{\nabla}_{sph} V \nonumber. \nonumber \]
Solution
Performing this calculation gives us
\[\begin{align*} \vec{E} &= - \left( \hat{r}\dfrac{\partial}{\partial r} + \hat{\theta}\dfrac{1}{r} \dfrac{\partial}{\partial \theta} + \hat{\varphi}\dfrac{1}{r \sin\theta} \dfrac{\partial}{\partial \varphi}\right) k\dfrac{q}{r} \\[4pt] &= - kq\left( \hat{r}\dfrac{\partial}{\partial r}\dfrac{1}{r} + \hat{\theta}\dfrac{1}{r} \dfrac{\partial}{\partial \theta}\dfrac{1}{r} + \hat{\varphi}\dfrac{1}{r \sin \theta} \dfrac{\partial}{\partial \varphi}\dfrac{1}{r}\right).\end{align*} \]
This equation simplifies to
\[\vec{E} = - kq\left(\hat{r}\dfrac{-1}{r^2} + \hat{\theta}0 + \hat{\varphi}0 \right) = k\dfrac{q}{r^2}\hat{r} \nonumber \]
as expected.
Significance
We not only obtained the equation for the electric field of a point particle that we’ve seen before, we also have a demonstration that \(\vec{E}\) points in the direction of decreasing potential, as shown in Figure \(\PageIndex{2}\).
Example \(\PageIndex{2}\): Electric Field of a Ring of Charge
Use the potential found previously to calculate the electric field along the axis of a ring of charge (Figure \(\PageIndex{3}\)).
Strategy
In this case, we are only interested in one dimension, the \(z\)-axis. Therefore, we use
\[E_z = - \dfrac{\partial V}{\partial z} \nonumber \]
with the potential
\[V = k \dfrac{q_{tot}}{\sqrt{z^2 + R^2}} \nonumber \]
found previously.
Solution
Taking the derivative of the potential yields
\[ \begin{align*} E_z &= - \dfrac{\partial}{\partial z} \dfrac{kq_{tot}}{\sqrt{z^2 + R^2}} \\[4pt] &= k \dfrac{q_{tot}z}{(z^2 + R^2)^{3/2}}. \end{align*} \]
Significance
Again, this matches the equation for the electric field found previously. It also demonstrates a system in which using the full del operator is not necessary.
Exercise \(\PageIndex{1}\)
Which coordinate system would you use to calculate the electric field of a dipole?
- Answer
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Any, but cylindrical is closest to the symmetry of a dipole.


38.7: Review Questions
Which teaching point should the nurse include when educating clients about the administration of trifluridine?
- Answer
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“Wear gloves when handling and instilling this medication.”
Which medication should the nurse anticipate being prescribed to prevent ophthalmia neonatorum due to Neisseria gonorrhea ?
- Answer
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Erythromycin
A client states that they have been told they are a “glaucoma suspect.” The nurse understands this means the client has one or more risk factors from which of the following groups?
- Answer
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Elevated intraocular pressure, optic nerve damage, visual field deficits
Which complication of untreated “pink eye” should the nurse teach to a parent of a child with eye redness and a thick discharge?
- Answer
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Meningitis
During an intake history, a client tells the nurse that they are taking an eye drop that changed the color of their eyelids and caused their eyelashes to thicken and grow, but they cannot remember the name of the drug. Which ophthalmic drug class should the nurse anticipate the client is taking?
- Answer
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Prostaglandin analogues
Which prostaglandin analogue does the nurse understand to be more effective in Black clients?
- Answer
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Travoprost
A client tells the clinic nurse that after administering timolol drops in their right eye, they noticed that their heart rate decreased to 50 beats per minute. Which question by the nurse best determines the most likely cause of bradycardia?
- Answer
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“How did you instill the medication?”
A client is taking prednisolone drops for uveitis. Which condition that can occur with prolonged use should the nurse warn the client about?
- Answer
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Cataracts
A client has been prescribed an ocular antiviral, trifluridine, for ocular herpes. The nurse instructs the client to seek immediate treatment if they experience which condition?
- Answer
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Floaters and flashes of light
Which potential systemic effect should the nurse anticipate when administering a nonselective ophthalmic beta blocker?
- Answer
-
It may mask signs of hypoglycemia.


11.7: References
References
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Anderson, R. (2016, September 6). The ugly truth about online dating. Psychology Today. https://www.psychologytoday.com/intl/blog/the-mating-game/201609/the-ugly-truth-about-online-dating
Buller, D. B. & Burgoon, J. K. (2006, March). Interpersonal deception theory. Communication Theory, 6 (3), 203-242.
Statista Research Department. (2021). Daily time spent on social networking by internet users worldwide from 2012 to 2022. Statistica. https://www.statista.com/statistics/433871/daily-social-media-usage-worldwide/
Davis, K. (2012). Friendship 2.0: Adolescents' experiences of belonging and self-disclosure online. Journal of Adolescence, 35 (6), 1527-1536.
De Choudhury, M., Counts, S., & Horvitz, E. (2013, May 3). Social media as a measurement tool of depression in populations. In H. Davis, H. Halprin, A. Pentland, M. Bernstein, L. Adamic, H. Alani, A. Monnin, & R. Rogers (Chairs.), WebSci '13: Proceedings of the 5th annual ACM web science conference (pp. 47-56). Association for Computing Machinery.
Office on Women’s Health. (2018, September 13). Emotional and verbal abuse. US Department of Health & Human Services. https://www.womenshealth.gov/relationships-and-safety/other-types/emotional-and-verbal-abuse
Evans, P. (2010). The verbally abusive relationship: How to recognize it and how to respond (3rd eds.). Simon and Schuster.
Evans, M. L., Lindauer, M., & Farrell, M. E. (2020). A pandemic within a pandemic—Intimate partner violence during Covid-19. New England Journal of Medicine, 383 (24), 2302-2304.
Facebook - Meld je aan of registreer je. (2021). Facebook. Retrieved January 7, 2022, from https://www.facebook.com/unsupportedbrowser#content-actio
Farkas, J., Schou, J., & Neumayer, C. (2018). Cloaked Facebook pages: Exploring fake Islamist propaganda in social media. New Media & Society, 20 (5), 1850-1867.
Internet Crime Complaint Center. (2020). Internet Crime Report 2020. Federal Bureau of Investigation. https://www.ic3.gov/Media/PDF/AnnualReport/2020_IC3Report.pdf
Feinberg, M., Willer, R., Stellar, J., & Keltner, D. (2012). The virtues of gossip: reputational information sharing as prosocial behavior. Journal of Personality and Social Psychology, 102 (5), 1015-1030. doi.org/10.1037/a0026650
Freyd, J.J. (1997) Violations of power, adaptive blindness, and betrayal trauma theory. Feminism & Psychology, 7, 22-32.
Galen, B. R., & Underwood, M. K. (1997). A developmental investigation of social aggression among children. Developmental Psychology, 33 (4), 589.
Gottfried, S. (2019, September 25). The science behind why people gossip—and when it can be a good thing. Time. https://time.com/5680457/why-do-people-gossip/
Gottman, J. M. (1994). What predicts divorce? The relationship between marital processes and marital outcomes. Lawrence Erlbaum Associates.
Gottman, J. M., & Driver, J. L. (2005). Dysfunctional marital conflict and everyday marital interaction. Journal of Divorce & Remarriage, 43 (3-4), 63-77.
Gottman, J. M. (2014). What predicts divorce?: The relationship between marital processes and marital outcomes. Psychology Press.
Graves, C. G., & Spencer, L. G. (2022). Rethinking the rhetorical epistemics of gaslighting. Communication Theory, 32 (1), 48-67. https://doi.org/10.1093/ct/qtab013
Guerrero, L. K., & Andersen, P. A. (1996). Jealousy experience and expression in romantic relationships. In P. A. Anderson & L. K. Guerrero (Eds.), Handbook of communication and emotion (pp. 155-188). Academic Press.
Guerrero, L. K., Trost, M. R., & Yoshimura, S. M. (2005). Romantic jealousy: Emotions and communicative responses. Personal Relationships, 12(2), 233-252.
Harsey, S., Zurbriggen, E., & Freyd, J. J. (2017). Perpetrator responses to victim confrontation: DARVO and victim self-blame. Journal of Aggression, Maltreatment, & Trauma, 26, 644-663.
Henniger, N. E., & Harris, C. R. (2014). Can negative social emotions have positive consequences?: An examination of embarrassment, shame, guilt, jealousy, and envy. In W. G. Parrott (Ed.), The positive side of negative emotions (pp. 76–97). The Guilford Press.
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When we talk about making the Web accessible for people with disabilities, who are the people we are talking about? Before we can learn what to do with our web pages, we need to understand what we are doing and who we are doing it for.
Note
Simulations
To help you understand what web navigation is like for people with disabilities, some organizations have developed simulations:
The first group that most people think of when considering accessibility for the Web is the blind and visually impaired.
Blind: Users have little or no usable vision. While a few users may use Braille, the majority use a screen reader—software that reads text out loud. Some people listen to the Web at speeds that sighted users find completely incomprehensible—the audio equivalent of “skimming” a page. Keep in mind that screen readers read everything that they encounter, and that they read it in the order they find it. In some cases, users with screen readers encounter online multimedia elements that start playing without warning. They must contend with two audio sources at the same time: the screen reader reading the web page text and the multimedia audio.
Visually impaired: Users may have some sight, but difficulty focusing or distinguishing small text. They may use a screen magnifier—software that enlarges everything on the screen to a more manageable size.
Colour blind: Most colour blindness involves difficulties distinguishing red and green. A smaller percentage of people have difficulties with the blue-purple portion of the colour spectrum. Still others are completely colour blind. There is a misconception that accessibility means using only black and white text, and that colour should be avoided. This is not true. The point is not to rely on the requirement of colour perception to reveal information. For example: asking readers (or learners) to “use only words in red to compose a paragraph”, or telling readers while filling in the form, that only “blue” fields are required.
As we will find, making the Web’s highly visual content accessible is not as daunting a task as it might seem. There are methods in place for providing alternatives for nearly every type of web content, and for making sure your content works well with the specialized hardware and software used.
Note
Ever wondered what the world looks like to colour-blind people? Test out Vischeck, a colourblindness simulator, on your site or any image.
http://www.vischeck.com
Since the majority of content on the Web is visual, students who are deaf or hard-of-hearing are not as likely to be affected. However, they often have communication and comprehension difficulties. If audio files or videos are a part of the curriculum, a text alternative should be provided. Many users will also benefit from easily understandable icons and clear terminology.
Ideally, videos should be captioned. Professional captioning can be costly, though for course materials requiring extremely high accuracy (such as math and physics equations), it may be the best choice. Software is also available to allow you to include captions in your videos yourself. If captioning is simply not an option, a text transcript of the video would be a reasonable alternative.
Note
Hearing people might assume that hard-of-hearing or deaf students would be reluctant to watch a video clip. But on the contrary, many find video and multimedia material entertaining and especially valuable because of all the other non-verbal communication that they convey. Samuel, a hardof-hearing ESL student in our focus group, greatly preferred videos or webcam interactions to text so that he could see the emotions and gestures of the other person. For students who can lip-read, video is still helpful if the speaker’s face is clearly visible at all times.
Mobility
Students with physical disabilities may be affected if their impairment hinders their ability to use a mouse or keyboard. This could be due to having little or no muscle control, nerve damage, or trembling; it could be a temporary problem, a lifelong condition, or the result of aging. Fine motor movements can pose a challenge, such as clicking on a very small icon.
Some users with mobility impairments will use a typical keyboard or mouse, but may take more time to perform tasks. Others use assistive input devices instead or in addition to a keyboard or mouse.
A standard trackball is often easier to control than a mouse. Some students use a standard graphics tablet since touching locations directly with a pen is easier for him than sliding a mouse.
Alternative keyboards allow users to position their hands more comfortably, or to press keys more accurately.
For people who cannot use their hands at all, headtracking allows the user to control the pointer through head movements. Mouse clicks can be replaced with a breath-controlled sip/puff switch or tappable headswitch.
Learning and Cognitive
While visual, hearing and physical disabilities are the most familiar forms of disability, the majority of students you may encounter who have a registered disability may in fact be learning disabled. Learning disability or “learning difficulty” is a broad term that includes dyslexia, brain injury, and aphasia.
“Dyslexia is the most commonly registered disability within the University and always features in the most commonly asked questions on accessibility issues by staff.” (Jeffels & Marston, 2003)
Students affected by learning disabilities may encounter difficulties with some of the following activities, among others:
spelling
reading aloud; stuttering
mathematical calculations
comprehension of large passages of text
effective time management or organization
rote memorization
concentration and focus
Try to read the passage in Figure \(\PageIndex{1}\). It may give you an idea of the difficulty and frustration experienced by many dyslexic readers, as seemingly normal text requires extra effort and concentration to parse.
Learning and cognitive disabilities are a challenging group to address, as there is no one approach that will suit everyone. Some students may learn just as quickly or more quickly than typical students when information is presented in a different medium. Some use the same technologies used by the visually impaired, such as screen readers and speech recognition software. Nevertheless, clear presentation and good navigation is critical. A variety of multimedia options will apply to different visual, auditory and learning skills.
Table \(\PageIndex{1}\): Content developed using traditional approach and suggestions for adaptations
Traditional approach
Adapted
Lecture type content
Chunks, include questions, statements of clarification and key points
Text-based content
Alternative presentation: audio, video, hands-on interaction; scaffold for various resources (preselect them)
Reading from a textbook
Offer vocabulary, issues to discuss in the forum, encourage note-taking, using graphic organizers, offer information prompts (selft-ests with open ended questions)
Assignments: written essay
Offer a choice: written, oral, video or visual presentation
Assessment Offer variety in responses: open-ended questions, oral response
Give clear scoring rubrics, be prompt and detailed in giving feedback
Aging Users
When considering accessibility in education, most people assume they will need to prepare for a few isolated examples of students with disabilities: one blind student in a class, or a handful of young students with learning issues. As we age, we may be affected by any of these types of disabilities to various degrees. Instructors should be aware that some of their older students may also have problems such as fading eyesight, or difficulty with fine mouse movements.
Assistive Technology
We have touched briefly on the idea of assistive technology, which is essentially any software or hardware that can be used to help overcome a disability.
Note
A pair of glasses could be considered assistive technology, as it helps the user overcome poor vision.
Instead of thinking about assistive technology in terms of types of disabilities it assists, let’s look at it from the point of view what kind of help it offers. Assistive technology could provide:


5.14: IV Infusion by Gravity
IV drip tubing comes in a variety of sizes called drop factors. The drop factor is the number of drops in one mL of solution using gravity IV tubing. The drop factor is printed on the IV tubing package. Macrodrip tubing includes tubing with drop factors of 10, 15, or 20 drops per mililiter and is typically used to deliver general IV solutions to adults. Microdrip tubing includes tubing with a drop factor of 60 drops per milliliter. It is typically used to deliver precise amounts of medication in small drops to children and infants. See Figure \(\PageIndex{1}\) [1] for an image of macrodrip and microdrip tubing.
In the sample problems below, you will be calculating the number of drops needed per minute to achieve the quantity of fluid to be used per hour in the provider order.
Practice Problem: IV Infusion (Example 1)
Let’s practice calculating the rate of IV infusion by gravity using macrodrip tubing.
Patient Information:
Name: Elaina Dimas, DOB: 12/09/19xx, Age 6, Allergies: NKDA, Weight: 15 kg
Diagnosis: Dehydration
Prescription: 0.9% Sodium Chloride 500 mL IV bolus over four hours
See Figure \(\PageIndex{2}\) [3] for the IV fluid supplied.
See Figure \(\PageIndex{3}\) [4] for IV tubing supplied.
Solve using dimensional analysis.
1. To set up the problem, begin by identifying the goal unit(s) for which we are solving. When calculating infusion rates by gravity, we need to calculate how many drops (gtts) of the solution will be infused each minute. This will allow us to count the actual drops dripping from the tubing and regulate the rate so the volume of infusion is delivered in the time ordered. Set up the problem by identifying the goal unit to solve, but instead of solving for one unit, we will be solving for two units: drops (gtts) and minutes:
\[ \frac{gtts}{min}~=~? \]
2. Set up the first fraction by matching drops (gtts) in the numerator to the goal unit. Review the tubing provided to determine how many drops per minute are administered with this type of tubing. In this example, the tubing is labelled as 60 gtts/mL. Plug in 60 in the numerator for how many drops are administered by the tubing in 1 mL, and then add 1 mL to the denominator:
\[\frac{gtts}{min}~=~\frac{60~gtts}{1~mL} \]
3.  Look for information provided in the problem related to mL. The prescription is for 500 mL of solution over four hours. Plug in 500 into the numerator, and place 4 hours in the denominator, and then cross off the mL units:
\[ \frac{gtts}{min}~=~\frac{60~gtts}{1~\cancel{mL}}~\times~\frac{500~\cancel{mL}}{4~hours} \]
4. Calculate the minutes to achieve our goal unit of drops per minute. Create the third fraction with hour in the numerator with the intent to cross off hour units. Using equivalencies, we know that 1 hour is equivalent to 60 minutes, so plug in 60 minutes in the denominator. Cancel out hours. Evaluate if we have reached our goal units. This equation now matches the goal of units/min and can be solved:
\[ \frac{gtts}{min}~=~\frac{60~gtts}{1~\cancel{mL}}~\times~\frac{500~\cancel{mL}}{4~\cancel{hours}}~\times~\frac{1~\cancel{hour}}{60~min} \]
5. Multiply across the numerators and denominators, and then divide the final fraction.
\[ \frac{gtts}{min}~=~\frac{60~gtts}{1~\cancel{mL}}~\times~\frac{500~\cancel{mL}}{4~\cancel{hours}}~\times~\frac{1~\cancel{hour}}{60~min}~=~\frac{60~gtts~\times~500~\times~1}{1~\times~4~\times~60~min}~=~{125~gtts/min} \]
6. The final answer is 125 drops/minute.
Practice Problem: IV Infusion (Example 2)
Let’s practice a second problem using different types of IV drip tubing and a different time to be infused.
Patient Information:
Name: Amber Gomez, DOB: 08/26/19xx, Age 26, Allergies: NKDA, Weight: 50 kg
Provider Order: Lactated Ringers 250 mL IV bolus over 2 hours
Fluid Supplied: See Figure \(\PageIndex{4}\) [5] for the fluid supplied.
Tubing Supplied: See Figure \(\PageIndex{5}\) [6] for the tubing available.
Solve using dimensional analysis:
\[\frac{gtts}{min}~\times~\frac{10~gtts}{1~\cancel{mL}}~\times~\frac{250~\cancel{mL}}{2~\cancel{hour}}~\times~\frac{1~\cancel{hour}}{60~min}~=~\frac{10~gtts~\times~250~\times~1}{1~\times~2~\times~60~min}~=~{20.83~gtts/min} \]
Round drops to the nearest whole number for a final answer of 21 drops/minute.
Review the following module within
SWTC’s Dimensional Analysis in Nursing
page for more information about solving drop rates.
Module 1.11
Please practice flow rate calculations with the interactive activity below.
Query \(\PageIndex{1}\)
- “Microdrip Tubing” and “Macrodrip Tubing” by Deanna Hoyord, Chippewa Valley Technical College HPS Lab is licensed under CC BY 4.0 ↵
- Open RN Project . (2020, December 9). Micro Tubing Drip Chamber Video. [Video]. YouTube. Licensed under CC BY 3.0 . https://youtu.be/reH50eHpmSQ ↵
- “0.9% Sodium Chloride 500 ml” by Deanna Hoyord, Chippewa Valley Technical College HPS Lab is licensed under CC BY 4.0 ↵
- “Microdrip Tubing” by Deanna Hoyord, Chippewa Valley Technical College is licensed under CC BY 4.0 ↵
- “0.9% Sodium Chloride 250 ml” by Deanna Hoyord, Chippewa Valley Technical College HPS Lab is licensed under CC BY 4.0 ↵
- “Macrodrip Tubing” by Deanna Hoyord, Chippewa Valley Technical College is licensed under CC BY 4.0 ↵
- RegisteredNurseRN. (2015, September 28). Dosage calculations for nursing students on IV drip rate factors made easy (Video 4) . [Video]. YouTube. All rights reserved. Video used with permission. https://youtu.be/W5VIc6f0fBA ↵


6.4: References
Centers for Disease Control and Prevention. (2014a, August 8). Tourette syndrome overview. Retrieved from http :// www . cdc . gov / ncbddd / tourette / documents / tourette – fact – sheet _ photo . pdf . pdf
Centers for Disease Control and Prevention. (2014b, April 17). Alcohol use in pregnancy. Retrieved from http :// www . cdc . gov / ncbddd / fasd / alcohol – use . html
Centers for Disease Control and Prevention. (2014c, September 12). An alcohol-free pregnancy is the best choice for your baby. Retrieved from http :// www . cdc . gov / ncbddd / fasd / documents / fasdbrochure _ final . pdf
Centers for Disease Control and Prevention. (2015a, January 12). Facts about cerebral palsy. Retrieved from http :// www . cdc . gov / ncbddd / cp / facts . html
Centers for Disease Control and Prevention. (2015b, February 11). Milestone moments. Retrieved from http :// www . cdc . gov / ncbddd / actearly / pdf / parents _ pdfs / milestonemomentseng 508. pdf
Centers for Disease Control and Prevention. (2015c, February 12). Developmental disabilities. Retrieved from http :// www . cdc . gov / ncbddd / developmentaldisabilities /
Centers for Disease Control and Prevention. (2015d, February 12). Facts about intellectual disability. Retrieved from http :// www . cdc . gov / ncbddd / actearly / pdf / parents _ pdfs / IntellectualDisability . pdf
Centers for Disease Control and Prevention. (2015e, February 24). Facts about ASD. Retrieved from http :// www . cdc . gov / ncbddd / autism / facts . html
Centers for Disease Control and Prevention. (2015f, March 9). Facts about fragile X syndrome. Retrieved from http :// www . cdc . gov / ncbddd / actearly / pdf / parents _ pdfs / fragile _ x . pdf
Centers for Disease Control and Prevention. (2015g, April 8). ADHD fact sheet. Retrieved from http :// www . cdc . gov / ncbddd / adhd / documents / adhdfactsheetenglish . pdf
Leahy, W., Fuzy, J., & Grafe, J. (2013). Providing home care : A textbook for home health aides (4th ed.). Albuquerque, NM: Hartman.


1.24.4: Window Management
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- Last updated
- Save as PDF
Windows 10 has several features that make it easier to multi-task and work with multiple windows at the same time.
Windows Display options
Windows organizes every running application into its namesake- windows. Windows may overlap each other. Windows 10 provides many ways to size your windows accordingly. The top right corner of every running window provides display options of minimizing the window, resize window/maximize the window, and close window/. Minimizing a window removes the window from view. It can be returned to its current view by selecting the running application in the taskbar. Maximizing a Window makes it Fullscreen. The resize feature returns it to an open running window. Closing a window removes the program from active memory.
Windows Snap
Windows Snap allows you to resize windows quickly and display two windows side by side. To perform Windows Snap, click and drag the desired window to the left or right until the cursor reaches the edge of the screen, then release the mouse. The window will snap into place. To unsnap a window, simply click and drag the window down If you have more than one window open when snapping, you'll be able to choose a window to snap to the other side of the screen automatically.
Windows Flip
Windows Flip can be used to switch between open windows. To do this, press and hold the Alt key on your keyboard, then press the Tab key. Continue pressing the Tab key until the desired window is selected.
Displaying the Windows 10 Desktop
Sometimes, a user can have so many windows open at once, they have trouble finding their desktop! To quickly see your desktop, click the bottom-right corner of the taskbar to minimize all open windows. You can do click the bottom right corner again to restore your windows (which were minimized) to their current location. In addition, there is a keyboard shortcut to do this. Simply press the Windows key and the letter “D” on the keyboard to toggle between your open applications and the Windows Desktop.


9.4: Spacetime Curvature
not one but two particles witness to gravitation
Splendid! And also simple! But isn’t Einstein’s view of motion too simple? We started out interested in the motion of a spaceship around Earth and in "gravitation." We seem to have ended up talking only about the motion of the satellite - or the proof mass - relative to a strictly local inertial reference frame, a trivially simple straightline motion. Where is there any evidence of "gravitation" to be seen in that? Nowhere.
This is the great lesson of Einstein: Spacetime is always and everywhere locally Lorentzian. No evidence of gravitation whatsoever is to be seen by following the motion of a single particle in a free-float frame.
One has to observe the relative acceleration of two particles slightly separated from each other to have any proper measure of a gravitational effect. Separated by how much? That depends on the region of spacetime and the sensitivity of the measuring equipment. Two ball bearings with a horizontal separation of 20 meters, dropped from a height of 315 meters above Earth’s surface with 0 initial relative velocity, hit the ground 8 seconds later ( \(24 \times 10^{8}\) meters of light-travel time later) with a separation that has been reduced by \(10^{-3}\) meter (Section 2.3). Two ball bearings with a vertical separation of 20 meters, dropped from a height of 315 meters with 0 initial relative velocity, in the same 8 seconds increase their separation by \(2 \times 10^{-3}\) meter. To measuring equipment unable to detect such small relative displacements the ball bearings count as moving in one and the same free-float reference frame. No evidence for gravitation is to be seen. More sensitive apparatus detects the tide-producing action of gravity - the accelerated shortening of horizontal separations parallel to Earth’s surface, the accelerated lengthening of vertical separations. Each tiny ball bearing still moves in a straight line in its own local free-float reference frame. But now - with the new precision - the region of validity of the one free-float reference frame does not reach out far enough to give a proper account of the motion of the other steel ball. The millimeter or two discrepancy is the way "gravity" manifests itself.
Tidal acceleration displays gravity as a local phenomenon. No mention here of the distance of the steel balls from the center of Earth! No mention here of acceleration relative to that center! The only accelerations that come into consideration are those of nearby particles relative to each other, the tidal accelerations described in the preceding paragraph.
These relative accelerations double when the separations are doubled. The true measure of the tide-producing effect has therefore the character of an acceleration per unit of separation. Let the acceleration be measured in meters of distance per meter of light-travel time per meter of light-travel time; that is, in units meters/meter \({ }^{2}\) or \(1 /\) meter \(\left[x=(1 / 2) a t^{2}\right.\) , so \(\left.a=2 x / t^{2}\right]\) . Then the measure of the tide-producing effect (different for different directions) has the units (acceleration/distance) or \(\left(1 / \mathrm{meter}^{2}\right)\) . In the example, in the two horizontal directions this quantity has the value \(\{2(-0.001\) meter \(\left.) /\left(24 \times 10^{8} \text { meter }\right)^{2}\right] / 20\) meter \(=-17.36 \times 10^{-24}\) meter \(^{-2}\) and in the vertical direction twice the value and the opposite sign: \(+34.72 \times 10^{-24}\) meter \(^{-2}\) . The tide-producing effect is small but it is real and it is observable. Further, it is a locally defined quantity. And Einstein tells us that we must focus our attention on locally defined quantities if we want a simple description of nature.
Einstein says more: This tide-producing effect does not require for its explanation some mysterious force of gravitation, propagated through spacetime and additional to the structure of spacetime. Instead, it can and should be described in terms of the geometry of spacetime itself as the curvature of spacetime.
Though Einstein speaks of four-dimensional spacetime, his concepts of curvature can be illustrated in terms of two-dimensional geometry on the surface of a sphere.


2 .
What types of decelerations are caused by cord compression?
5 .
List the advantages and disadvantages of intermittent auscultation.
6 .
List the advantages and disadvantages of external FHR monitoring.
7 .
List the advantages and disadvantages of internal FHR monitoring.
9 .
Repetitive variable decelerations are noted on an FHR tracing. What condition might be the cause of this periodic change in the FHR pattern?
10 .
The fetus has been diagnosed with growth restriction. What should the nurse expect regarding fetal reserve?
11 .
Uterine contractions are occurring frequently. The nurse counts 7 contractions in a 10-minute period. What periodic change is expected, and what is the cause?
12 .
What nursing actions does the nurse perform when a prolapsed cord is suspected?
13 .
What nursing action does the nurse perform when the nurse observes a prolonged deceleration?
14 .
How does the nurse determine the success of nursing actions performed to address abnormal FHR tracings?
15 .
What is always the priority step in intrauterine resuscitation when oxytocin is in use?


27.7: Thin Film Interference
Learning Objectives
By the end of this section, you will be able to:
- Discuss the rainbow formation by thin films.
The bright colors seen in an oil slick floating on water or in a sunlit soap bubble are caused by interference. The brightest colors are those that interfere constructively. This interference is between light reflected from different surfaces of a thin film; thus, the effect is known as thin film interference . As noticed before, interference effects are most prominent when light interacts with something having a size similar to its wavelength. A thin film is one having a thickness \(t\) smaller than a few times the wavelength of light, \(\lambda\). Since color is associated indirectly with \(\lambda\) and since all interference depends in some way on the ratio of \(\lambda\) to the size of the object involved, we should expect to see different colors for different thicknesses of a film like in Figure \(\PageIndex{1}\).
What causes thin film interference? Figure 2 shows how light reflected from the top and bottom surfaces of a film can interfere. Incident light is only partially reflected from the top surface of the film (ray 1). The remainder enters the film and is itself partially reflected from the bottom surface. Part of the light reflected from the bottom surface can emerge from the top of the film (ray 2) and interfere with light reflected from the top (ray 1). Since the ray that enters the film travels a greater distance, it may be in or out of phase with the ray reflected from the top. However, consider for a moment, again, the bubbles in Figure \(\PageIndex{1}\). The bubbles are darkest where they are thinnest. Furthermore, if you observe a soap bubble carefully, you will note it gets dark at the point where it breaks. For very thin films, the difference in path lengths of ray 1 and ray 2 in Figure \(\PageIndex{2}\) is negligible; so why should they interfere destructively and not constructively? The answer is that a phase change can occur upon reflection. The rule is as follows:
When light reflects from a medium having an index of refraction greater than that of the medium in which it is traveling, a \(180^{\circ}\) phase change (or a \(\lambda /2\)) occurs.
In the film in Figure \(\PageIndex{2}\) is a soap bubble (essentially water with air on both sides), then there is a \(\lambda / 2\) shift for ray 1 and none for ray 2. Thus, when the film is very thin, the path length difference between the two rays is negligible, they are exactly out of phase, and destructive interference will occur at all wavelengths and so the soap bubble will be dark here.
The thickness of the film relative to the wavelength of light is the other crucial factor in thin film interference. Ray 2 in Figure \(\PageIndex{2}\) travels a greater distance than ray 1. For light incident perpendicular to the surface, ray 2 travels a distance approximately \(2t\) farther than Ray 1. When this distance is an integral or half-integral multiple of the wavelength in the medium (\(\lambda_{n} = \lambda / n\), where \(\lambda\) is the wavelength in vacuum and \(n\) is the index of refraction), constructive or destructive interference occurs, depending also on whether there is a phase change in either ray.
Example \(\PageIndex{1}\): Calculating Non-reflective Lens Coating Using Thin Film Interference
Sophisticated cameras use a series of several lenses. Light can reflect from the surfaces of these various lenses and degrade image clarity. To limit these reflections, lenses are coated with a thin layer of magnesium fluoride that causes destructive thin film interference. What is the thinnest this film can be, if its index of refraction is 1.38 and it is designed to limit the reflection of 550-nm light, normally the most intense visible wavelength? The index of refraction of glass is 1.52.
Strategy
Refer to Figure \(\PageIndex{2}\) and use \(n_{1} = 1.00\) for air, \(n_{2} = 1.38\), and \(n_{3} = 1.52\). Both ray 1 and ray 2 will have a \(\lambda / 2\) shift upon reflection. Thus, to obtain destructive interference, ray 2 will need to travel a half wavelength farther than ray 1. For rays incident perpendicularly, the path length difference is \(2t\).
Solution
To obtain destructive interference here,
\[2t = \frac{\lambda_{n_{2}}}{2}, \nonumber\]
where \(\lambda_{n_{2}}\) is the wavelength in the film and is given by \(\lambda_{n_{2}} = \frac{\lambda}{n_{2}}\).
Thus,
\[2t = \frac{\lambda / n_{2}}{2}.\nonumber\]
Solving for \(t\) and entering known values yields
\[ \begin{align*} t &= \frac{\lambda / n_{2}}{4} \\[4pt] &= \frac{\left(550 nm \right) / 1.38}{4} \\[4pt] &= 99.6 nm. \end{align*}\]
Discussion
Films such as the one in this example are most effective in producing destructive interference when the thinnest layer is used, since light over a broader range of incident angles will be reduced in intensity. These films are called non-reflective coatings; this is only an approximately correct description, though, since other wavelengths will only be partially cancelled. Non-reflective coatings are used in car windows and sunglasses.
Thin film interference is most constructive or most destructive when the path length difference for the two rays is an integral or half-integral wavelength, respectively. That is, for rays incident perpendicularly, \(2t = \lambda_{n}, 2\lambda_{n}, 3\lambda_{n},...\) or \(2t = \lambda_{n}/2, 3\lambda_{n}/2, 5\lambda_{n}/2,...\) To know whether interference is constructive or destructive, you must also determine if there is a phase change upon reflection. Thin film interference thus depends on film thickness, the wavelength of light, and the refractive indices. For white light incident on a film that varies in thickness, you will observe rainbow colors of constructive interference for various wavelengths as the thickness varies.
Example \(\PageIndex{2}\): Soap Bubbles: More Than One Thickness can be Constructive
- What are the three smallest thicknesses of a soap bubble that produce constructive interference for red light with a wavelength of 650 nm? The index of refraction of soap is taken to be the same as that of water.
- What three smallest thicknesses will give destructive interference?
Strategy and Concept:
Use Figure \(\PageIndex{2}\) to visualize the bubble. Note that \(n_{1} = n_{3} = 1.00\) for air, and \(n_{2} = 1.333\) for soap (equivalent to water). There is a \(\lambda / 2\) shift for ray 1 reflected from the top surface of the bubble, and no shift for ray 2 reflected from the bottom surface. To get constructive interference, then, the path length difference (\(2t\)) must be a half-integral multiple of the wavelength -- the first three being \(\lambda_{n}/2, 3\lambda_{n}/2\), and \(5\lambda_{n}/2\). To get destructive interference, the path length difference must be an integral multiple of the wavelength -- the first three being \(0, \lambda_{n},\) and \(2\lambda_{n}\).
Solution (a):
Constructive interference occurs here when
\[2t_{c} = \frac{\lambda_{n}}{2}, \frac{3\lambda_{n}}{2}, \frac{5\lambda_{n}}{2},...\label{27.8.4}\]
The smallest constructive thickness \(t_{c}\) thus is
\[ \begin{align*} t_{c} &= \frac{\lambda_{n}}{4} = \frac{\lambda / n}{4} \\[4pt] &= \frac{\left(650 nm\right) / 1.333}{4} \label{27.8.5} \\[4pt] &= 122 nm. \end{align*}\]
The next thickness that gives constructive interference is \(t'_{c} = 3\lambda_{n}/4\), so that
\[t'_{c} = 366 nm. \label{27.8.6} \nonumber\]
Finally, the third thickness producing constructive interference is \(t''_{c} \lt 5\lambda_{n} / 4\), so that
\[t''_{c} = 610 nm. \label{27.8.7} \nonumber\]
Solution (b):
For destructive interference, the path length difference here is an integral multiple of the wavelength. The first occurs for zero thickness, since there is a phase change at the top surface. That is,
\[t_{d} = 0. \label{27.8.8}\]
The first non-zero thickness producing destructive interference is
\[2t'_{d} = \lambda_{n}. \label{27.8.9}\]
Substituting known values gives
\[t'_{d} = \frac{\lambda_{n}}{2} = \frac{\lambda / n}{2} = \frac{\left(650 nm \right) / 1.333}{2} \label{27.8.10}\] \[= 244 nm.\] Finally, the third destructive thickness is \(2t''_{d} = 2\lambda_{n}\), so that
\[t''_{d} = \lambda_{n} = \frac{\lambda}{n} = \frac{650 nm}{1.333} \label{27.8.11}\] \[= 488 nm.\]
Discussion:
If the bubble was illuminated with pure red light, we would see bright and dark bands at very uniform increases in thickness. First would be a dark band at 0 thickness, then bright at 122 nm thickness, then dark at 244 nm, bright at 366 nm, dark at 488 nm, and bright at 610 nm. If the bubble varied smoothly in thickness, like a smooth wedge, then the bands would be evenly spaced.
Another example of thin film interference can be seen when microscope slides are separated (Figure \(\PageIndex{3}\)). The slides are very flat, so that the wedge of air between them increases in thickness very uniformly. A phase change occurs at the second surface but not the first, and so there is a dark band where the slides touch. The rainbow colors of constructive interference repeat, going from violet to red again and again as the distance between the slides increases. As the layer of air increases, the bands become more difficult to see, because slight changes in incident angle have greater effects on path length differences. If pure-wavelength light instead of white light is used, then bright and dark bands are obtained rather than repeating rainbow colors.
An important application of thin film interference is found in the manufacturing of optical instruments. A lens or mirror can be compared with a master as it is being ground, allowing it to be shaped to an accuracy of less than a wavelength over its entire surface. Figure \(\PageIndex{4}\) illustrates the phenomenon called Newton’s rings, which occurs when the plane surfaces of two lenses are placed together. (The circular bands are called Newton’s rings because Isaac Newton described them and their use in detail. Newton did not discover them; Robert Hooke did, and Newton did not believe they were due to the wave character of light.) Each successive ring of a given color indicates an increase of only one wavelength in the distance between the lens and the blank, so that great precision can be obtained. Once the lens is perfect, there will be no rings.
The wings of certain moths and butterflies have nearly iridescent colors due to thin film interference. In addition to pigmentation, the wing’s color is affected greatly by constructive interference of certain wavelengths reflected from its film-coated surface. Car manufacturers are offering special paint jobs that use thin film interference to produce colors that change with angle. This expensive option is based on variation of thin film path length differences with angle. Security features on credit cards, banknotes, driving licenses and similar items prone to forgery use thin film interference, diffraction gratings, or holograms. Australia led the way with dollar bills printed on polymer with a diffraction grating security feature making the currency difficult to forge. Other countries such as New Zealand and Taiwan are using similar technologies, while the United States currency includes a thin film interference effect.
MAKING CONNECTIONS: TAKE-HOME EXPERIMENT -- THIN FILM INTERFERENCE
One feature of thin film interference and diffraction gratings is that the pattern shifts as you change the angle at which you look or move your head. Find examples of thin film interference and gratings around you. Explain how the patterns change for each specific example. Find examples where the thickness changes giving rise to changing colors. If you can find two microscope slides, then try observing the effect shown in Figure 3. Try separating one end of the two slides with a hair or maybe a thin piece of paper and observe the effect.
Problem-Solving Strategies for Wave Optics
- Step 1. Examine the situation to determine that interference is involved. Identify whether slits or thin film interference are considered in the problem.
- Step 2. If slits are involved, note that diffraction gratings and double slits produce very similar interference patterns, but that gratings have narrower (sharper) maxima. Single slit patterns are characterized by a large central maximum and smaller maxima to the sides.
- Step 3. If thin film interference is involved, take note of the path length difference between the two rays that interfere. Be certain to use the wavelength in the medium involved, since it differs from the wavelength in vacuum. Note also that there is an additional \(\lambda / 2\) phase shift when light reflects from a medium with a greater index of refraction.
- Step 4. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. Draw a diagram of the situation. Labeling the diagram is useful.
- Step 5. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
- Step 6. Solve the appropriate equation for the quantity to be determined (the unknown), and enter the knowns. Slits, gratings, and the Rayleigh limit involve equations.
- Step 7. For thin film interference, you will have constructive interference for a total shift that is an integral number of wavelengths. You will have destructive interference for a total shift of a half-integral number of wavelengths. Always keep in mind that crest to crest is constructive whereas crest to trough is destructive.
- Step 8. Check to see if the answer is reasonable: Does it make sense? Angles in interference patterns cannot be greater than \(90^{\circ}\), for example.
Summary
- Thin film interference occurs between the light reflected from the top and bottom surfaces of a film. In addition to the path length difference, there can be a phase change.
- When light reflects from a medium having an index of refraction greater than that of the medium in which it is traveling, a \(180^{\circ}\) phase change (or a \(\lambda /2\) shift) occurs.
Glossary
- thin film interference
- interference between light reflected from different surfaces of a thin film