Subsection 5.4.1 Proportion and Variation
Two variables are directly proportional (or just proportional) if the ratios of their corresponding values are always equal. Consider the functions described in the tables below. The first table shows the price of gasoline as a function of the number of gallons purchased.
Gallons of gasoline 
Total price 
Price/ Gallons 
\(4\) 
\(\$9.60\) 
\(\dfrac{9.60}{4}=2.40\) 
\(6\) 
\(\$14.40\) 
\(\dfrac{14.40}{6}=2.40\) 
\(8\) 
\(\$19.20\) 
\(\dfrac{19.20}{8}=2.40\) 
\(12\) 
\(\$28.80\) 
\(\dfrac{28.80}{12}=2.40\) 
\(15\) 
\(\$36.00\) 
\(\dfrac{36.00}{15}=2.40\) 
Years 
Population 
People/Years 
\(10\) 
\(432\) 
\(\dfrac{432}{10}\approx 43\) 
\(20\) 
\(932\) 
\(\dfrac{932}{20}\approx 47\) 
\(30\) 
\(2013\) 
\(\dfrac{2013}{30}\approx 67\) 
\(40\) 
\(4345\) 
\(\dfrac{4345}{40}\approx 109\) 
\(50\) 
\(9380\) 
\(\dfrac{9380}{50}\approx 188\) 
\(60\) 
\(20,251\) 
\(\dfrac{20,251}{60}\approx 338\) 
The ratio \(\dfrac{\text{total price}}{\text{number of gallons}}\text{,}\) or price per gallon, is the same for each pair of values in the first table. This agrees with everyday experience: The price per gallon of gasoline is the same no matter how many gallons you buy. Thus, the total price of a gasoline purchase is directly proportional to the number of gallons purchased.
The second table shows the population of a small town as a function of the town’s age. The ratio \(\dfrac{\text{number of people}}{\text{number of years}}\) gives the average rate of growth of the population in people per year. You can see that this ratio is not constant; in fact, it increases as time goes on. Thus, the population of the town is not proportional to its age.
Checkpoint 5.4.1. QuickCheck 1.
When does a table represent proportional variables?
If it has a constant slope
If it includes the point \((0,0)\)
If the ratio output/input is constant
If each output is double the previous one
The graphs of the two functions described above by tables (the gasoline price and the small town population tables) are shown below.
We see that the price, \(P\text{,}\) of a fillup is a linear function of the number of gallons, \(g\text{,}\) purchased. This should not be surprising if we write an equation relating the variables \(g\) and \(P\text{.}\) Because the ratio of their values is constant, we can write
\begin{equation*}
\frac{P}{g}= k
\end{equation*}
where \(k\) is a constant. In this example, the constant \(k\) is \(2.40\text{,}\) the price of gasoline per gallon. Solving for \(P\) in terms of \(g\text{,}\) we have
\begin{equation*}
P = kg = 2.40g
\end{equation*}
which we recognize as the equation of a line through the origin.
If \(y\) is proportional to \(x\text{,}\) we also say that \(y\) varies directly with \(x\text{,}\) and we make the following definition.
Direct Variation.
\(y\) varies directly with \(x\) if
\begin{equation*}
y = kx
\end{equation*}
where \(k\) is a positive constant called the constant of variation.
Example 5.4.2.
The circumference, \(C\text{,}\) of a circle varies directly with its radius, \(r\text{,}\) because
\begin{equation*}
C = 2\pi r
\end{equation*}
The constant of variation is \(2\pi\text{,}\) or about 6.28.
The amount of interest, \(I\text{,}\) earned in one year on an account paying 7% simple interest, varies directly with the principal, \(P\text{,}\) invested, because
\begin{equation*}
I = 0.07P
\end{equation*}
In the opening example, we saw that the price of gasoline, \(P\text{,}\) is a linear function of the number of gallons purchased. From the definition above, we see that any direct variation defines a linear function of the form
\begin{equation*}
y = f(x) = kx
\end{equation*}
Compareing this equation to the standard form for a linear function, \(y = b + mx\text{,}\) we see that the constant term, \(b\text{,}\) is zero, so the graph of a direct variation passes through the origin. The positive constant \(k\) in the equation \(y = kx\) is just the slope of the graph, so it tells us how rapidly the graph increases.
Checkpoint 5.4.3. Practice 1.
Which of the graphs above could represent direct variation? Explain why.
(a). The graph is a straight line that increases.
(b). The graph is a straight line through the origin.
(c). The graph is a straight line that decreases.
None of the above
Solution.
(b): The graph is a straight line through the origin.
Checkpoint 5.4.4. QuickCheck 2.
A table describes direct variation if the ratio of corresponding entries is .
A graph describes direct variation if it is a .
An equation describes direct variation if it has the form .
If two variables vary directly, we may also say that they are .
Subsection 5.4.2 The Scaling Property of Direct Variation
Because the graph of \(y=kx\) passes through the origin, direct variation has the following scaling property: if we double the value of \(x\text{,}\) then the value of \(y\) will double also. In fact, increasing \(x\) by any factor causes \(y\) to increase by the same factor. Look again at the table for the price of buying gasoline. Doubling the number of gallons of gas purchased, say, from \(4\) gallons to \(8\) gallons or from \(6\) gallons to \(12\) gallons, causes the total price to double also.
Checkpoint 5.4.5. QuickCheck 3.
You invest $800 for one year at 7% simple interest. The interest earned is
\begin{equation*}
I=0.07(800)= \fillinmath{XXXXXXXXXX}
\end{equation*}
If you increase the investment by a factor of, say, 1.6, to 1.6(800) or $1280, the interest will be
\begin{equation*}
I=0.07(1280)=\fillinmath{XXXXXXXXXX}
\end{equation*}
The original interest is increased by a factor of .
Example 5.4.6.
Tuition at Woodrow University is $400 plus $30 per unit. Is the tuition proportional to the number of units you take?
Imogen makes a 15% commission on her sales of environmentally friendly products marketed by her coop. Do her earnings vary directly with her sales?
Solution.

Let \(u\) represent the number of units you take, and let \(T(u)\) represent your tuition. Then
\begin{equation*}
T(u) = 400 + 30u
\end{equation*}
Thus, \(T(u)\) is a linear function of \(u\text{,}\) but the \(T\)intercept is \(400\text{,}\) not \(0\text{.}\) Your tuition is not proportional to the number of units you take, so this is not an example of direct variation. You can check that doubling the number of units does not double the tuition. For example,
\begin{equation*}
T(6) = 400 + 30(6) = 580
\end{equation*}
and
\begin{equation*}
T(12) = 400 + 30(12) = 760
\end{equation*}
Tuition for \(12\) units is not double the tuition for \(6\) units. The graph of \(T(u)\) in figure (a) does not pass through the origin.
Let \(S\) represent Imogen’s sales, and let \(C(S)\) represent her commission. Then
\begin{equation*}
C(S) = 0.15S
\end{equation*}
Thus, \(C(S)\) is a linear function of \(S\) with a \(C\)intercept of zero, so Imogen’s earnings do vary directly with her sales.
Checkpoint 5.4.7. Practice 2.
Which table could represent direct variation? Explain why. (Hint: What happens to \(y\) if you multiply \(x\) by a constant?)
\(x\) 
\(1\) 
\(2\) 
\(3\) 
\(6\) 
\(8\) 
\(9\) 
\(y\) 
\(2.5\) 
\(5\) 
\(7.5\) 
\(15\) 
\(20\) 
\(22.5\) 
\(x\) 
\(2\) 
\(3\) 
\(4\) 
\(6\) 
\(8\) 
\(9\) 
\(y\) 
\(2\) 
\(3.5\) 
\(5\) 
\(7\) 
\(8.5\) 
\(10\) 
Solution.
(a): If we multiply \(x\) by a constant, \(y\) is also multiplied by the same constant.
Subsection 5.4.4 Direct Variation with a Power of \(x\)
In many situations, \(y\) is proportional to a power of \(x\text{,}\) instead of \(x\) itself.
Direct Variation with a Power.
\(y\) varies directly with a power of \(x\) if
\begin{equation*}
y = kx^n
\end{equation*}
where \(k\) and \(n\) are positive constants.
Example 5.4.11.
The surface area of a sphere varies directly with the square of its radius. A balloon of radius 5 centimeters has surface area \(100\pi\) square centimeters, or about 314 square centimeters.
Find a formula \(S=f(r)\) for the surface area of a sphere as a function of its radius \(~\alert{\text{[TK]}}\text{.}\)
What is the radius of a sphere whose surface area is \(200\pi\) square centimeters,or about 628 square centimeters?
Sketch a graph of the function.
Solution.
If \(S\) stands for the surface area of a sphere of radius \(r\text{,}\) then
\begin{equation*}
S = f(r) = kr^2
\end{equation*}
To find the constant of variation, \(k\text{,}\) we substitute the values of \(S\) and \(r\text{.}\)
\begin{align*}
\alert{100\pi} \amp = k(\alert{5})^2\\
4\pi \amp= k
\end{align*}
Thus, \(~S = f(r) = 4\pi r^2\text{.}\)
We solve the variation equation for \(r\) when \(S=200\pi\text{.}\)
\begin{align*}
\alert{200\pi} \amp = 4\pi r^2\\
50 \amp= r^2\\
\sqrt{50} \amp=r
\end{align*}
The radius is \(\sqrt{50}\) centimeters, or about 7.1 centimeters.

The graph has the shape of the basic function \(y=x^2\text{,}\) so all we need are a few points to "anchor" the graph. We know that \(f(5)=314\text{,}\) and \(f(1)= 4\pi \cdot 1^2\) or about \(12.6\text{.}\) A graph is shown below.
(Need graph here)
Checkpoint 5.4.12. Practice 4.
The volume of a sphere varies directly with the cube of its radius. A balloon of radius 5 centimeters has volume \(\dfrac{500\pi}{3}\) cubic centimeters, or about 524 cubic centimeters.
Find a formula for the volume, \(V\text{,}\) of a sphere as a function of its radius, \(r\text{.}\)
What is the radius of a balloon whose volume is \(\dfrac{1000\pi}{3}\) cubic centimeters, or about 1028 cubic centimeters?
Sketch a graph of the function.
Solution.
\(\displaystyle V=\dfrac{4}{3}\pi r^3\)
We solve the variation equation for \(r\) when \(V=dfrac{1000\pi}{3}\text{.}\)
\begin{align*}
\alert{dfrac{1000\pi}{3}} \amp = \dfrac{4}{3}\pi r^3\\
250 \amp= r^3\\
\sqrt[3]{250} \amp=r
\end{align*}
The radius is \(\sqrt[3]{250}\) centimeters, or about 6.3 centimeters.
(Need graph here)
In any example of direct variation, as the input variable increases through positive values, the output variable increases also. Thus, a direct variation is an increasing function, as we can see when we consider the graphs of some typical direct variations shown below.
Checkpoint 5.4.14. QuickCheck 5.
Decide whether each statement is true or false.
Every increasing function is a direct variation.
Every direct variation is an increasing function.
The graph of every direct variation is a straight line.
The graph of every direct variation passes through the origin.
Even without an equation, we can check whether a table of data describes direct variation or merely an increasing function. If \(y\) varies directly with \(x^n\text{,}\) then \(~y = kx^n~\text{,}\) or, equivalently,\(~\dfrac{y}{x^n} = k.~\)
Test for Direct Variation.
If the ratio \(~\dfrac{y}{x^n}~\) is constant, then \(y\) varies directly with \(x^n\text{.}\)
Example 5.4.15.
Delbert collects the following data and would like to know if \(y\) varies directly with the square of \(x\text{.}\) What should he calculate?
\(x\) 
\(2\) 
\(5\) 
\(8\) 
\(10\) 
\(12\) 
\(y\) 
\(6\) 
\(16.5\) 
\(36\) 
\(54\) 
\(76\) 
Solution.
If \(y\) varies directly with \(x^2\text{,}\) then \(y = kx^2\text{,}\) or \(\dfrac{y}{x^2}= k\text{.}\) Delbert should calculate the ratio \(\dfrac{y}{x^2}\) for each data point.
\(x\) 
\(2\) 
\(5\) 
\(8\) 
\(10\) 
\(12\) 
\(y\) 
\(6\) 
\(16.5\) 
\(36\) 
\(54\) 
\(76\) 
\(\dfrac{y}{x^2}\) 
\(1.5\) 
\(0.66\) 
\(0.56\) 
\(0.54\) 
\(0.53\) 
Because the ratio \(\dfrac{y}{x^2}\) is not constant, \(y\) does not vary directly with \(x^2\text{.}\)
Subsection 5.4.5 Scaling
If \(y\) varies directly with \(x\text{,}\) then doubling \(x\) causes \(y\) to double also. But what if \(y\) varies directly with a power of \(x\) :
Is the area of a \(16\)inch circular pizza double the area of an \(8\)inch pizza?
If you double the dimensions of a model of a skyscraper, will its weight double also?
You probably know that the answer to both of these questions is No. The area of a circle is proportional to the square of its radius, and the volume (and hence the weight) of an object is proportional to the cube of its linear dimension. Variation with a power of \(x\) produces a different scaling effect.
Example 5.4.16.
The Taipei 101 building is 1671 feet tall, and in 2006 it was the tallest skyscraper in the world. Show that doubling the dimensions of a model of the Taipei 101 building produces a model that weighs 8 times as much.
Solution.
The Taipei 101 skyscraper is approximately box shaped, so its volume is given by the product of its linear dimensions, \(V = lwh\text{.}\) The weight of an object is proportional to its volume, so the weight, \(W\text{,}\) of the model is
\begin{equation*}
W = klwh
\end{equation*}
where the constant \(k\) depends on the material of the model. If we double the length, width, and height of the model, then
\begin{align*}
W_\text{new} \amp = k(2l)(2w)(2h)\\
\amp = 2^3(klwh) = 8W_\text{old}
\end{align*}
The weight of the new model is \(2^3 = 8\) times the weight of the original model.
Checkpoint 5.4.17. Practice 6.
Use the formula for the area of a circle to show that doubling the diameter of a pizza quadruples its area.
Step 1: The formula for the area of a circle of radius \(r\) is \(A=\fillinmath{XXXXXX}\)
Step 2: If we double the diameter, the new radius is \(\fillinmath{XXXXXX}\)
Step 3: Substitute the new expression for the radius into the area formula to get the area of the new circle.
Solution.
\(A= \pi r^2\)
The new radius is \(2r\text{.}\)
The new area is \(\pi (2r)^2 = 4 \pi r^2 = 4A\)
In general, if \(y\) varies directly with a power of \(x\text{,}\) that is, if \(y = kx^n\text{,}\) then doubling the value of \(x\) causes \(y\) to increase by a factor of \(2^n\text{.}\) In fact, if we multiply \(x\) by any positive number \(c\text{,}\) then
\begin{align*}
y_\text{new} \amp =k(cx)^n\\
\amp =c^n(kx^n) = c^n(y_\text{old})
\end{align*}
so the value of \(y\) is multiplied by \(c^n\text{.}\)
We will call \(n\) the scaling exponent, and you will often see variation described in terms of scaling. For example, we might say that "the area of a circle scales as the square of its radius." (In many applications, the power \(n\) is called the scale factor, even though it is not a factor but an exponent.)