Unit 5 Functions and Volume — Unit Plan

Let’s say we have an input-output rule that gives exactly one output for each allowable input. Then we say the output depends on the input, or the output is a function of the input.

For example, the area of a square is a function of the side length because the area can be found from the side length by squaring it. So when the input is 10 cm, the output is 100 cm².

Sometimes we might have two different rules that describe the same function. As long as we always get the same single output from any given input, the rules describe the same function.

We can sometimes represent functions with equations. For example, the area, $A$, of a circle is a function of the radius, $r$, and we can express this with this equation: $\displaystyle A=\pi r^2$

We can also draw a diagram to represent this function:

In this case, we think of the radius, $r$, as the input and the area of the circle, $A$, as the output. For example, if the input is a radius of 10 cm, then the output is an area of $100\pi$ cm², or about 314 cm². Because this is a function, we can find the area, $A$, for any given radius, $r$.

Since $r$ is the input, we say that it is the independent variable, and since $A$ is the output, we say that it is the dependent variable.

We sometimes get to choose which variable is the independent variable in the equation. For example, if we know that

$\displaystyle 10A-4B=120$

then we can think of $A$ as a function of $B$ and write

$\displaystyle A=0.4B+12$

or we can think of $B$ as a function of $A$ and write

$\displaystyle B=2.5A-30$

Here is the graph showing Noah's run.

A graph in the coordinate plane, horizontal, distance in meters, 0 to 24 by threes, vertical, time in seconds, 0 to 10 by ones. The graph begins at the origin and steadily increases as it moves right, passing through the labeled point at ( 18 comma 6 ).

The time in seconds since he started running is a function of the distance he has run. The point $(18,6)$ on the graph tells us that the time it takes him to run 18 meters is 6 seconds. The input is 18 and the output is 6.

The graph of a function is all the coordinate pairs, (input, output), plotted in the coordinate plane. By convention, we always put the input first, which means that the inputs are represented on the horizontal axis, and the outputs are represented on the vertical axis.

Functions are all about getting outputs from inputs. For each way of representing a function—equation, graph, table, or verbal description—we can determine the output for a given input.

Let’s say we have a function represented by the equation $y = 3x +2$, where $y$ is the dependent variable and $x$ is the independent variable. If we wanted to find the output that goes with 2, we could input 2 into the equation for $x$ and find the corresponding value of $y$. In this case, when $x$ is 2, $y$ is 8 since $3\boldcdot 2 + 2=8$.

If we had a graph of this function instead, then the coordinates of points on the graph would be the input-output pairs.

So we would read the $y$-coordinate of the point on the graph that corresponds to a value of 2 for $x$. Looking at the following graph of a function, we can see the point $(2,8)$ on it, so the output is 8 when the input is 2.

The graph of a line in the coordinate plane with the origin labeled “O”. The horizontal axis has the numbers negative 1 through 2 indicated and there are vertical gridlines between each integer. The vertical axis has the numbers negative 2 through 8, in increments of 2, indicated, and there are horizontal grid lines in between each integer. The line begins to the right of the y axis and below the x axis. It slants upward and to the right passing through the point with coordinates negative 1 comma negative 1, crosses the y axis at 2, and passes through the indicated point labeled 2 comma 8.

Suppose a car is traveling at 30 miles per hour. The relationship between the time in hours and the distance in miles is a proportional relationship.

More generally, if we represent a linear function with an equation like $y = mx + b$, then $b$ is the initial value (which is 0 for proportional relationships), and $m$ is the rate of change of the function.

If we represent a linear function in a different way, say with a graph, we can use what we know about graphs of lines to find the $m$ and $b$ values and, if needed, write an equation.

Water has different boiling points at different elevations. At 0 m above sea level, the boiling point is $100^\circ$ C. At 2,500 m above sea level, the boiling point is $91.3^\circ$ C. If we assume the boiling point of water is a linear function of elevation, we can use these two data points to calculate the slope of the line: $\displaystyle m=\frac{91.3-100}{2,500-0}=\frac{\text-8.7}{2,500}$

This slope means that for each increase of 2,500 m, the boiling point of water decreases by $8.7^\circ$ C.

Next, we already know the $y$-intercept is $100^\circ$ C from the first point, so a linear equation representing the data is $\displaystyle y=\frac{\text-8.7}{2,500}x+100$

This equation is an example of a mathematical model. A mathematical model is a mathematical object, like an equation, a function, or a geometric figure, that we use to represent a real-life situation. Sometimes a situation can be modeled by a linear function. We have to analyze the information we are given and use judgment about whether using a linear model is a reasonable thing to do. We must also be aware that the model may make imprecise predictions or may only be appropriate for certain ranges of values.

Testing our model for the boiling point of water, it accurately predicts that at an elevation of 1,000 m above sea level (when $x=1,000$), water will boil at $96.5^\circ$ C (since $y=\frac{\text-8.7}{2,500}\boldcdot 1000+100=96.5$). For higher elevations, the model is not as accurate, but it is still close. At 5,000 m above sea level, it predicts $82.6^\circ$ C, which is $0.6^\circ$ C off the actual value of $83.2^\circ$ C. At 9,000 m above sea level, it predicts $68.7^\circ$ C, which is about $3^\circ$ C less than the actual value of $71.5^\circ$ C. The model continues to be less accurate at even higher elevations since the relationship between the boiling point of water and elevation isn’t linear, but for the elevations in which most people live, it’s pretty good.

The volume of a three-dimensional figure, like a jar or a room, is the amount of space the shape encloses. We can measure volume by finding the number of equal-size volume units that fill the figure without gaps or overlaps.

For example, we might say that a room has a volume of 1,000 cubic feet or that a pitcher can carry 5 gallons of water. We could even measure volume of a jar by the number of beans it could hold, though a bean count is not really a measure of the volume in the same way that a cubic centimeter is because there is space between the beans. (The number of beans that fit in the jar do depend on the volume of the jar, so it is an okay estimate when judging the relative sizes of containers.)

In earlier grades, we studied three-dimensional figures with flat faces that are polygons. We learned how to calculate the volumes of rectangular prisms. Now we will study three-dimensional figures with circular faces and curved surfaces: cones, cylinders, and spheres.

To help us see the shapes better, we can use dotted lines to represent parts that we wouldn't be able to see if a solid physical object were in front of us. For example, if we think of the cylinder in this picture as representing a tin can, the dotted arc in the bottom half of that cylinder represents the back side of the circular base of the can. What objects could the other figures in the picture represent?

We can find the volume of a cylinder with radius $r$ and height $h$ using two ideas we've seen before:

• The volume of a rectangular prism is the result of multiplying the area of its base by its height.
• The base of the cylinder is a circle with radius $r$, so the base area is $\pi r^2$.

Remember that $\pi$ is the number we get when we divide the circumference of any circle by its diameter. The value of $\pi$ is approximately 3.14.

Just like a rectangular prism, the volume of a cylinder is the area of the base times the height. For example, consider a cylinder whose radius is 2 cm and whose height is 5 cm.

​The base has an area of $4\pi$ cm² (since $\pi\boldcdot 2^2=4\pi$), so the volume is $20\pi$ cm³ (since $4\pi \boldcdot 5 = 20\pi$). Using 3.14 as an approximation for $\pi$, we can say that the volume of the cylinder is approximately 62.8 cm³.

In general, the base of a cylinder with radius $r$ units has area $\pi r^2$ square units. If the height is $h$ units, then the volume $V$ in cubic units is $V=\pi r^2h.$

If a cone and a cylinder have the same base and the same height, then the volume of the cone is $\frac{1}{3}$ of the volume of the cylinder.

For example, the cylinder and cone shown here both have a height of 7 feet and a base with radius 3 feet.

The cylinder has a volume of $63\pi$ cubic feet since $\pi \boldcdot 3^2 \boldcdot 7 = 63\pi$.

The cone has a volume that is $\frac13$ of that, or $21\pi$ cubic feet.

If the radius for both solids is $r$ and the height for both solids is $h$, then the volume of the cylinder is $\pi r^2h$. That means that the equation to give the volume, $V$, of the cone is $\displaystyle V=\frac{1}{3}\pi r^2h.$

As we saw with cylinders, the volume $V$ of a cone depends on the radius $r$ of the base and the height $h$:

$\displaystyle V=\frac13 \pi r^2h$

If we know the radius and height, we can find the volume. If we know the volume and one of the dimensions (either radius or height), we can find the other dimension.

For example, imagine a cone with a volume of $64\pi$ cm³, a height of 3 cm, and an unknown radius $r$. From the volume formula, we know:

$\displaystyle 64 \pi = \frac{1}{3}\pi r^2 \boldcdot 3$

Looking at the structure of the equation, we can see that $r^2 = 64$, so the radius must be 8 cm.

Now imagine a different cone with a volume of $18 \pi$ cm³, a radius of 3 cm, and an unknown height $h$. Using the formula for the volume of the cone, we know:

$\displaystyle 18 \pi = \frac{1}{3} \pi 3^2h$

So, the height must be 6 cm. Can you see why?

Think about a sphere with radius $r$ units that fits snugly inside a cylinder. The cylinder must then also have a radius of $r$ units and a height of $2r$ units. Using what we have learned about volume, the cylinder has a volume of $\pi r^2 h = \pi r^2 \boldcdot (2r)$, which is equal to $2\pi r^3$ cubic units.

We know from an earlier lesson that the volume of a cone with the same base and height as a cylinder has $\frac{1}{3}$ of the volume. In this example, such a cone has a volume of $\frac{1}{3} \boldcdot \pi r^2 \boldcdot 2r$, or $\frac{2}{3} \pi r^3$ cubic units.

If we filled the cone and sphere with water and then poured that water into the cylinder, the cylinder would be completely filled. That means the volume of the sphere and the volume of the cone add up to the volume of the cylinder. In other words, if $V$ is the volume of the sphere, then

$\displaystyle V +\frac{2}{3}\pi r^3= 2 \pi r^3$

This leads to the formula for the volume of the sphere,

$\displaystyle V = \frac{4}{3} \pi r^3$

The formula $V=\frac43 \pi r^3$ gives the volume of a sphere with radius $r$.

We can use the formula to find the volume of a sphere with a known radius. For example, if the radius of a sphere is 6 units, then the volume would be $\frac{4}{3} \pi (6)^3 = 288\pi$, or approximately 905 cubic units. 

We can also use the formula to find the radius of a sphere if we only know its volume. For example, if we know that the volume of a sphere is $36 \pi$ cubic units but we don't know the radius, then this equation is true: 

$\displaystyle 36\pi=\frac43\pi r^3$

That means that $r^3 = 27$, so the radius $r$ has to be 3 units in order for both sides of the equation to have the same value.

Many common objects, from water bottles to buildings to balloons, are similar in shape to rectangular prisms, cylinders, cones, or spheres—or even combinations of these shapes! Using the volume formulas for these shapes allows us to compare the volume of different types of objects, sometimes with surprising results.

For example, a cube-shaped box with side length 3 centimeters holds less than a sphere with radius 2 centimeters because the volume of the cube is 27 cubic centimeters ($3^3 = 27$) and the volume of the sphere is around 33.51 cubic centimeters ($\frac43\pi \boldcdot  2^3 \approx 33.51$).