4.4: Slope and Rate of Change
Learning Objectives
- Find positive and negative slopes.
- Recognize and find slopes for horizontal and vertical lines.
- Understand rates of change.
- Interpret graphs and compare rates of change.
Introduction
Wheelchair ramps at building entrances must have a slope between \begin{align*}\frac{1}{16}\end{align*}
We come across many examples of slope in everyday life. For example, a slope is in the pitch of a roof, the grade or incline of a road, or the slant of a ladder leaning on a wall. In math, we use the word slope to define steepness in a particular way.
\begin{align*}\text{Slope} = \frac{\text{distance moved vertically}}{\text{distance moved horizontally}}\end{align*}
To make it easier to remember, we often word it like this:
\begin{align*}\text{Slope} = \frac{\text{rise}}{\text{run}}\end{align*}
In the picture above, the slope would be the ratio of the height of the hill to the horizontal length of the hill. In other words, it would be \begin{align*}\frac{3}{4}\end{align*}
If the car were driving to the right it would climb the hill - we say this is a positive slope. Any time you see the graph of a line that goes up as you move to the right, the slope is positive.
If the car kept driving after it reached the top of the hill, it might go down the other side. If the car is driving to the right and descending, then we would say that the slope is negative.
Here’s where it gets tricky: If the car turned around instead and drove back down the left side of the hill, the slope of that side would still be positive. This is because the rise would be -3, but the run would be -4 (think of the \begin{align*}x-\end{align*}
Find the Slope of a Line
A simple way to find a value for the slope of a line is to draw a right triangle whose hypotenuse runs along the line. Then we just need to measure the distances on the triangle that correspond to the rise (the vertical dimension) and the run (the horizontal dimension).
Example 1
Find the slopes for the three graphs shown.
Solution
There are already right triangles drawn for each of the lines - in future problems you’ll do this part yourself. Note that it is easiest to make triangles whose vertices are lattice points (i.e. points whose coordinates are all integers).
a) The rise shown in this triangle is 4 units; the run is 2 units. The slope is \begin{align*}\frac{4}{2}= 2\end{align*}
b) The rise shown in this triangle is 4 units, and the run is also 4 units. The slope is \begin{align*}\frac{4}{4}= 1\end{align*}
c) The rise shown in this triangle is 2 units, and the run is 4 units. The slope is \begin{align*}\frac{2}{4}= \frac{1}{2}\end{align*}
Example 2
Find the slope of the line that passes through the points (1, 2) and (4, 7).
Solution
We already know how to graph a line if we’re given two points: we simply plot the points and connect them with a line. Here’s the graph:
Since we already have coordinates for the vertices of our right triangle, we can quickly work out that the rise is \begin{align*}7 - 2 = 5\end{align*}
If you look again at the calculations for the slope, you’ll notice that the 7 and 2 are the \begin{align*}y-\end{align*}
Slope between \begin{align*}(x_1, y_1)\end{align*}
or \begin{align*}m= \frac{\Delta y}{\Delta x}\end{align*}
In the second equation the letter \begin{align*}m\end{align*}
Example 3
Find the slopes of the lines on the graph below.
Solution
Look at the lines - they both slant down (or decrease) as we move from left to right. Both these lines have negative slope.
The lines don’t pass through very many convenient lattice points, but by looking carefully you can see a few points that look to have integer coordinates. These points have been circled on the graph, and we’ll use them to determine the slope. We’ll also do our calculations twice, to show that we get the same slope whichever way we choose point 1 and point 2.
For Line \begin{align*}A\end{align*}
\begin{align*}&(x_1, y_1) = (-6, 3) \qquad (x_2, y_2) = (5, -1) && (x_1, y_1) = (5, -1) \qquad (x_2, y_2) = (-6, 3)\\
&m = \frac{y_2-y_1}{x_2-x_1} = \frac{(-1)-(3)}{(5)-(-6)} = \frac{-4}{11} \approx -0.364 && m = \frac{y_2-y_1}{x_2-x_1} = \frac{(3)-(-1)}{(-6)-(5)} = \frac{4}{-11} \approx -0.364\end{align*}
For Line \begin{align*}B\end{align*}
\begin{align*}&(x_1, y_1) = (-4, 6) \qquad (x_2, y_2) = (4, -5) && (x_1, y_1) = (4, -5) \qquad (x_2, y_2) = (-4, 6)\\
& m = \frac{y_2-y_1}{x_2-x_1} = \frac{(-5)-(6)}{(4)-(-4)} = \frac{-11}{8} = -1.375 && m = \frac{y_2-y_1}{x_2-x_1} = \frac{(6)-(-5)}{(-4)-(4)} = \frac{11}{-8} = -1.375\end{align*}
You can see that whichever way round you pick the points, the answers are the same. Either way, Line \begin{align*}A\end{align*}
Khan Academy has a series of videos on finding the slope of a line, starting at http://www.youtube.com/watch?v=hXP1Gv9IMBo.
Find the Slopes of Horizontal and Vertical lines
Example 4
Determine the slopes of the two lines on the graph below.
Solution
There are 2 lines on the graph: \begin{align*}A (y = 3)\end{align*}
Let’s pick 2 points on line \begin{align*}A\end{align*}
\begin{align*}m= \frac{y_2 - y_1} {x_2 - x_1} = \frac{(3)-(3)}{(5)-(-4)} = \frac {0}{9} = 0.\end{align*}
If you think about it, this makes sense - if \begin{align*}y\end{align*} doesn’t change as \begin{align*}x\end{align*} increases then there is no slope, or rather, the slope is zero. You can see that this must be true for all horizontal lines.
Horizontal lines (\begin{align*}y\end{align*} = constant) all have a slope of 0.
Now let’s consider line \begin{align*}B\end{align*}. If we pick the points \begin{align*}(x_1 , y_1) = (5, -3)\end{align*} and \begin{align*}(x_2 , y_2) = (5, 4)\end{align*}, our slope equation is \begin{align*}m = \frac{y_2 -y_1}{x_2 - x_1} = \frac{(4)-(-3)}{(5)-(5)} = \frac{7}{0}\end{align*}. But dividing by zero isn’t allowed!
In math we often say that a term which involves division by zero is undefined. (Technically, the answer can also be said to be infinitely large—or infinitely small, depending on the problem.)
Vertical lines \begin{align*}(x =\end{align*} constant) all have an infinite (or undefined) slope.
Find a Rate of Change
The slope of a function that describes real, measurable quantities is often called a rate of change. In that case the slope refers to a change in one quantity \begin{align*}(y)\end{align*} per unit change in another quantity \begin{align*}(x)\end{align*}. (This is where the equation \begin{align*}m= \frac{\Delta y}{\Delta x}\end{align*} comes in—remember that \begin{align*}\Delta y\end{align*} and \begin{align*}\Delta x\end{align*} represent the change in \begin{align*}y\end{align*} and \begin{align*}x\end{align*} respectively.)
Example 5
A candle has a starting length of 10 inches. 30 minutes after lighting it, the length is 7 inches. Determine the rate of change in length of the candle as it burns. Determine how long the candle takes to completely burn to nothing.
Solution
First we’ll graph the function to visualize what is happening. We have 2 points to start with: we know that at the moment the candle is lit (\begin{align*}time = 0\end{align*}) the length of the candle is 10 inches, and after 30 minutes (\begin{align*}time = 30\end{align*}) the length is 7 inches. Since the candle length depends on the time, we’ll plot time on the horizontal axis, and candle length on the vertical axis.
The rate of change of the candle’s length is simply the slope of the line. Since we have our 2 points \begin{align*}(x_1, y_1) = (0, 10)\end{align*} and \begin{align*}(x_2, y_2) = (30, 7)\end{align*}, we can use the familiar version of the slope formula:
\begin{align*}\text{Rate of change} = \frac{y_2 - y_1}{x_2 -x_1} = \frac{(7 \ \text{inches})-(10 \ \text{inches})}{(30 \ \text{minutes})-(0 \ \text{minutes})} = \frac{-3 \ \text{inches}}{30 \ \text{minutes}}= -0.1 \ \text{inches per minute}\end{align*}
Note that the slope is negative. A negative rate of change means that the quantity is decreasing with time—just as we would expect the length of a burning candle to do.
To find the point when the candle reaches zero length, we can simply read the \begin{align*}x-\end{align*}intercept off the graph (100 minutes). We can use the rate equation to verify this algebraically:
\begin{align*}\text{Length burned} & = \text{rate} \times \text{time}\\ 10 & = 0.1 \times 100\end{align*}
Since the candle length was originally 10 inches, our equation confirms that 100 minutes is the time taken.
Example 6
The population of fish in a certain lake increased from 370 to 420 over the months of March and April. At what rate is the population increasing?
Solution
Here we don’t have two points from which we can get \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates for the slope formula. Instead, we’ll need to use the alternate formula, \begin{align*}m=\frac{\Delta y}{\Delta x}\end{align*}.
The change in \begin{align*}y-\end{align*}values, or \begin{align*}\Delta y\end{align*}, is the change in the number of fish, which is \begin{align*}420 - 370 = 50\end{align*}. The change in \begin{align*}x-\end{align*}values, \begin{align*}\Delta x\end{align*}, is the amount of time over which this change took place: two months. So \begin{align*}\frac{\Delta y}{\Delta x} = \frac{50 \ \text{fish}}{2 \ \text{months}}\end{align*}, or 25 fish per month.
Interpret a Graph to Compare Rates of Change
Example 7
The graph below represents a trip made by a large delivery truck on a particular day. During the day the truck made two deliveries, one taking an hour and the other taking two hours. Identify what is happening at each stage of the trip (stages A through \begin{align*}E\end{align*}).
Solution
Here are the stages of the trip:
a) The truck sets off and travels 80 miles in 2 hours.
b) The truck covers no distance for 2 hours.
c) The truck covers \begin{align*}(120 - 80) = 40 \ \text{miles}\end{align*} in 1 hour.
d) The truck covers no distance for 1 hour.
e) The truck covers -120 miles in 2 hours.
Let’s look at each section more closely.
A. \begin{align*}\text{Rate of change} = \frac{\Delta y}{\Delta x}=\frac{80 \ \text{miles}}{2 \ \text{hours}}=40\ \text{miles per hour}\end{align*}
Notice that the rate of change is a speed—or rather, a velocity. (The difference between the two is that velocity has a direction, and speed does not. In other words, velocity can be either positive or negative, with negative velocity representing travel in the opposite direction. You’ll see the difference more clearly in part E.)
Since velocity equals distance divided by time, the slope (or rate of change) of a distance-time graph is always a velocity.
So during the first part of the trip, the truck travels at a constant speed of 40 mph for 2 hours, covering a distance of 80 miles.
B. The slope here is 0, so the rate of change is 0 mph. The truck is stationary for one hour. This is the first delivery stop.
C. \begin{align*}\text{Rate of change} = \frac{\Delta y}{\Delta x}=\frac{(120 - 80) \ \text{miles}}{(4-3) \ \text{hours}} = 40 \ \text{miles per hour}\end{align*}. The truck is traveling at 40 mph.
D. Once again the slope is 0, so the rate of change is 0 mph. The truck is stationary for two hours. This is the second delivery stop. At this point the truck is 120 miles from the start position.
E. \begin{align*}\text{Rate of change} = \frac{\Delta y}{\Delta x}=\frac{(0-120) \ \text{miles}}{(8-6) \ \text{hours}}=\frac{-120 \ \text{miles}}{2 \ \text{hours}}=-60 \ \text{miles per hour}\end{align*}. The truck is traveling at negative 60 mph.
Wait – a negative speed? Does that mean that the truck is reversing? Well, probably not. It’s actually the velocity and not the speed that is negative, and a negative velocity simply means that the distance from the starting position is decreasing with time. The truck is driving in the opposite direction – back to where it started from. Since it no longer has 2 heavy loads, it travels faster (60 mph instead of 40 mph), covering the 120 mile return trip in 2 hours. Its speed is 60 mph, and its velocity is -60 mph, because it is traveling in the opposite direction from when it started out.
Lesson Summary
- Slope is a measure of change in the vertical direction for each step in the horizontal direction. Slope is often represented as “\begin{align*}m\end{align*}”.
- Slope can be expressed as \begin{align*}\frac{\text{rise}}{\text{run}}\end{align*}, or \begin{align*}\frac{\Delta y}{\Delta x}\end{align*}.
- The slope between two points \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2)\end{align*} is equal to \begin{align*}\frac{y_2 - y_1}{x_2 -x_1}\end{align*}.
- Horizontal lines (where \begin{align*}y = a\end{align*} constant) all have a slope of 0.
- Vertical lines (where \begin{align*}x = a\end{align*} constant) all have an infinite (or undefined) slope.
- The slope (or rate of change) of a distance-time graph is a velocity.
Review Questions
- Use the slope formula to find the slope of the line that passes through each pair of points.
- (-5, 7) and (0, 0)
- (-3, -5) and (3, 11)
- (3, -5) and (-2, 9)
- (-5, 7) and (-5, 11)
- (9, 9) and (-9, -9)
- (3, 5) and (-2, 7)
- (2.5, 3) and (8, 3.5)
- For each line in the graphs below, use the points indicated to determine the slope.
- For each line in the graphs above, imagine another line with the same slope that passes through the point (1, 1), and name one more point on that line.
- The graph below is a distance-time graph for Mark’s three and a half mile cycle ride to school. During this ride, he rode on cycle paths but the terrain was hilly. He rode slower up hills and faster down them. He stopped once at a traffic light and at one point he stopped to mend a punctured tire. The graph shows his distance from home at any given time. Identify each section of the graph accordingly.
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