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| | 17 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Here are a few practice exercises before getting started with this lesson.
Which of the tables represents the linear equation in two variables y=2x+4?
Dominika is curious about the amount of time she spends going to and from school. One day, she decides to measure the time it takes her to get to school on foot and compare it to the amount of time it takes her to arrive home by bus. She drew the following graph for her trip to school.
In order to understand the relationship between two quantities and make predictions about their behavior, it is important to be able to compare and quantify how one changes with respect to the other. These ideas define the concept of rate of change.
The rate of change ROC is a ratio used to compare how a variable changes in relation to another variable. It is determined by dividing the change in the output variable y by the change in the input variable x. For any ordered pairs (x_1,y_1) and (x_2,y_2), the rate of change is calculated using the following formula.
Rate of Change&= Change in y/Change inx [1em] Rate of Change&= y_2-y_1/x_2-x_1
The Greek letter Δ (Delta) is commonly used to represent a difference. This leads to an alternative way of writing the formula for ROC.
Rate of Change = Δ y/Δ x
| Output Units | Input Units | Rate of Change Units | Possible Interpretation |
|---|---|---|---|
| meters | seconds | meters/second | Car speed over a certain period of time |
| bacteria | hours | bacteria/hour | Growth rate of bacteria in an experiment |
| U.S. dollars | hours | U.S. dollars/hour | Worker's hourly wage |
As mentioned above, the rate of change is a quantity that compares the change in the output variable to the input variable. These are also known as the dependent and independent variables, respectively.
In the context of functions, the input is often referred to as the independent variable because it can be chosen arbitrarily from the domain. Conversely, the output is called the dependent variable because its value depends on the value of the independent variable. For instance, if the price of oranges is $ 2.50 per pound, the total cost is determined by the product of the unit price and the weight in pounds. ccccc Cost & & Unit Price & & Weight [0.4em] y & = & 2.50 & * & x
As shown, the total cost of oranges depends on how many pounds of fruit are purchased. Therefore, the cost of oranges y is the dependent variable and the number of pounds purchased x is the independent variable.Diego is taking the bus back to his hometown. During the trip, he sees that there are signs by the side of the road that indicate the distance from the city.
He keeps track of some of the signs and notes the time taken to reach them. He puts the information together using a table of values.
| Time (minutes) | Distance (kilometers) |
|---|---|
| 0 | 0 |
| 10 | 15 |
| 20 | 30 |
| 30 | 45 |
| 40 | 60 |
| 50 | 75 |
| 60 | 90 |
Find the rate of change between (0,0) and (2,50).
y= 50x+30 The y-variable represents the distance in kilometers that the car can travel and the x-variable the gas in gallons. Find the rate of change between x=1 and x=3.
The rate of change is defined as the ratio of the change in the dependent variable to the change of the independent variable. Rate of Change = Δ y/Δ x Note that the distance covered by the car depends on the time elapsed. On the other hand, time passes by without relation to the the car and can take any non-negative value without restrictions. Therefore, the dependent variable in this case is the distance, and time is independent. Let t represent the time and d the distance. Rate of Change = Δ d/Δ t = d_2-d_1/t_2-t_1 Recall that the unit of the rate of change is the ratio of the unit of the dependent variable to the unit of the independent variable. Rate of Change Units = kilometers/minute Therefore, the unit of the rate of change is kmmin, which quantifies kilometers traveled per minute. The rate of change can be interpreted as the speed of the bus. Interpretation for the Rate of Change [0.1cm] The speed of the car during a specific period of time
Now, by using the rate of change formula together with the information from the table, the rate of change of the car between 10 and 20 minutes will be calculated. Notice that the corresponding distances are 15 and 30 kilometers, respectively.
Substitute ( 10,15) & ( 20,30)
The rate of change between 10 and 20 minutes is 1.5 kmmin. By following the same procedure it is possible to find the rate of change between the other points of interest. The following table summarizes these results for the rate of change corresponding to the required time intervals.
| Rate of Change = d_2-d_1/t_2-t_1 | ||
|---|---|---|
| Times | Substitute | Evaluate |
| 10- 20 min | 30 km- 15 km/20 min- 10 min | 1.5 kmmin |
| 0- 60 min | 90 km- 0 km/60 min- 0 min | 1.5 kmmin |
Rate of Change = Δ s/Δ t Notice that in a coordinate pair, the first coordinate represents the value of the independent variable, while the second coordinate corresponds to the dependent variable. Using this information, the rate of change between the points (0,0) and (2,50) will be calculated for the given graph.
Substitute ( 0,0) & ( 2,50)
The acceleration in the first 2 seconds is 25 kmhr * s. By following the same procedure, the rate of change for the other pair of points can be found. The following table summarizes the results for the rate of change of the graph between the specified pair of points.
| Rate of Change = y_2-y_1/x_2-x_1 | ||
|---|---|---|
| Points | Substitute | Evaluate |
| ( 0, 0),( 2, 50) | 50- 0/2- 0 | 25 |
| ( 2, 50),( 4, 100) | 100- 50/4- 2 | 25 |
It was found that the y-value corresponding to x=1 is y=80. Therefore, the coordinate pair (1,80) is a point on the graph of the line representing the given linear equation. By repeating this process, the specific points of interest can be found.
| Linear Equation: y = 2x+3 | ||
|---|---|---|
| x | Substitute | Point |
| x = 1 | y &= 50( 1)+30 y &= 80 | ( 1, 80) |
| x = 3 | y &= 50( 3)+30 y &= 180 | ( 3, 180) |
| x = 4 | y &= 50( 4)+30 y &= 230 | ( 4, 230) |
| x = 7 | y &= 50( 7)+30 y &= 380 | ( 7, 380) |
Now that the points are known, all that is left to do is to repeat the same procedure as in Part B. The following table summarizes the results for the pair of points of interest.
| Rate of Change = y_2-y_1/x_2-x_1 | ||
|---|---|---|
| Points | Substitute | Evaluate |
| ( 1, 80),( 3, 180) | 180- 80/3- 1 | 50 |
| ( 4, 230),( 7, 380) | 380- 230/7- 4 | 50 |
| ( 1, 80),( 7, 380) | 380- 80/7- 1 | 50 |
It is interesting to note that the rate of change is the same between all the pairs of points used. This occurred in Parts A and B as well.
The following applet shows different representations for the data of two variables. Practice finding the rate of change between the indicated values. If the answer is not an integer number, round it to two decimal places.
Any line in the coordinate plane has a constant rate of change between any pair of its points. This can be checked by moving the points in the following applet.
Any linear equation in two variables can written in the following form where y is the dependent variable, x is the independent variable, and m and b can be any real numbers. y=mx+b Now, since this is a linear equation in two variables, all of its solutions lie on a line in the coordinate plane. In what follows, two arbitrary points (x_1,y_1) and (x_2,y_2) will be used, and the rate of change between them will be found by using the rate of change formula. Rate of Change = y_2-y_1/x_2-x_1 Because these arbitrary points are solutions to the linear equation, they satisfy the equation. Therefore, it is possible to find an explicit expression for y_1 and y_2 in terms of x_1 and x_2, respectively. y_1 = mx_1+b y_2 = mx_2+b Now that the explicit form for y_1 and y_2 is known, the rate of change between these points can be calculated.
y_1= mx_1+b, y_2= mx_2+b
As stated before, m is a real number and, therefore, a constant. It has been found that, no matter which two points on the line are used, the rate of change between them will always be a constant value. This constant value m is called the slope of the line.
Consider a set of data points for which the rate of change between every pair of consecutive points is a constant value m. Consider two consecutive points (x_1,y_1) and (x_2,y_2).
| Rate of Change | Rewrite |
|---|---|
| y_2-y_1/x_2-x_1= m | y_2-y_1 = m (x_2-x_1) |
Note that since the rate of change between every pair of consecutive points is constant, they all can be rewritten in a similar way.
| Rate of Change | Rewrite |
|---|---|
| y_2-y_1/x_2-x_1= m | y_2-y_1 = m (x_2-x_1) |
| y_3-y_2/x_3-x_2= m | y_3-y_2 = m (x_3-x_2) |
| ... | ... |
| y_n-y_(n-1)/x_n-x_(n-1)= m | y_n-y_(n-1) = m (x_n-x_(n-1)) |
Next, it will be shown that every point from the data set satisfies the following linear equation in two variables. y-y_1 = m (x-x_1) This will be shown first for the third point from the data set, (x_3,y_3). The expression on the right-hand side of the equation will be rewritten by adding and subtracting y_2 to obtain an identity.
Since (x_3,y_3) and (x_2,y_2), and (x_2,y_2) and (x_1,y_1), are pairs of consecutive points, the differences y_3-y_2 and y_2-y_1 can be rewritten in terms of the constant rate of change m and the corresponding x-values.
y_3-y_2= m(x_3-x_2), y_2-y_1= m(x_2-x_1)
Therefore, the third point (x_3,y_3) satisfies the linear equation y-y_1=m(x-x_1). Now it will be shown that the fourth point of the data set, (y_4,x_4), also satisfies the equation.
Recall that it is known that y_4-y_3=m(x_4-x_3) and, from the previous result, it is also known that y_3-y_1 = m (x_3-x_1).
y_4-y_3= m (x_4-x_3), y_3-y_1= m (x_3-x_1)
As shown, (x_4,y_4) satisfies the equation as well. This process can be repeated with all the points, and all will satisfy the linear equation y-y_1=m(x-x_1). And since every point satisfies the equation, every point is a solution. Therefore, every point of the data set lies on the line representing this linear equation.
Diego's brother is very interested in crypotcurrencies and is telling Diego about how their values changed in the first weeks of the last year. One of them is of special interest to him because its graph is a straight line. He tried to print some information to show Diego, but his printer broke before it could print the complete tables of values for the cryptocurrencies.
Diego's brother wants to identify the table in which the values have a linear relationship. For each table, the left column represents the week number and the right column represents the value of the cryptocurrency in dollars. Which of the three tables best describes a linear relation?
Since the relation represented in Table A increases by equal differences in equal intervals, the data can be modeled with a linear relation. Now, the same procedure will be applied to other tables.
Since the differences between consecutive y-values are not constant for equal intervals, Table B and Table C do not satisfy the condition. As such, they do not represent linear relations.
The following table shows the relationship between two variables, x and y. Look at the data carefully and determine whether the relationship between x and y can be described by linear equation.
In a previous example, it was shown how substituting a specific x-value into a linear equation in two variables determined an associated y-value. The set of all the points obtained in this way represent a line in the coordinate plane. This idea is depicted below by showing some points for a particular linear equation.
This type of relation in which an input is assigned only one specific output is known as a function. To each input, typically represented by the independent variable x, the function assigns an output represented as f(x). In the example above y depends on x. This can be denoted explicitly by using function notation. ccc Equation & & Function y=2x+1 & & f(x)=2x+1 In general, functions can assign an output value to the input value in many different ways and their graphs can be very different. A particular function whose graph is a line is known as a linear function. Therefore, a linear equation can represent a linear function.
A linear function is a function with a constant rate of change. A linear function can be represented by a linear equation in two variables. Graphically, a linear function is a nonvertical line.
When working with linear functions, there are two points of special interest. These are the points where the line crosses the axes, known as intercepts.
The x-intercept of a line is the x-coordinate of the point where the line crosses the x-axis. The y-intercept of a line is the y-coordinate of the point where the line crosses the y-axis. The y-intercept of an equation is also known as its initial value.
Diego wants to make sure he has enough data available on his smartphone to stream a series while taking a trip with his family. He visits a local branch of his service provider and updates to the Unlimited Plan,
which costs an initial service fee and a special price per gigabyte of data used. The graph represents the costs of the service plan.
What does the y-intercept represent?
y = - 10 x +100 In this function, y is the percentage of charge remaining in the battery after x number of hours using the phone for streaming videos. What is the x-intercept for this linear function? Write the answer as a single number.
Therefore, the y-intercept is 4. Since this is the cost of the service when 0 gigabytes of data are used, x=0, it can be concluded that $4 is the initial service fee.
y= 0
LHS-100=RHS-100
.LHS /(-10).=.RHS /(-10).
Rearrange equation
It has been found that the x-intercept is 10. The given linear equation relates the number of hours of streaming x to the percentage of battery charge left on the phone y. This means that the x-intercept — the number of hours after which y=0 — represents the time it takes for the battery to drain completely.
The following applet alternates between a graph and a linear equation, both of which represent linear functions. Identify the indicated intercept of the given function and write the answer as a single number.
The rate of change of a nonlinear function is not constant — it may even change from point to point. To measure the change for this type of function, an average rate of change is defined by averaging the total change of the function over a specific interval. The following graph illustrates how the average rate of change depends on the interval considered.
The slope of the linear function can be calculated by using the points common to both functions, (x_1,y_1) and (x_2,y_2). slope = y_2-y_1/x_2-x_1 The y-values that correspond to x_1 and x_2 are also the nonlinear function's values f(x_1) and f(x_2), respectively. Then, the slope formula can be rewritten and identified as the average rate of change of the function f(x) over the interval [x_1, x_2].
Average Rate of Change = f(x_2)-f(x_1)/x_2-x_1Diego decides to fly back home after visiting his family in his hometown. He is very excited as the airplane readies for takeoff. The plane gets into position at the beginning of a long, straight runaway surface and starts accelerating. Diego enjoys the view through the window.
| Time (seconds) | Distance (meters) |
|---|---|
| 0 | 0 |
| 3 | 18 |
| 6 | 72 |
| 9 | 162 |
| 12 | 288 |
| 15 | 450 |
| 18 | 648 |
| 21 | 882 |
According to this table, which of the following options best approximates the speed of the plane at takeoff?
What is the average speed during the first 28 seconds of the landing?
Average Rate of Change = d_2-d_1/t_2-t_1 Note that since distance covered depends on the time value but time has no restriction, distance d is the dependent variable and time t is the independent variable. Now, to best approximate the speed at takeoff, the interval used for finding the average rate of change should be as short and as close to takeoff as possible.
By doing this, the calculation will not average the takeoff speed with the speeds at the beginning and the middle of the motion, and a better approximation for the speed at time of takeoff alone will be obtained. As indicated by the arrows on the table above, the coordinate pairs to be used are (18,648) and (21,882).
Substitute ( 18,648) & ( 21,882)
It was found that in the interval of time just before takeoff, the average speed of the plane was 78 ms.
Substitute ( 0,0) & ( 28,1176)
The average speed during the first 28 seconds of the landing process is 42 ms. The average speed during the smaller interval, t=24 seconds to t=28 seconds, can be found in the same way.
Substitute ( 24,1152) & ( 28,1176)
During this period, the average speed of the plane was just 6 ms, as this calculation included only a short period of time closer to when the plane stopped moving. It does not include the data corresponding to times when the plane had a greater speed, as the average speed for the first 28 seconds of the landing process does.
The following applet shows different representations for the data of two variables. Practice finding the average rate of change between the indicated values. If the answer is not an integer number, round it to two decimal places.
It is important to mention that the graphs of some functions can be comprised of straight lines with different slopes. The average rate of change is used for these functions as well, since the graph is not a single straight line and the rate of change is not constant.
Dominika is curious about the amount of time she spends going to and from school. One day, she decides to measure the time it takes her to get to school on foot and compare it to the amount of time it takes her to arrive home by bus. She drew the following graph for her trip to school.
Average Speed: 100 meters per minute
Rate of Change Units = meters/minute Because the rate of change is quantifying how much distance is covered per minute, it can be interpreted as Dominika's speed. Note that the rate of change is not constant. For this reason, the average speed can then be found by using the the average rate of change formula. Average Rate of Change = d_2-d_1/t_2-t_1 Because the average speed during the whole trip is desired, the interval of interest is t_1 = 0 to t_2 = 10 minutes. From the graph, the corresponding distances at times t_1 and t_2 can be found as follows. t_1= 0 & ⇒ d_1 = 0 t_2= 10 & ⇒ d_2 = 1200 The average speed will now be calculated.
Substitute ( 0,0) & ( 10,1200)
Therefore, Dominika's average speed during the her walk to school was 120 meters per minute. Note that this speed does not correspond to the rate of change of any of the lines on the graph. Instead, this is the rate of change of the line passing through the initial and final points.
After this time, she got on the bus and arrived back home at t=12 minutes. Recall that the distance between her home and her school is 1200 meters.
Now that the graph is complete, the average speed for the return trip will be calculated. This time, the interval of interest is t_1 = 0 to t_2 = 12 minutes. From the graph, the corresponding distances at times t_1 and t_2 can be found as follows. t_1= 0 & ⇒ d_1 = 0 t_2= 12 & ⇒ d_2 = 1200 The average speed will now be calculated.
Substitute ( 0,0) & ( 12,1200)
The average speed on the trip home was 100 meters per minute. Once more, note that this speed is the rate of change of the line passing through the initial and final points of the graph.
Average speed going to school on foot 120 mmin [1.5em] Average speed returning from school by bus 100 mmin
This does not mean that Dominika's average speed was greater than the bus's. For the calculation of the return trip, the 9 minutes that Dominika waited — and therefore did not move — were averaged together with the motion of the bus. The following graph shows the waiting period and the bus ride in different colors.The average speed of the bus can be found by repeating the same calculation but considering only the last 3 minutes of the return trip — that is, using only the interval from t_1=9 to t_2=12. t_1= 9 & ⇒ d_1 =0 t_2=12 & ⇒ d_2 =1200 Now the calculation will be repeated using these values.
Substitute ( 9,0) & ( 12,1200)
This result shows that the actual average speed of the bus is 400 meters per minute. This value corresponds to the rate of change of the last segment in the previous graph. It is also considerably faster than Dominika's average walking speed of 120 meters per minute.
In this lesson, the concepts of x- and y-intercepts, as well as linear functions, were introduced. However, it is important to note that the concept of a function is more general and that there are also nonlinear functions.
Other main ideas of the lesson were the rate of change and the average rate of change. These basic concepts lead to the definition of instantaneous rate of change. Unlike the average rate of change, which is an approximation for an interval, the instantaneous rate of change gives the exact value of the rate of change at a specific point.
The following table models Tadeo's income y, in dollars, for working x hours.
Knowing that Tadeo's earnings follow a linear relationship, find the value of B.
We want to find the missing values in our table of values. Let's begin by trying to identify a pattern of change between consecutive rows.
We can see that the change in x from one row to the next one is 1, but we cannot determine the change in the y-values for consecutive rows. However, because the values follow a liner relationship, the rate of change is constant. This means that the rate of change between x=4 and x=5 is the same as the rate of change between x=4 and x=7. Rate of change = 81.55-46.60/7-4 Let's solve for the rate of change.
The rate of change is 11.65, which means that the y-values increase by 11.65 between consecutive rows. In context, this means that Tadeo earns $11.65 per hour. His wage is constant, so we can find the value of A by adding 11.65 to 44.60. Then we can increase A by 11.65 again to find the value of B.
We found that A= 58.25 and B= 69.90.