This is the second post in the series on working with linear models. Here is the link to the first post: Using Linear Models – Graphing a Linear Model Using the x- and y-Intercepts
In this post I will discuss writing a linear model or linear equation to represent a problem situation. Once the equation is written, I will identify and interpret the y-intercept. The embedded video will model all of the steps necessary to solve the problem. Here is the problem.
A spring has a length of 8 cm when a 20-g mass is hanging at the bottom end. Each additional gram stretches the spring another 0.15 cm. Write an equation for the length y of the spring as a function of the mass x of the attached weight. Graph the equation. Interpret the y-intercept.
Understanding the first sentence of the problem and how the values are related is key to writing the equation. From the first sentence it can be concluded that the length of the spring will depend on the amount of weight attached to the spring. This means the weight is the independent variable and the length of the spring is the dependent variable. The problem even goes as far as stating that length should be represented by the variable y and the weight is represented with the variable x. x traditionally is the independent variable and y is the dependent variable, as is the case in this problem.
The second sentence reads, “Each additional gram stretches the spring another 0.15 cm.” This relationship between the additional weight and the amount of stretch is a rate of change. The spring will expand 0.15 cm for every 1 gram added. Remember, the slope of a line is also a rate comparing the change in y-values over the change in x-values and because the y-values are representing the length of the spring and the x-values are representing the weight added, we can ascertain that the slope is 0.15cm/1gram or m = 0.15.
It is given in the problem, the spring has a length of 8 cm when a 20-g mass is hanging at the bottom end. This can be written as the ordered pair (20, 8 ) and can be used with the slope m = 0.15 to write a linear equation for this problem. To write this equation, it will be easiest to start with the point-slope form of a linear equation:
y – y1 = m(x – x1) –> Point-Slope Form of a Linear Equation (1)
Substituting the point (20,8) and the slope m = 0.15 into equation 1,
y – 8 = 0.15(x – 20). (2)
To you can make a graph from this form, but since interpreting the y-intercept is a part of the problem using some algebra to put the equation into slope-intercept form will be useful. Remember, the slope-intercept form of a linear equation is given by y = mx + b, where m is the slope and b is the y-intercept. To put equation 2 into slope-intercept form, first 0.15 must be distributed to give:
y – 8 = 0.15x – 3. (3)
Finally, adding 8 to both sides of equation 3 will give the slope-intercept form:
Y = 0.15x + 5. (4)
The y-intercept is (0,5) and means that when the spring has now weight attached , it is 5 cm long. As before, see the video for the graphing portion of this problem.
A related example to this spring problem, what mass would be needed to stretch the spring to a length of 9.5 cm? This is very simple to complete if you understand the meaning of the variables. Y represents the length of the spring and x represents the amount of weight attached to the spring. For this problem, you must find HOW MUCH WEIGHT must be attached to stretch the spring 9.5 cm. Thus, you are given the length of the spring or the y-value and you must solve for the x-value.
To complete this task, substitute 9.5 for y in equation 4 to get:
9.5 = 0.15x + 5. (5)
To solve for x, subtract 5 from both sides,
4.5 = 0.15x (6)
And divide both sides by 0.15 to find,
x = 30 (7)
Relating this answer to the problem, it will take 30 grams to stretch the spring to 9.5 cm.
Filed under: Algebra 2, Equations and Graphs, Using Linear Models | Tagged: Algebra 2, graphing a linear model using x- and y- intercepts, interpreting the y-intercept, point-slope form of a linear equation, problem solving, slope-intercept form of a linear equation, writing a linear equation, writing a linear model | 3 Comments »