1 Reminders 1. A quiz will be given every other week starting next week. So a quiz will be given on January 0, February 1, February 27, March 1, March 27, April 10, and April 24. 2. Section Grade (a) Homework (50%) with lowest 2 scores dropped. i. Homework will be graded out of 1 point. You will receive 1 point if you make a good effort (attempt at least /4 of the assignment) and 0 points if not. ii. Turn in homework every Tuesday in discussion. (b) Quizzes (50%) with lowest score dropped. i. Quizzes will be graded out of 10 points. ii. Each quiz will cover the math you practiced in the homework since the last quiz. For example, the February 1th quiz will test material covered in the homework turned in on February 4th and February 1th. iii. On a quiz day, we will start class with any questions you have. Once all questions are answered, I will hand out the quiz and you ll have some specified amount of time to finish (between 20-0 minutes). Since the quiz will be given at the beginning of discussion on the Tuesdays which have quizzes (see schedule below), don t be late to these discussions! iv. If you require DSP accommodations please come discuss with me at least a couple days before the first quiz so we can make such accommodations. (c) Section grade will be curved so that the median score receives a B+.. It is up the student to notify the instructor or his/her GSI in writing before the end of the second week of the term of all foreseeable conflicts between the syllabus and scheduled extra-curricular activities. -Prof Slaman 4. All this information is on my website at mikaylalynnkelley.weebly.com 2 Goal of Today s Worksheet Think about lines! Slope is the most important characteristic of a line and so we ll spend time talking about what slope is, how to calculate it, and what you can do with it. Conceptual Questions 1. Define the following: (a) Linear equation: Page 1 of 7
Solution: An equation in two first-degree (i.e. no powers involved) variables which has a line as its graph. (b) The slope of a non-vertical line: Solution: The slope of a non-vertical line is the vertical change divided by the horizontal change as one travels alone the line. An important aspect of a line is that its slope is constant along the line. (c) The slope of a vertical line: Solution: The slope of a vertical line is undefined. (d) Slope-intercept form: Solution: A line with slope m and y-intercept b has the equation in slopeintercept form y = mx + b. (e) Point-slope form Solution: A line with slope m which passes through the point (x 1, y 1 ) has an equation in point slope form given by y y 1 = m(x x 1 ) 2. Answer the following questions. (a) Explain the benefits of point-slope form vs slope intercept form vs ax+by = c form. Solution: Point-slope form: a point on the line and the slope of the line is readily available to you. Slope intercept form: the y-intercept and the slope of the line is readily available to you. ax + by = c: this is the most general equation of line. This means that any line (including vertical lines) can be written in this form with a, b, c chosen appropriately. Verticle lines can be written by setting b = 0 (notice that the other two forms require b = 1). (b) What is the general equation of a vertical and horizontal line? Solution: The general equation of a vertical line is x = c (where c is some real number). The general equation of a horizontal line is y = c (where c is some Page 2 of 7
real number). A general equation for a line (resp. parabola, cubic, etc.) is one where there are variables like a, b, c,... that can be replaced to pick out any line (resp. parabola, cubic, etc). (c) Describe what a slope of 5/4 means. Solution: A line having slope 5/4 means that for every 4 x-units the line travels to the right, the line travels 5 y-units upwards. 4 Mathematical Questions 1. Linear Equations and Slope (a) Find an equation in slope-intercept form, ax + by = c form, and point-slope form of a line through ( 8, 1) with undefined slope. Solution: This is a bad/trick question. Only one person caught the issue! Vertical lines cannot be written in slope intercept or point-intercept form (see conceptual question (b) above). Set a = 1, b = 0, c = 8 and you have the equation of the line in ax + by = c form. (b) Find k so that the line through (4, 1) and (k, 2) is (a) perpendicular to 5x 2y = 1 Solution: We put the line in slope-intercept form: 5x 2y = 1 2y = 5x 1 y = 5 2 x + 1 2 So the line 5x 2y = 1 has slope 5. A line perpendicular to this line will 2 therefore have slope 2 (opposite reciprical of 5 ). Thus the slope of the 5 2 line which connects (4, 1) and (k, 2) must have slope 2. We look for k 5 such that this is the case: 2 ( 1) k 4 = 2 5 2(k 4) = 15 2k = 7 k = 7 2 (b) parallel to 2x + y = 6. Solution: We put the line in slope-intercept form: 2x + y = 6 y = 2x + 6 y = 2 x + 2 Page of 7
So the line 2x + y = 6 has slope 2. A line parallel to this line will have the same slope of 2. Thus the slope of the line which connects (4, 1) and (k, 2) must have slope 2. We look for k such that this is the case: 2 ( 1) k 4 = 2 2(k 4) = 9 2k = 1 k = 1 2 (c) Use slopes to show that the square with vertices ( 2, 5), (4, 5), (4, 1), ( 2, 1) has diagonals that are perpendicular. Solution: Draw a picture! Diagonal 1 has endpoints ( 2, 1) and (4, 5). The slope of this diagonal is therefore 5 ( 1) 4 ( 2) = 1 Diagonal 2 has endpoints ( 2, 5) and (4, 1). therefore 1 5 4 ( 2) = 1 The slope of this diagonal is Since the slopes of these diagonals are opposite reciprocals the diagonals are perpendicular to eachother. (d) Consider the equation x a + y b = 1. (a) Show that this equation represents a line by writing it in the form y = mx + b. Solution: Multiply both sides by b to get x a + y b = 1 b a x + y = b y = b a x + b (b) Find the x and y intercepts of this line. Solution: The y-intercept is (0, b) and the x-intercept is (a, 0). (c) Explain in your own words why the equation in this exercise is known as the intercept form of a line. Solution: The x-intercept is given as the coefficient of x and the y-intercept is given as the coefficient of y in the equation. Page 4 of 7
2. Graphing Lines. Graph the following equations. (a) y + 8 = 0 Solution: Use Wolfram Alpha. (b) A line perpendicular to x 5y = 0 and goes through the origin. Solution: Use Wolfram Alpha to graph y = 5x. Interpreting Linear Equations (a) Some scientists believe there is a limit to how long humans can live. One supporting argument is that during the past century, life expectancy from age 65 has increased more slowly than life expectancy from birth, so eventually these two will be equal, at which point, according to these scientists, life expectancy should increase no further. In 1900, life expectancy at birth was 46 yr, and life expectancy at age 65 was 76 yr. In 2010, these figures had risen to 78.7 and 84.1 respectively. In both cases, the increase in life expectancy has been linear. Using these assumptions and the data given, find the maximum life expectancy for humans. Solution: *****We ll round decimals throughout. This was a hard problem as we saw in class! The key to linear equation word problems is to figure out what value is changing in response to another variable. That is, you must first isolate your independent variable (x-axis) and your dependent variable (y-axis) from a paragraph of words and numbers like above. In this problem the independent variable is years since 1900 and the dependent variable is life expectancy. We want to write equations for two lines: one which captures how life expectancy from birth changed with time and another which captures how life expectancy from age 65 changed with time. Let us first look at the line y 1 representing life expectancy from birth. At t = 0 yrs from 1900 (i.e. in year 1900) the life expectancy from birth was 46. So one point on our line (in fact the y-intercept) is (0, 46). At t = 110 yrs from 1900 (i.e. in year 2010) the life expectancy from birth was 78.7. Thus a second point on our line is (110, 78.7). We now have enough information to calculate the equation of our first line y 1 = m 1 t + b 1 : and so m 1 = 78.7 46 110 0 =.288 y 1 =.288t + 46 is the equation representing how life expectancy from birth has changed since 1900. Page 5 of 7
Next we look at the line y 2 representing life expectancy from age 65. At t = 0 yrs from 1900 (i.e. in year 1900) the life expectancy from age 65 was 76. So one point on our line (in fact the y-intercept) is (0, 76). At t = 110 yrs from 1900 (i.e. in year 2010) the life expectancy from age 65 84.1. Thus a second point on our line is (110, 78.7). We now have enough information to calculate the equation of our first line y 2 = m 2 t + b 2 : and so m 2 = 84.1 76 110 0 =.076 y 2 =.076t + 76 is the equation representing how life expectancy from age 65 has changed since 1900. According to the word problem the maximum life expectancy will occur at t such that life expectancy from birth=life expectancy from age 65. We now have equations representing both of these values as functions of t (years from 1900) and so we set the two equations equal to one another to find t when the maximum will occur. So we have y 1 =.288t + 46 =.076t + 76 = y 2 = t = 19.92 yrs since 1900 and maximum life expectancy will occur in year 209.92. We plug t = 19.92 back into either equation (since at this time t the equations output the same value by construction!) to calculate the value of the maximum life expectancy:.288 (19.92) + 46 = 86.29 yrs old and so our answer for the maximum life expectancy is 86.29 yrs old. (b) The mortality rate for children under 5 yrs of age around the world has been declining in a roughly linear fashion in recent years. The rate per 1000 live births was 90 in 1990 and 48 in 2012. (a) Determine a linear equation that approximates the mortality rate in terms of time t, where t represents the number of years since 1990. Solution: *****We ll round decimals throughout. The y-intercept is given in the problem, namely (0, 90) since the rate was 90 at t = 0 years since 1990. To calculate the slope we use (22, 48) as second point given in the problem on our line representing change in mortality rate in terms of time. We have enough information to calculate the slope of our line now: 48 90 m = 22 0 = 21 11 Page 6 of 7
Thus let y=mortality rate and we have y = 21 11 t + 90 as our linear equation that approximates mortality rate in terms of years since 1900. (b) One of the Milennium Development Goals (MDG) of the WHO is to reduce the mortality rate for children under 5 years of age to 0 by 2015. If this trend were to continue, in what year would this goal be reached? Solution: *****We ll round decimals throughout. Here we are given a certain mortality rate y = 0. We want to calculate t such that (t, 0) is on the line described in part a since we re assuming this equation describes the trend of mortality rate as a function of time. Thus we solve 0 = 21 11 t + 90 for t to find the year the goal of a mortality rate of 0 will be achieved. A bit of algebra shows t = 1.42 and so the goal will be reached in 2021. Page 7 of 7