MAT1193 1f. Linear functions (most closely related to section 1.4) Linear functions are some of the simplest functions we ll consider. They have special properties and play an important role in many areas of mathematics. Most importantly for this class, linear functions are the most direct way to get a handle on the idea of rate of change. A final reason that linear functions are so important is that almost every function is a linear function, at least in small bits. We ll see what that means later in the class. But for now, we introduce the most important equation in this class: rate of change = Δy Δx Mathematicians often use the Greek letter Δ (capital delta) to mean change in. So Δy/Δx is the change in y divided by the change in x. We can rewrite this equation as Δy = rate of change * Δx In other words, Δy can be obtained by multiplying Δx by a fixed number the rate of change. We say that Δy is proportional to Δx and that Δy and Δx form a proportional relationship. The most common application for this equation relates to the formula: rate time = distance As example, suppose you re traveling at 72 m.p.h. for 20 minutes. How far have you gone? Well 20 minutes is 1/3 of an hour, and 72 m.p.h. means exactly that: if you travel one hour, you ll go 72 miles. So if you travel for 1/3 of an hour, you should travel 72/3 miles = 24 miles. Now consider a similar problem. At 10:30 a.m. you re 10 miles north of town driving north at 72 m.p.h. Where will you be at 10:45 a.m. if keep up your speed. We want to use the formula rate time = distance but obviously we can t just use 10:30 a.m. as the time. Actually the formula means that rate times the change in time is equal to distance, so we want the change in time, Δt. We can figure Δt out (in our head) by taking 10:45 10:30 = 15 minutes = ¼ hour. Now we plug into the formula (72 mi/hr)* (1/4 hr) = 72/4 mi = 18 mi. So we re 18 miles north of town, right? Wrong! The distance in the formula refers to the change in position, which we ll call Δp. So to get the final position, we need to add Δp to our original position (10 miles north) to get the final answer that we end up 28 miles north of town at 10:45. So we can rewrite our formula
rate Δtime = Δposition or rate = Δposition Δtime This is just one version of a type of problem that we ll be discussing over and over in this class. So even though this particular problem is really easy, let s be careful and convert it all the way into mathematical notation. The problem says At 10:30 a.m. you re 10 miles north of town. So let s let p be the variable of position (in miles north of town) and let t be the variable representing time (in hours). p and t are both variables and we think of p as a function of time, p(t). That is p(t) is the position you re at when time = t. Now we need some notation to be able to talk about the specific value of the position at the specific time given at the start of the problem. A typical way that mathematician do this is to use a subscript, which is a number attached with a variable name. So if you want to talk about two specific values of the variable x you might call them x1 and x2. Many times if there is a particular base value or value at the start of the problem we use the subscript 0. So for this problem we could write the beginning position and time as p0 = 10 mi and t0 = 10.5 hours (the 30 minutes in 10:30 is 1/2 or.5 hours). If we let r = 72 mi/hr be our rate then we have the following steps to find p(10.75) which is a way of writing the position at t =10:45 or 10 and ¾ hours: Δt = t - t 0 =10.75 10.5 =.25hr Δp = r * Δt = 72 *.25 = 18 mi Δp = p - p 0 p = p 0 + Δp = 10 mi +18 mi = 28 mi Now let s turn away from applications and consider the different forms of the equation for a line. To figure out the equation of a line you need to know one of two pairs of things: (1) the slope and one point on the line or (2) two points on the line. You all have learned (and maybe forgotten) about the different forms for the equation of a line: slope intercept, point slope, two- point. These forms are also described in the book on p. 43. But if you simply memorize these you loose out on a fundamental concept of what it means for a function to be linear. For this reason, you will not be tested on the definitions of different formulas for a line. Instead, I want you to trace everything back to the most important equation in this class: slope = Δy Δx which is equivalent to Δy = slope * Δx
From there, all the different equations of a line can be obtained by some simple algebra. So to find the equation of a line, there are three simple steps: 1. Find the slope m and one point on the line (x0, y0). We ll call the point (x0, y0) the base point. 2. Substitute into the proportional relationship Δy = m* Δx where Δy is the distance from y to the base point value y0, and Δx is the distance from y to the base point value x0, 3. If you need to, do some algebra if necessary to write y as a function of x As an example, let s find the equation of the line that goes through the point (- 3,5) and has a slope of - 2. We re already done with step #1: the base point is (- 3,5) and the slope was given in the problem. For step 2, we find the equation to be satisfied by an unknown point on the line, call it (x,y). Now we just need to substitute into Δy = m* Δx. Δy ia just the change in the y value from our unknown point (x,y) to the base point (- 3,5). The distance from 5 to y is just y- 5, so Δy = y- 5. In the same way we find Δx = x- (- 3) = x+3. Substituting in we find y - 5 = (-2)*(x + 3) Unless somebody asks you differently, that is a fine equation for a line. But if we want to make it more like the standard definition for a function, we can do some algebra (step 2) to get the output variable y alone on the left hand side of the equation. That s easy. Add 5 to both sides and you get y = (-2) *(x + 3) +5
Now if you want to simplify it even more, a bit more algebra gives the slope intercept form: y = 2x - 6 +5 = 2x -1 This equation might be the simplest equation if you just want to calculate y, but the equation y - 5 = (-2)*(x + 3) contains the history of the problem right in the formula: you can easily pick out the base point and the slope. Sometimes leaving things in the form Δy = m* Δx actually makes things easier. Converting between degrees Celsius ( o C ) and degrees Fahrenheit ( o F ) gives one example. How do we do it? Suppose we re traveling in Europe and we want to convert their temperature report into the more familiar degrees Fahrenheit. To find the conversion, the first step is to find one point on the line and the slope. We aren t given the slope in this problem, but we can figure out the slope if we know two points on the line. In this case, knowing the freezing and boiling points of water gives us two points. Water freezes at 0 o C and 32 o F. Boiling is at 100 o C or 212 o F,. To get the slope remember that o C is the input and o F is the output. So slope = Δo F Δ o C = 212o F 32 o F 100 o C 0 o C = 180o F 100 o C =1.8 o F o C The units here help to see how this slope is a rate of change. In English this says that the temperature in Fahrenheit changes 1.8 degrees per degree of change in Celsius. We can also rewrite the equation to get a formula that we can use to calculate new temperatures: ( ) * Δo C Δ o F = 1.8 o F o C Now suppose we want to convert 25 o C to Fahrenheit. Let s give the unknown temperature that we re looking for a name, say f. We need to pick a base point in order to find Δ o F and Δ o C We could use either freezing or boiling, they both would work. Lets take freezing. If we re at 25 o C then Δ o C = 25 o C - 0 o C = 25 o C. In English: our temperature is 25 o C above freezing. Using our conversion, Δ o F = 1.8 ( o F/ o C)* 25 o C = 1.8*25 o F = 45 o F. In other words, the temperature we re looking for is 45 o F above freezing. Since freezing is 32 o F, the temperature we re looking for equals 32 o F+45 o F=77 o F. If we stick more with the mathematical way of writing, we would write 45 o F = Δ o F = f 32 o F Now we solve for f by adding 32 o F to both sides need to find 45 o F + 32 o F = f Now suppose you re traveling in Europe in the summer and are trying to convert the forecasted temperature back into the more familiar Fahrenheit numbers, and you
can remember to use (25 o C, 77 o F ) as a base point. Now suppose the forecast is 23 o C. From our calculations above we know that the change in degrees Fahrenheit is 1.8 times the change in Celsius. The change in Celsius is - 2 o C (2 o C below the base point) so the change is Fahrenheit is 1.8*(- 2) = - 3.6 o F below our basepoint of 77 o F, so the predicted temperature is 73.4 in o F. If you can remember the rate of change of 1.8 and the basepoint (25 o C, 77 o F ), then all your conversions will use pretty small numbers, at least in the summer time. If you stay for the winter, you might want to choose another convenient basepoint, so that the temperatures that you want can be found from that base point plus the rate of change times a small change in temperature. Viewing functions as rates of change times small fluctuations around a base point is one of the fundamental concepts in this course. Let s work another example. Suppose you are working for the water company and are trying to regulate the contaminants in the city s drinking water. The new target concentration for active live contaminants is 35 live organisms per 100 gallons of drinking water. To keep things shorter, let s abbreviate Live Organisms Per Hundred Gallons as lophg. You know that in the past that use of 1500 pounds of chlorine per week (ppw) was sufficient to keep the contaminants at 50 lophg. So you tried increasing the chlorine to 2000 ppw and you then measured 25 lophg. So obviously, you put in too much chlorine. How much chlorine should you try to hit the target of 35 live organisms per 100 gallons? This is a problem in interpolation: we have a relation between two variables for two different points; given a value of one of the variables that is in between the two points, what is the corresponding value of the other variable? The most common form of interpolation is linear interpolation where a linear function is used to interpolate between the values. It s easier to show in a picture:
To solve this problem, lets give our variables names: let o be the number of live organisms in the drinking water (measured in lophg) and let c be the amount of chlorine that you use to treat the water (measured in ppw). Noitce that in terms of the application described, it s most natural to think of o(c): the number of organisms is a function of how much chlorine you use to treat the water. But in the problem we re solving we want to put a target number of organisms in and get the appropriate amount of chlorine as our answer: we want c(o). Note that these are two perspectives on the same relationship, one is just running the relationship backward. In other words c(o) is the inverse function for o(c), and o(c) is the inverse function for c(o). To get the answer in the problem, we want to derive a linear function for c(o). We have two points on the line, so we need to derive the slope. From our formula slope = Δc 2000 1500 = Δo 25 50 = 500 25 = 20 ppw lophg Let s interpret this equation viewing the slope as a rate of change: we have to add 20 pounds of chlorine per week (ppw) for every live organism per hundred gallon (lophg) reduction in drinking water contaminants. (Notice that the negative sign in the slope means that adding chlorine corresponds to reducing contaminants.) OK. Now we have the slope, let s choose the base point. We can take either point given in the problem, but lets use the historical point (50 lophg, 1500 ppw). So we just write down the equation for the line
Δc = m* Δo (c 1500) = 20*(o 50) That s all we really need to do for our equation. Now we want to know the value c(35). So we plug in 35 lophg for o and solve for c. (c 1500) = 20 * (35 50) (c 1500) = 20 * ( 15) = 300 Before we finish off the algebra, lets stop and interpret this equation at this point. It says that if we want to reduce the number of live organisms per hundred gallons by 15, we need to increase the amount of chlorine by c- 1500= Δc = (- 20)*(- 15) = 300 pounds per week. Now if we start at 1500 ppw the amount of chlorine we need is c 1500 = 300 c = 300 +1500 =1800 So we need c = 1800 ppw of chlorine to give us 35 lophg of contaminants. Extrapolation is essentially the same as interpolation, except the predicted value comes outside of the range of the known data. For example, suppose that a glacier in the Canadian rockies is receding due to global warming. In 2006 it extended 20 miles into the valley, but by 2010 is only extended 14 miles into the valley. 2 miles into the valley the glacier forms a natural damn holding back the summer runoff from another valley. If the glacier recedes too far and no longer damns the runoff, it will result in flooding downstream washing away the summer wheat crop that is grown in the valley. How many years would you estimate the residents of the valley have to come up with a solution for preventing the expected flood? Again we need to give our variables names: let d be the distance that the glacier extends into the valley (in miles) and let y be the year. As before for the application it s natural to view the distance down the valley as a function of time, d(t). But we re interested in predicting the year given a specific distance. That is we want to know t(d) for d=2 miles. We have two points on the line, and we need the slope slope = Δt 2010 2006 = = 4 Δd 14 20 6 =.67 years mile Interpreting slope as a rate of change, for every mile further up the valley it takes.67 years for the glacier to melt that much. (Notice that the negative sign in the slope means that waiting a positive number of years means that the distance up the valley is decreasing.).we can choose either point given in the problem. Let s choose the most recent one, (14 miles, 2010 years). So we just write down the equation for the line
Δt = m* Δd (t 2010) =.67*(d 14) Now we want to know the value t(2). So we plug in 2 miles for d and solve for t. (t 2010) =.67*(2 14) (t 2010) =.67*( 12) = 8years Note that this equations says that the residents have 8 years from 2010 before the glacier recedes to a distance of 2 miles. That s really what we want to know. If actually want to solve it all the way and get the year, we find that t = 8 + 2010 = 2018years Here s the graph: