Applications of Linear Functions

Motion at constant speed

The same thing from a different point of view

Now consider a car travelling say from Vancouver to Abbotsford along Hwy1. If we measure time from when it crosses the Port Mann bridge, then at time zero (when it's on the bridge) the distance that it still has to travel is about  90km, so at a speed of 90km/h, it should take just about one hour to complete its trip. If we now let y be the number of km left to travel (as opposed to the distance travelled so far), and x be the number of minutes since the car crossed the bridge, then when x=0 we have y=90, and when x=60 we have y=0. More generally, after x minutes, the distance travelled will be x/60 times the 90km distance travelled in an hour, and the distance remaining will be 90km minus that. So now (with our new definitions of x and y),  y and x will be related by  y = 90 - (x/60)90 = -1.5x+90, (which is mx+b with m=-1.5 and b=90).

In general, by changing scale and/or reference points, we can find different equations describing the same situation - but on the other hand, we shall also often see the same equations arising in completely different applications. For example, the same equation that we have just seen in a motion problem might also arise in a business situation, as follows.

Another situation giving the same equation

When people are selling things, they often find that the number of customers goes down if they raise the price (and vice versa). Imagine that a seller of gizmos makes on average 30 sales a week at a price of  $40 per gizmo, but that for each $2 price increase she loses 3 sales per week (and for each $2 price reduction she increases her sales by the same amount). So if she charges $42 per gizmo she might expect to sell only 27 gizmos per week, but if she charges just $36 each she might sell 36 gizmos in a typical week. What would you expect to happen if she raised the price by just $1? Since this is just half of the $2 we would expect the number of sales to drop just half as much, ie from 30 down to 28.5. (Although it may not make sense to talk about a fractional number of sales in any particular week, it is perfectly possible for the average weekly sales volume to be a fractional number). So, for a price increase of just $1, we'd expect a sales drop of 3/2 or 1.5 sales per week.

Now consider how many sales to expect if the price charged is $x.  This represents an increase of $(x-40) from the $40 price that us gave 30 sales, and each $1 of increase should lose us 1.5 sales, so altogether we'd expect the sales to be reduced from the original 30 by x-40 times 1.5. In other words, if we let y be the expected average number of sales per week, then when the price is $x we should get y=30-(x-40)(1.5)=90-1.5x= (-1.5)x+90 (which is again mx+b with m=-1.5 and b=90).

(Of course this all depends on the number of sales lost per $ of price increase being independent of the actual price, which might not always be the case, but is a reasonable assumption if the price change is not too great.)

Summary

The same kind of formula applies (at least approximately) in many other applications, and one of the great powers of mathematics is the fact that the same equations and techniques can be used over and over again in many different situations.