Real+World+Discontinuity+Examples

===My students were curious about what real life example might be represented by a piecewise function of a graph with a "hole" in it and then a function value at that discontinuity that is above or below the function's graph. Can anyone think of one?===

Answers:
1) Here's one that has a hole at an endpoint. The price of a container with milk in it varies linearly with the volume of milk. Something like: P(V) = 72V + 21 Where P is measured in cents and V is measured in gallons. Now, the P intercept (the price you'd pay for 0 gallons of milk) is not 0 (we usually say you're paying for the container). Most people wouldn't pay 21 cents for no milk, so it would make a lot of sense (or cents if you're feeling punny) to define P like this: P(V) = { 0, V = 0 ; 72V+21 , V > 0 } Here's a somewhat contrived one. In many sports, you can bet on particular things happening during the game. For example, you might bet that Reggie Bush will rush for 152 yards on Sunday. If he rushes for less than 152 yards, you lose; if he rushes for more than 152 yards, you lose. If he rushes for exactly 152 yards, he wins. Your profit on the bet is: P(Y) = { 0, Y < 152 ; B, Y = 152 ; 0, Y > 152 } Where Y is the number of yards for which RB rushes and B is the amount that you bet. In soccer, you have to kick the ball at exactly the right angle to pass between the goalie's hands and the bar of the goal. If the ball goes wide, that's really bad for your team. If the ball hits the goalie's hands, that's not as bad as the ball will still be in play. If the ball goes in the goal, that's great. So, the utility function for angles at which you can kick the ball is" U(A) = { 0, A > P ; 100, A = P ; 25 A < P } Where P is the perfect angle.

2) There is a class of problems where a function behaves quite "normally," except at one point where something special happens. An example is the "River Crossing Problem" in the Key Curriculum Presss text (page 52 in the second edition). Calvin is crossing the river. The length of time it takes depends on the angle his path makes with the river bank. For angles close to 0 degrees or 180 degrees the time is long because of the great length of the path. The shortest path is for an angle of 90 degrees. But at 90 degrees there is a bridge he can use! So there is a removable discontinuity at 90 degrees with a function value that does not equal the limit as the angle approaches 90 degrees.

3) Real-world examples of discontinuous functions that students will understand are hard to come by. That's one reason that calculus was successful for modeling the physical world for 100 years before we began worrying about discontinuities and other functional misbehavior. This problem may be too soon for them to understand, so you might want to hold it for a month or so. Here is an example of at least a semi-real life situation. It is taken from Dan Kalman's wonderful book "Polynomia and Related Realms", page 137. A hiker wants to go from point A to point B as shown below: CP-B A He can travel along the road by way of point C, or he can take a shortcut through the woods on a heading of angle T east of north, emerging at some point, P, along the road between C and B. On the road the hiker can cover 4.5 miles per hour. Going through the woods he can travel only 3 miles per hour. (We've all seen this kind of problem before, but the twist is the angle T is our independent variable) If the distance from A to C is 6 miles, and from C to B is 8 miles, at what angle should the hiker go to reach point B as soon as possible? You need to work out the details to see what happens. The total time for the route is f(T) = 2*sec(T) - 4*tan(T)/3 + 16/9. For T between 0 and pi/2, there is one zero for f', which occurs at T* = Arcsin(2/3). Computing f'' shows this to be a local minimum. The time f(T*) = 3.27. If we consider the endpoints, we have f(0) = 34/9 = 3.78 and at the angle pointing straight at B we have f(T_B) = 10/3 = 3.33. So, it looks like we are done. Take an angle of Arcsin(2/3) or about 41.8. But not so fast..... The formula f(T) = 2*sec(T) - 4*tan(T)/3 + 16/9 is not valid for T = 0. At that angle the hiker doesn't walk through the woods at all, but travels around the periphery. This is a distance of 14 miles at 4.5 miles/hr, or only 28/9 = 3.11 hours. The correct time function is discontinuous at 0, with a jump discontinuity. (28/9 if T = 0 f(T) = ( (2*sec(T) - 4*tan(T)/3 + 16/9, if 0 < T <= T_B. It isn't exactly the hole you've asked for, but it is the only example I know.

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