MATH 131--Calculus for Physical and Life Sciences

Measuring Rates of Change

September 20, 2004

The following values show my position on the Mass Pike as

a function of time as I drove to Holy Cross this morning:

  time t (minutes)         0         5           10          20             30              35             40             45

  distance x (miles)         0          2.5        4.5         15.5           26.5           32             35             38

>    with(plots): 

Warning, the name changecoords has been redefined

>    trip := [[0,0],[5,2.25],[10,4.5],[20,15.5],[30,26.5],[35,32],[40,35],[45,38]];

trip := [[0, 0], [5, 2.25], [10, 4.5], [20, 15.5], [30, 26.5], [35, 32], [40, 35], [45, 38]]

>    plot(trip,x=0..50,y=0..40,style=point,symbol=circle,labels=[t,d]);

[Maple Plot]

>    pplot:=%:

"Connecting the dots" with straight line segments:

>    lplot:=plot(trip,x=0..50,y=0..40,labels=[t,d],color=black):

>    display({pplot,lplot});

[Maple Plot]

For instance, over the 10 minutes from   t = 10,   to   t = 20,   my average velocity was

                                  (x(20)-x(10))/(20-10)  = (15.5-4.5)/10  = 1.15 miles/minute = 69 miles per hour

(OK, I was pushing it a bit!  Luckily there weren't any state troopers out right then!)

Similarly, over the 5 minutes from t = 40 to t = 45,   my average velocity was

 

(x(45)-x(40))/(45-40)  = (38-35)/5  = .6 miles/minute = 36 miles per hour.   My average velocity for

the whole trip was

(x(45)-x(0))/(45-0)  = (38-0)/45  = .84 miles/minute = 50 .7 miles per hour

>