Thursday, September 24, 2009

Secants, and Average Rates of Change

Hello everyone,

Today, we did further investigation into secants, and how they determine average rates of change in a given interval of a function.

For linear polynomial functions, it is easy to determine the average rate of change; you simply look at the leading coefficient of the function.

For any other polynomial function, (and ill use the example of f(x) = x^2, x e [0,10]) however, the rate of change is different when x=5, than it is when x=6. It changes constantly throughout.

What can be determined however, is the average rate of change from two points along the graph. This can be done using the formula, where m= slope of secant:
m=(y2-y1)/(x2-x1)

(my appologies; the formula maker is not agreeing with me, so just bear with my format)

for example, in the given polynomial function above, if i wanted to find the average rate of change, for the given domain, I would calculate the slope of the secant that passes through the points, (0,0) and (10,100).

By plugging the points in to the formula:

m=(y2-y1)/(x2-x1)
m=(100-0)/(10-0)
m=10
So there you have it; the average slope of the stated domain of the function x^2, is 10. Sure the slope of the secant of the whole domain tells me the average rate of change, but it does not tell me the slope at more specific areas on the graph.
I am going to split the domain of this function into two seperate domains: x e [0,5] and x e [5,10]. This will result in seperate secants for each new domain.
m, x e [0,5]
m=(y2-y1)/(x2-x1)
m=(25-0)/(5-0)
m=5
So the slope for this secant is 5
m, x e [5,10]
m=(y2-y1)/(x2-x1)
m=(100-25)/(10-5)
m=15
The slope of this secant is 15
What if, however, i want to find the slope, not for the average of an interval, but for a specific point on the graph? For the example function that i provided above, what if i wanted to find the slope when x=7? What i will do first, is i will calculate the slope of the secant between the x values of 7 and 8, (7,49) and (8,64)
m, x e [7,8]
m=(y2-y1)/(x2-x1)
m=(64-49)/(8-7)
m=15
Then between the values of 7 and 7.1
m, x e [7,7.1]
m=(y2-y1)/(x2-x1)
m=(50.41-49)/(7.1-7)
m=14.1
You can now observe that the secant is becoming a tangent, and y2 approaches 49, and as x2 approaches 7.
Now lets go between the x values of 7, and 7.00001
m, x e [7,7.00001]
m=(y2-y1)/(x2-x1)
m=(49.00014-49)/(7.00001-7)
m=14.00001
As y2 approaches y1, and as x2 approaches x1, secant approaches tangent, the instantanious slope, which we will probably cover in the next lesson. The homework touched base on this, kind fo forshadowing in a way, so i figured id make the connection: secants can become more specific, untill the tell the slope at an exact point.
Enjoy the University fair tomorrow,
David