Optional Project 25: Instantaneous Rate of Change

Instantaneous Rate of Change

Suppose the height of an object (in feet) after t seconds is represented by the function below:

Height = -.5t^2+4t 

A graph of the function is shown above.

It takes 8 seconds for the object to travel in the air and fall back to the ground (height = 0).  The object finishes where it started (on the ground) so the object’s average rate of change for the 8 seconds is 0.  

Now what if we were to calculate the average rate of change (average velocity) for the object on the first four seconds.

At t = 0, h = 0  

At t = 4, h = 8  

The object travels 8 feet (8-0) in 4 seconds (4-0).  Therefore, the average rate of change (average velocity) of the object is 2 feet per second.

Graphically, the instantaneous rate of change is the slope of the tangent line at a certain time.  In the above example, the tangent line (a line that only touches a curve at one point) is shown.  The slope of this line will yield how fast the object was traveling at t = 1 seconds.  

However, to calculate the slope of a line however we need two points.

We can choose two points that lie on the function that are “really close” to t = 1.

For this example:

At t = 1, h = -.5(1)^2 + 4(1) = -.5 + 4 = 3.5

At t = 1.00001, h = -.5(1.00001)^2 + 4(1.00001) = 3.50003

The slope of the line that passes (1, 3.5) and (1.00001, 3.50003) is:

(3.50003-3.5)(1.00001-1)=.00003.0001=3

Therefore, a good approximation for the instantaneous velocity of the object at t = 1 seconds is 3 feet per second.

Optional Project 25: Variables 'co' and 'x' have been initialized.  There is also a working method called getYPoly.

co represents the coefficients of a polynomial from least to greatest. ({0, 4, -.5} in the example above)

x represents the x-value in which we are finding the instantaneous rate of change at (1 in the previous example)

getYPoly(co, x) returns the y-value for a polynomial with coefficients 'co' at x = 'x'.

Example: If co = {0, 4, -.5}, then getYPoly(co, 4) returns 8.0

Task:  Appropriately assign the value for the variable called 'dydx' that represents the instantaneous rate of change for the polynomial at the given x.

**If your code works for 5 test cases, you can enter your e-mail address

Universal Computational Math Methods:

pow(5,2) returns 25.0

abs(-3.0) returns 3

sqrt(49.0) returns 7.0