We started today's class by deriving the
difference quotient. It is the average rate of
change of y with respect to x the given interval
Graphically, it is the slope of the secant line connecting points
P and Q.
Title: Sep 27-7:52 PM (1 of 10)
The slope of the secant line through points P and Q is given by
Title: Sep 27-8:14 PM (2 of 10)
We then moved on to do Example 1 on the book which is on page 90.
Let the function f be defined by
a) Find the average rate of change of f over the interval
Title: Sep 27-8:24 PM (3 of 10)
b) Find the equation of the corresponding secant line.
Title: Sep 27-8:41 PM (4 of 10)
c) Plot the graph of f and the secant line
Title: Sep 27-8:48 PM (5 of 10)
We then talked about the instantaneous rate of
change. If the average rate of change has a
limiting value as the interval decreases in size
then it is called the instantaneous rate of change
of outputs with respect to inputs. Graphically, it
is represented by a tangent line.
Title: Sep 27-8:57 PM (6 of 10)
This is the traditional notation for the
limiting value which is read as `` the limit
as x approaches zero of the difference
Title: Sep 27-9:05 PM (7 of 10)
We then worked on another example.
a) Find the slope of the tangent line to the graph of f at (2,12)
Now, as h gets smaller 3h approaches zero
and the limiting value is 12.
Title: Sep 27-9:11 PM (8 of 10)
b) Find the equation of the tangent line
The tangent line passes through the point (2,12)
with the slope of 12.
Title: Sep 27-9:28 PM (9 of 10)
Thanks everyone that is the scribe for today. Remember to study
tonight for tomorrow`s test on the first unit. Our homework for tonight
is Exercise 2.2 questions number 2,10,12,24 and all odd numbers.
Title: Sep 27-9:33 PM (10 of 10)