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Monday 8/16
The first day of class and we were doing math! The door opening problem reminded us of several concepts we had talked about in Pre-calculus and Analysis; definitions of the average rate of change and the instantaneous rate of a function and how to calculate or estimate each of them.

The average rate of change of function f on the interval[a,b] is

The instantaneous rate of change of function f at x = a is The investigation is shown in the slide show below.



Following the exploration we talked about how an instantaneous rate of change could be calculated algebraically. The set-up given in the second expression is the starting point, followed by substitution and algebraic manipulation until the limit is changed from indeterminate form i.e. 0 / 0 into a limit expression which can be evaluated. The example done in class is shown below. The IRC is calculated at x = 1

Tuesday 8/17
Having learned how to find, or estimate, the instantaneous rate of change of a function defined algebraically, we moved on to consider how to do this when the function is defined graphically. A tangent line is the limit of a secant line as the two points get closer and closer to each other. Since the IRC of a function is defined as the limit of the ARC of a function as the width of the interval approaches zero, we can estimate the IRC by sketching the tangent line to the curve at the point of interest and calculating the slope of the tangent line. To see an animation of this limiting process, click on the link[| here]

The problem we worked during class are shown below. Notice that the mathematical word for the instantaneous rate of change of a function is the DERIVATIVE.

Wednesday 8/18
Exploration 1.2 ,part of our homework last night, asked us to graph five different functions and copies of these graphs are shown below.

The function is decreasing when x = 1. The rate of change at x = 1 is slow because the y values are decreasing by relatively small amounts for each increase in x.



This trigonometric function is neither decreasing nor increasing when x = 1.



This quadratic function is increasing quickly when x = 1



This function is increasing quickly when x = 1.



This function is slowly decreasing when x = 1.

Next we finished discussing how to estimate the instantaneous rate of change of a function which is presented as a table of data. The idea of finding an average rate of change over an interval, then allow the width of the interval to approach zero is still the fundamental idea but, in the absence of an equation or the graph of the function, it is not possible to let the width of the interval approach zero so we do the best we can! This means we use the **smallest possible** interval which includes the x value of interest. Example



We have three choices if we wish to estimate the IRC at, say, t = 0.3 and each of the possible calculations are shown below.

The last task was an introduction to a second major topic of Calculus; the Definite Integral. Unfortunately we didn't have time to complete the exploration and discuss our findings but I'm sure we will do so tomorrow! The exploration is shown in the slide show below.

Thursday 8/19
We began with our third instantaneous rate of change warm up problem.

Our next problem was #15 from page 13 of our textbook, an assigned problem which some of the class had difficulty with. The problem was designed to make connections between the numerical and graphical approaches to finding the derivative of a function at a particular point.

Bacteria in a laboratory culture are multiplying in such a way that the surface area of the culture, a(t), in mm², is given bya (t) = 200(1.2)^t, where t is the number of hours since the culture was started.

a) Find the average rate of increase of bacteria from t = 2 to t = 2.1.

So. the surface area of bacteria is increasing at a rate of 52.99mm² / hr

b) Write an equation for r(t), the average rate of change of a(t), from 2 hours to t hours. Plot the graph of r using a friendly window that includes t = 2 as a grid point. What do you notice when you trace the graph to t = 2? The average rate of change of any function f on the interval[a,b] is

so for our function a(t), r(t) = (a(t) - a(2)) / (t - 2)

Look carefully and you may notice the hole in the graph of r(t).; it occurs at the point where x = 2 indicating that r(2) is undefined when x = 2. However, the values of r(t) appear to have a limit at t = 2 because as t approaches 2 from either side, the r(t) values appear to be approaching the same number.

c) The instantaneous rate of change (the derivative of a(t)) is 52.508608...mm²/hr. How close to this value is r(2.01)? so r(2.01) is .04789... units away from the derivative of a(t) at t = 2

How close must t be kept to 2 on the right side so that the average rate is within 0.01 unit of this derivative? We can see from the intersection of the graphs of r(t) and y = 52.518608, that x must be kept with .0020888 units of 2 to keep the r(t) values within 0.01 units of the derivative of a(t) when t = 2.

We ran out of time so stay tuned for tomorrow's lesson on the Trapezoidal rule; a method to find the definite integral of a function without having to count squares!

Friday 8/20
First task was our problem of the day; to find the IRC of f(x)= 1/(2x-1)² at x = 1. Solution is below.

Following this, we talked about our two homework problems.

Page 17 #7 Estimate the instantaneous rate of change of f(x) = tan(x) at x = 1. The numerical method is the most efficient method to use and screen shots from the calculator show how we can draw the conclusion that the IRC of tan x at x = 1 is approximately 3.426.

Page 17 # 9 Question and answers are shown below. Lastly we worked on an investigation to develop the Trapezoidal Rule. The exploration is shown in the slide show below.