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Monday 8/15
The first day of school and we did math, learning how to use the TiNSpire CAS calculator and at the same time reviewing the average rate of change of a function over an interval and its relationship with the instantaneous rate of change of a function at one value of the independent variable.



We concluded that the instantaneous rate of change is the limit of the average rate of change as the width of the interval gets increasingly small. We calculated the average rate of change numerically on slide 1.10 above; we need to think about how we can calculate this limit algebraically and graphically. Stay tuned for tomorrow's lesson; it should be exciting!

Tuesday 8/16
We began class with our first Problem of the Day. Most of the class chose to approach this problem graphically, thinking of the instantaneous rate of change in terms of the slope of a tangent line to the curve at a particular point. The graph of function f is a downward opening parabola so a) the tangent line with the greatest positive slope would occur at the left end of its domain .i.e. in this case when t = 0.

b) the tangent line with the least VALUE of its slope would occur at the left end of the appropriate domain. We used a calculator to find the x intercept which is approximately 8.24 so the least instantaneous rate of change occurs when t = 8.24

c) The tangent line with zero slope must be horizontal and this occurs at t = 4, the point at which the function attains its maximum value. Some students calculated this value of t by using -b/(2a); others graphed and traced; others graphed and used the maximum command.

Then there was part d). This question asked us to FIND not ESTIMATE, the instantaneous rate of change of the function when t = 1. We discussed the concept of finding the limit of the slope of a secant line as the length of the secant line approached zero. A series of screen-shots from the calculator, illustrating this idea, can be seen below.





In this last slide, the right end of the secant line coincides with the left end so it is impossible to see the secant line. However, the slope of the secant line and the slope of the tangent line are equal to each other, and both are equal to 6. We can conclude that the Instantaneous rate of change of f at x = 1 might be 6.

The 90 million dollar question is, how can we figure this out algebraically? Let's use a variable, say b, for the x coordinate of the point at the right end of the secant line. Set up the expression for the average rate of change of f from x = 1 to x = b, and allow b to approach 1. The work will look something like this.

Wednesday 8/17
One of the problems of the day was to find the slope of the tangent line to the function f(x) = x 3 - 1 at the point where x = 1. some students grabbed an Nspire and worked the problem graphically as shown in the screen shot below.

The problem with this method is that we are not really sure the measurement of the slope of the tangent line is exact.Fortunately we have an algebraic method to confirm the answer. See below.

Here area couple of other similar Problems of the Day.

Following these problems, we discussed the graphs we had sketched using exploration 1.2. These problems and their solutions can be seen in the pdf file below. We decided a function is increasing quickly if its instantaneous rate of change is greater than 1, and a function is decreasing quickly if the IRC is less than -1. We also spent some time working on the graph of the secant function. The asymptotes occur at x = pi/2 because cos(pi/2) = 0, secant(pi/2) is the reciprocal of cos(pi/2), so sec(pi/2) is undefined. This would result in either a vertical asymptote or a hole in the graph, but as x approaches pi/2 from the left side, the cosine values are small and positive, making the secant values increasingly large and positive. As x approaches pi/2 from the right side, the cosine values are small and negative, making the secant values increasingly large and negative. This behavior indicates a vertical asymptote rather than a hole in the graph at x = pi/2. The last thing we worked on was Exploration 1.3, an introduction to the DEFINITE INTEGRAL. For the time being, this is being defined graphically rather than algebraically and its "definition" can be seen in the graphic below. If the function happens to be a velocity function, the definite integral represents the distance traveled over a given time interval. More about this in tomorrow's post!

Thursday 8/18
Wednesday's homework was Exploration 1.3, an introduction to the DEFINITE INTEGRAL. For the time being, this is being defined graphically rather than algebraically and its "definition", for functions with y values which are positive, can be seen in the graphic below. If the function happens to be a velocity function, the definite integral represents the distance traveled over a given time interval. We began class with two problems of the day.

To answer part c) we sectioned the area into six trapezoids each of width 1 unit. This idea can be extended to using n trapezoids of equal width h, creating a general formula which can be used to estimate the area between a curve and the x axis for any function with positive y values. This is the idea behind the TRAPEZOIDAL RULE shown in the graphic below.



Friday 8/19
Today we had a "mock test"; review problems to prepare for our non-threatening test on Monday. View the slideshow below to see the problems and their solutions. media type="custom" key="10236755"