Scribe Post - October 12, 2010

On Friday we took a Quiz which covered all of Unit 2 to begin the 80 minute period. Following the quiz Mr. O'Brien -update- introduced us to new material that will be gone over more extensively in upcoming chapters. This material consists of proofs relative to the 'derivatives of the trigonometric functions.'

We began this lecture by using Geogebra to graph sin(x). We then plotted a single point "a" onto the graph (can be seen at origin). We then graphed a line of tangency at point "a" on graph "sin(x)".



The next step was finding the slope of the tangent line. We found that when "a" was at the origin the slope of the tangent line was 1. However, when "a" climbed to the maximum height of sin(x) at (1,1) the slope was 0.



We then plotted another point, "b", which would consistently have the same x-value as "a", and the slope of the line of tangency for the y-value. In order to plot this new point you would plug-in: (x(a), m). m, being the slope of the tangent line. Knowing that the derivative of a point is the slope of the tangent line at that point on the graph, this point "b" will express the derivative values of sin(x). In order to capture a visual of the derivative function, we applied the tool, "trace" to this point, "b". Then, we dragged point "a" along the function "sin(x)", as point "b" followed tracing the graph of sin(x)' (in bold below)



We now see the significance of the slope at 1 and 0. When the slope of the tangent line is 1 sin(x)' is going to have a y-value of 0. However, when the slope of the tangent line is 0 sin(x)' is going to have a maximum y-value, and in this case 1.

This traced graph resembles the graph of cos(x).

This graph led us to the first proof of the day:



In order to prove this algebraically we used some help from an old trig. theorem:



This is applied to the general derivative form of sin(x+h), (as h --> 0). The steps to finding the derivative of sin and any other trig function can be found at this website.

we then wanted to find:



=





we knew that:

because,

Through the 'quotient rule' and solving : we found that:

and =

then we used the same quotient rule to prove:

, =

and:





-update (excess amounts of work)

the trick to remembering some of these derivatives is to remember that "sec" and "tan" hang out together and "csc" and "cot" hang out together.

if you feel that you don't understand this algebraic process, return to the link posted above to understand these procedures to a greater extent.

after these proofs Mr. O'Brien wrote this down on the board which is the double derivative of csc.









-update (of all work above)

and seeing that you cannot simplify this any more, it is fine to leave it in this form.

The homework for next class is on ical. Part of the assignment will be to prove csc(x) (above) and cot(x).

-update (much shorter ending)

The classes next scribe will be Nate.