Week+06+Sept+22-26

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Sept 22nd 2008
For the first fifteen minutes of class we practiced the use of power rule, after which time particle motion was the focus of the class period. We had our first AP problem presented to us and we used the problem to discuss, again, the relationship between position, velocity and acceleration and the use of the derivative to find these quantities. The problem was from 1983 and the question is as follows:

A particle moves along the x axis so that at time t its position is given by x(t) = t³ -6t² + 9t +11 The first thing we did was to graph the function in parametric mode so that we could see its motion. Tracing the graph indicated that the particle began moving to the right, changed direction and moved to the left, changing direction one more time then finished by moving right. a) What is the velocity of the particle at t = 0?  Since the velocity of the particle is given by the derivative of it's position v(t) = x'(t) = 3t² -12t +9 and v(0) = 9. Therefore, the velocity of the particle at t = 0 is 9.

b) During what time intervals is the particle moving left? Having looked at the parametric graph of the particle's position, we knew that the particle appeared to change direction when t = 1 and t = 3. Since a graph is never sufficient evidence to justify an answer, we looked for ways to solve this problem algebraically. If the particle is moving left, its velocity will be negative so we need to solve the quadratic inequality 3t² -12t +9 < 0 dividing by 3 t² - 4t + 3< 0 factoring (t - 1) (t - 3) < 0 The sign line indicates that the product of the two factors (t-1) and (t-3) will be negative if 1 < t < 3 so the particle is moving to the left when 1 < t < 3.

c) What is the total distance traveled by the particle from t = 0 to t = 2? We know that the particle changes direction when t = 1 and that x (t ) describes the position for any time t. If we find where the particle is for t = 0, t = 1 and t = 2, we can determine the distance traveled. x(0) = 9 x(1) = 13 so particle travels 4 units on[0,1] x(2) = 11 so particle travels 2 units on [1,2]

Total distance traveled from t = 0 to t = 2 is 6 units.

**__Tuesday, September 23__**
Today we went over several homework problems, shown below.



Page 97 3). WE did this problem on the calculator to figure out when the particle changes direction. x = -t^3 + 13t^2 - 35t + 27 t greater than or equal to 0

We plugged this in to y1 with our calculator in parametric and dot mode. WE used the right scroll key to watch the particle's motion. Next we found the derivative (velocity function):

v(t) = -3t^2 + 26t - 35 Changes of direction may occur when v(t) = 0. So next we found the zeros. This can be solved graphically or you can see if it factors. the zeros ended up being:

t = 1.666 and t = 7 At these times there was a change in direction.

We also concluded teh discussion of displacement versus distance with respect to the in-class problem:

x(t) = t^3 - 6t^2 + 9t + 11 y(t) = 0 v(t) = 3t^2 -12t +9 The question was involving total distance traveled on (0,2). If using the calculator for this (Trap Rule) you must either realize that a negative area cannot exist, and so correct for that, or just put the absolute value of the function in to begin with. If you do not correct for this, you will get displacement (in this case, 2) versus distance (in this case, 6). Ms. Gentry also showed us the "Area Finder" on the Calc Menu, Option 7 (you just have to input the lower and upper limits), but she said that you must show your work if you use this.

September 24th 2008
Today we reviewed for tomorrow's test using the Senteo answering system. The slide show below shows the problems we used. media type="custom" key="2050234" The solutions to the problems may appear shortly! Students, please post a solution.

Solution for question 4

question 5 by andrew spitz

question 6 by rachel neill

acceleration is m/s² which = the derivative of a velocity by time graph

take the derivative of the derivative of the distance by time graph

s(t)=t²-t use power rule s'(t)=2t-1 use power rule s"(t)=2 s"(2)=2

Question 7 by Emily Hixon

=**Thursday September 25, 2008**=

Test Day...

=**Friday September 26, 2008**= Today in class we worked on exploration 3.6. We made conjectures about the derivatives of certain functions and then found them graphically.We also unknowingly discovered the chain rule.

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