Fall+Week+05

toc

Monday 9/14
Today we had our tests returned to us and we spent a few minutes going over the problems. After that, we went over our homework; some thought-provoking problems on continuity and the Intermediate value theorem we learned on Friday. Solutions to these problems can be seen in the slide-show posted on last week's page under Friday 9/11.

Tuesday 9/15
The theme of class today was moving from considering the derivative as the slope of the tangent line to a curve at a single point to the understanding that the derivative of a function is a function in it's own right. We began with a problem of the day:



Each student was asked to choose an x value and use definition of the derivative of a function at a point to find the derivative algebraically. An example of such a calculation is shown here.



Ms. Gentry then told us about the new notation f '(2), which means the derivative of function f at x = 2. ( no, the ' is not a typo!), after which we were able to complete the table and graph the new set of ordered pairs.

We could see that the y coordinate of the ordered pairs of the derivative function matched up with the slopes of the tangent lines to the graph of the original function f(x)= 2/x.

Reading the graph of the derivative, y = f'(x), from left to right, the y values are negative and decreasing (-.125, -.222, -.5, -2..) meaning that the slopes of the tangent lines on f(x) = 2/x should be close to zero and negative at x = -4 but the tangent lines should get steeper as x approaches zero from the left side.

Our last task was to conjecture the equation of the derivative of function f(x) =2/x. We decided that the shape of the function was similar to a reflection of the function y = -1/x² reflected over the x axis. We tried putting this equation into our calculator to sketch it along wiht the stta plot of the derivative function but we found that it wasn't quite right. Our second attempt was the function y = -2/x² and this result is shown in the graphic below.



Of course, Ms. Gentry could not stop there! She had to ask the question, "How can we show that, for sure, the derivative of f(x) = 2/x is f '(x) = -2/x

Y es, we went on to look at a new way to define the derivative, this time the derivative of a function at every point x, rather than the derivative of a function at a particular point c. This involved an "ah-ha" moment for some of us as we recognized our friend the difference quotient! Remember, DQ does not stand for the Dairy Queen, folks!

Wednesday 9/16
A couple of problems of the day:

The second problem led us to graph both y = f(x) and y = f ' (x) on the same set of axes so we could discuss the features of each graph. We then talked about a couple of the homework problems which reinforced the concepts in the slide above. We were able to sketch the derivative of a function even though the function rule was not given. We looked for local maximum and minimum point s on the graph of function f because, at these same x values, the graph of the derivative would have a zero. If the graph of function f is increasing, on that same domain, the graph of the derivative would have positive function values. Conversely, if function f is decreasing, then corresponding points on the graph of the derivative would have negative function values. Lastly we turned our attention to #11 which gave us the graph of an exponential function to explore. We found by a method of guess and check that if f(x) = e^x, then f '(x) = e^x ! A function which is its own derivative! We discovered this by using the calculator- see the screen shots below.

Homework was a worksheet

**Thursday 9/17**
Today we began class with this problem

After that we all made up our own functions,with atleast one part containing a power, and found the derivatives to those functions.

The conjecture about finding the derivative of a power function is this: If f(x)= ax^n, then f '(x)= anx^n-1 We proved that by doing this next problem! We also learned a little more about notation. __Function__ __Derivative__ f(x)=3x^2 f '(x)= 6x y=4x-x^3 dy/dx= 4-3x^2 3v-1/v d/dv=3+1/v^2

Even though we will see the notation f(x) most of the time it is still important to know the others. **:)**

**Friday 9/18**
Problem of the day: Investigation 3.5a) Exploration 3.5a) Velocity and Acceleration.

1. What is the difference between displacement of an object from a given point and the distance it travels while it is in motion? According to the text, displacement tells how far an object is from a reference point, and in which direction. Distance tells how far an object is from the reference point, without regard to direction. If an object moves along the x axis, starting at x = 5, moves 10 units to the right, then moves 26 units to the left it will have traveled a total distance of 36 units. Its final position would be at x = -11 so its displacement would be -16.

2. What is the difference between the velocity of a moving object and the speed at which it is going? Velocity tells how fast an object is moving and in which direction. Speed tells only how fast an object moves without regard to direction.

3. If the displacement of a moving object from a fixed point is given by y = f(t), where t is time, write three different symbols for the velocity v. Velocity is the derivative (instantaneous rate of change) of the position of the particle so its velocity can by represented symbolically by y' or f '(t) or dy/dt

4. If the velocity of a moving object is given by v = g(t), write three different symbols for the acceleration, a. Acceleration is the derivative (instantaneous rate of change) of the velocity of the particle so its acceleration can by represented symbolically by v ' or g'(t) or dv/dt.

5. The acceleration, a , is the second derivative of the displacement, y = f(t). Write three different symbols for the acceleration as a second derivative. y  or f (t) or d²y/dt²

6. True or False? 'If the acceleration of a moving object is negative, then the object is slowing down". Justify your answer. This statement is false because the object could be moving in negatiove direction and speeding up. For example, when t = 3secs, v = -16m/sec and when t = 5sec, v = -28m/sec.  The average acceleration over this interval would be (-28 - -16)/(5-3) = -6 m/sec². Negative acceleration but the particle is speeding up.

7. A particle moves along the x axis with a displacement x feet from the origin given a s a function of t seconds by x = t³ - 9t² + 23t - 15. What is meant by "a particle"? A very small object the size of a point on a graph.

8. Write an equation for the velocity of the particle in problem 7 as a function of t. Velocity is the derivative (instantaneous rate of change) of the displacement function so, using the power rule, x '(t) = v(t) = 3t² -18t + 23

9. With your calculator in parametric mode, plot x as a function of t with y = 1. Use a t domain of [0, 7] and a window of [-20, 20] for x and [0, 2] for y. Set the grapher for path style. Describe the motion you see. The particle starts at x = -15, moving to the right for about 2 secs. The particle changed direction, moving to the left for another couple of seconds, changing direction again at about x = -3 after which it continues to the right for the remainder of the 7 second time period.

10. At t = 2, what are the velocity and acceleration of the particle? Is the particle speeding up or slowing down at this time? At what rate? Justify your answer. x '(t) = v(t) = 3t² -18t + 23 so v(2) = 3(2)² - 18(2) + 23 = -1 v '(t) = a(t) = 6t - 18 so a(2) = -6 Since v'(2) < 0 and a(t) < 0, the particle's velocity is negative and decreasing. Therefore the particle is speeding up.