Post #22

After reading Trina's post, I've decided to post my own blog and answer a few of her questions.

1. Table Problems

Questions 4 and 5 on the non-calculator portion are questions that use a table, which is given to you.

Question 4 asks what the average rate of change of the function F on [1,4] is. To solve this problem just take your values for F(1) and F(4) and plug them into the slope formula, (F(b)-F(a))/(b-a). Using your graph you find that F(1)=2 and F(4)=6.
Plug 4 in for b and 1 in for a, giving you (6-2)/(4-1)= 4/3

Question 5 asks you to find H'(3) when H(x)=G(F(x)). All you do is differentiate, or take the derivative, of H(x). In order to do this though, you need to realize that G(F(x)) can be differentiated similar to sin(2x). You differentiate sin(2x) by taking the derivative of the outside, sin, and keeping the same inside, then you multiply everything by the derivative of the inside, which would give you cos(2x)*(2), or 2cos(2x). Now that you realize this, you can differentiate H(x). First derivative of outside, G'(f(x)), then you multiply by the derivative of your inside, G'(F(x))*F'(x). After all of this you plug in 3 for x. G'(F(x))*F'(x); G'(F(3))*F'(3); G'(4)*(2); (1/2)*(2)=1.

3. Graph Problems

Question 10 asks at what time does the object attain it's maximum acceleration, giving the graph of the object's velocity. The key word here is the object's velocity. Since a derivative is just the slope of a graph, your actually looking for where on the graph is the slope the steepest, or highest y value. After looking at the graph, you notice that at 8 < t < 9 is where the slope is the greatest because it goes from -4 to 0 in one second, which gives you a slope=4.

Question 11 asks at what time is the object farthest from the starting point. To answer this question, you need to use the rules of interpreting graphs. Knowing that position is the original function, velocity the first derivative, and acceleration the second derivative. Now that you know your going from your first derivative to the original graph, you just draw it by using your rules: 1. Any part of the first derivative graph that is above the x axis is increasing on the original graph; 2. Any part of the first derivative graph that is below the x axis is decreasing on the original graph. So from 0 to 6 the original graph is increasing, and from 6 to 9 the original graph is decreasing. This creates a negative parabola negative absolute value graph. Once done drawing just find the maximum of that graph and that is your answer, which is 6.

Question 12 states that at t=8, the object was at position x=10 at t=5, the object's position was x=?. To be honest i just guessed on this question, knowing that my position graph was a parabola and at t=6 that was the farthest possible point that the object could travel from the starting point, so that ruled out answer choice E. Next i saw that at t=8, x=10, and since 8 is farther away from 6 than 5 was, I just chose 13, answer choice D, because it made sense, being that it was greater that 10, but I did not want to choose 15, choice E, because that seemed like it would too big.

I hope I can help with this blog Trina.