1. Suppose f(x) = √x.
a) Calculate the average rate of change between x = 9 and x = 9.1.
f(9.1) - f(9) / 9.1 - 9.0 = average rate of change
√9.1 - √9 / 0.1 = 0.1662 (Store the answer from your calc in A)
b) Calculate the average rate of change between x = 8.9 and x = 9.0.
f(9) - f(8.9) / 9 - 8.9 = average rate of change
√9 - √8.9 / 0.1 = 0.1671 (Store the answer from your calc in B)
c) Use your answers above to approximate the instantaneous rate of change at x = 9.0.
Take the answers from questions A and B:
(A+B) / 2 = 0.16666
NOTE: You must have your answers for A) and B) to four decimal places, you don't start seeing a change between the two answers until four decimal places. Marks normally aren't given out if your answers aren't to four decimal places. When calculating the answer to C) in your calculator, you must have brackets around A + B or the calculator will calculate B / 2 then + A into the solution.
2. A ball is dropped from a height of 400 ft and falls towards the earth in a straight line. In t seconds the ball drops a distance of d = 16t² feet.
a) How many seconds after release does the ball hit the ground?
The ball is dropped from a height of 400 ft, therefore the ball must travel 400 ft before it hits the ground:
400 = 16t²
400 / 16 = t²
t = 5
b) What is the average velocity of the ball during the time it is falling?
0 - 400 / 5 - 0 = -80 ft/sec
The reason why we take 400 and subtract it from 0 and not the other way around is because the ball is falling down not falling up, so the velocity is increasing in a negative manner.
c) Estimate the instantaneous velocity of the ball when it hits the ground.
160 ft/sec
There are many methods of finding this answer. You can graph it and keep zooming into that point and use the "SLOPE" program, you can estimate by using the symetric difference quotient and others. Unfortunately my mind is mashed right now and I can't recall any other from the top of my head. =S
3. An object travels in such a way that its position at various points in time is given in the table:
a) Find the average velocity of the object between t = 1.3 and t = 1.9.
5.3 - 5.1 / 1.9 - 1.3 =
0.2 / 0.6 = 1/3
b) Estimate the instantaneous velocity at t = 1.9. Show work.
5.3 - 5.6 / 1.9 - 1.8 =
-0.3 / 0.1 = -3
While Mr. K was giving us the solutions to this quiz he talked about the exam for the course. Here's a rough outline of the exam:
- consists of two sections, multiple choice and long answer.
- each section has a non-calculator part and a calculator part.
- multiple choice has 28 questions non-calculator (50 min) and 17 calculator questions (50 min).
- long answer has 3 calculator questions (50 min) and 3 questions non-calculator (50 min).
- the questions and sections come in the order I just explained.
- the exam itself is a MARATHON 4 HOURS from 8 a.m. to 12 p.m.
Now onto the topic we talked about in class....or at least what I managed to write down of it. =S "Limits"
We started with a function f(x) = 2x - 1: (Note: graph may be up tomorrow, it's too late for me honestly =S)
We were given lim x->2 f(x) which is, lim x->2 2x - 1. If you were to try and find x = 2 the answer would come up to be 3. But the lim x->2 means you have a value very close to 2 your input and a value very close to 3 your output, but it never actually gets to the point (2,3). I'm not entirely sure if what I just explained made any sense I made a lot of point form notes and I'm trying to work from it I apologize now if any of my information is incorrect or has no value to it.
We were then given (x² - 1) / (x - 1) and we graphed it on our calculators. When the graph was shown in a "Zoom Standard" window it looked like just an ordinary straight line. But when you look at it in a "Zoom Decimal" window, you see that the line in fact has a HOLE in it! The reason for this is because (x² - 1) factors and reduces. The hole in the graph is at x = 1 because there is no value for that. When you look at the table there seems to be a limit at x = 1. The y values get closer and closer to 2 but never really gets there. Therefore x can never equal 1 cause of the limit.
lim x->1 (x²- 1) / (x - 1) = (x - 1)(x + 1) / (x - 1) = (x + 1)
This is were my notes get a little confusing =S. Mr. K started talking about dividing by 0 and why it's not allowed. When you multiple 2 by 3 you get 6 and when you divide 6 by 3 you get 2, so when you multiple 2 by 0 you get 0 and when you divide 0 by 0 you get 2? So basically 0/0 can give you any number? We call this "indeterminate". If you divide 5 by 1 you get 5, if you divide 5 by 0.1 you get 50, if you divide 5 by 0.01 you get 500 and so on. As your denominator gets smaller and your numerator stays the same the value it generates increases until ∞, which isn't a number therefore you can't divide by 0.
lim x->0+ abs(x) / x lim x->0- abs(x) / x
If you were to have lim x->0 abs(x) / x, this would work if on the graph the above limits met at the same place, however they don't so the value does not exist.
Homework for tonight, or should I say earlier was I believe EXERCISE 2.7 on The Limit of a Function. I think it was just odd number questions only. There was also suppose to be a link or something of some sort to this derivative quiz on this website but last time I checked it hadn't been up yet.
Once again I apologize if any of this seems jibberish to you, please feel welcome to comment and tell me every I missed for this class (which is probably a lot) and the next scribe will be Lindsay. It was an easy choice, she told me to pick her and me being the nice guy I am had to agree =D
No comments:
Post a Comment