February 28, 2007
I am going to go and try to attempt the chapter 7 review one more time and hopefully I'll get some of my confusion cleared up.
February 27, 2007
We discussed how the pre-test went with the substitute and the multiple choice questions were pretty straight forward and no one really had any problems. What did leave us all puzzled was the long question in the back, the "open response".
The question was . . .
(1) Let f be a differentiable function such that f " is continuous and f and f ' have the values given in the table below.
(now I wished I had the smart board so I could draw the box like Mr. K had done in class..ah the wonders of technology, I honestly can't keep up with it.)
Don't forget the mnemonic " LIATE ".
I nverse trignometric
So an example problem that we did that even got Mr. K, who's a math gyro (uh I don't know if I am using that word right) thinking was . . .
So to review . . .
+ When the integral is a composite of functions, you are better off using substitution.
+ when the integral is a product of functions, you are better off using integration by parts.
Last but not least next scribe is Suzanne.
The excitement of an early Hall of Fame scribe
How much you supported each other
Linger’s "Won’t let you Fall"
"Lessons from the Geese"
Yes, it's second semester. Yes, the schedule is hard. And yes, there's so much to do--
Didn't you have something pretty special? Is it lost? Can it be found?
February 26, 2007
February 25, 2007
Chapter 7 pre-Test: Antidiffernetiation
The multiple choice were pretty much calculator work. I will just list the answers our group got.
#1) If the definite integral 1S3 (x2-1)dx is approximately using the Midpoint Rule with n=4, the error is:
#2) If -1Sk (3x2-2x)dx=6, then k=?
#3) If 0Sk (sec2x)/(1+tanx)dx=ln2, and k>0, then the value of k is:
#4) Let R be the region in the first quadrant enclosed by the graph of y=xcosx, x=0, y=0, and x=k for 0<(3.14/2). The area of R, in terms of k is:
This is where all of us (as a class) just became clueless. We looked at it for a long time and read it over again, but we were not 100% or even 90% sure what to do or where to even begin. We're not sure what that -2 is all about and if it's supposed to be in that missing spot in the table.
#1) Let f be a differentiable function such that f " is continous and f and f ' have the values given in the table below.
Use the information in the table to evaluate:
a) 0S2 xf ' (x2)dx
b)1S3 xf ' (x)dx
This was the answer our group got (aka manny figured it out :) ).
Well that's it for that class. As a class we are really confused about this section, we feel we're not ready for that test on tuesday. We just need a huge review Mr K., a HUGE review.
Oh the scribe for next class is ANH :)
February 18, 2007
, since we only have class like every second day. =T tomorrow we have a Jr High tour, so that means our test might be on Friday? or Next Tuesday? I'll admit i am finding it difficult to even review and practice for unit seven. The test dates keep moving farther away and i do my other subjects first. I am going to try my best to get some study time down for this unit. Everything is confusing... good luck everybody =).
A) given f(3)=8 and f'(3)=4 find f(3.02)
y - y1 = m ( x - x1)
y - 8 = -4 (x - 3)
y = -4x + 20
y(3.02) = -4 ( 3.02) + 20
Y(3.02) = -12.08 + 20
Y(3.02) = 7.92
B) given y=1 and xy2 - 2y 3 4y3 = 6
solve for x
x - 2 + 4 = 6
x = 4
y2 + 2xyy' - 2y' + 12y2y' = 0
y' ( 2xy -2 + 12y2 ) + -y2
y' = -y2<.sup> / ( 2xy - 2 + 12y2
y'(1 , 4) = -1 / 18
after reviewing some of the questions on the quiz mr.k gave us a table to fill out that consisted of the trapazoid sum, reiman sum, and the simpson sum
i wasn't allowed to upload the table so im just gonna type it on
N ..Tn ....... error ..... Mn ....... error ..... Sn .....error
4 ...0.2656. 0.0156 ... 0.2422 ..-0.0078 .. 0.25 .. 0
8 .. 0.2539 .0.0039 ....0.2480 ..-0.0200 ...0.25 ...0
16 .0.2510 .0.0010 ....0.2495 ..-0.0005... 0.25 ...0
Tn = (B + B)H / 2
Mn = REIMANN SUM
*use 0.5 as the x-choice
Sn = T + 2M / 3
* WE DO NOT HAVE TO KNOW THE SIMPSON SUM FOR ANY TESTS OR EXAM
ERROR = approximation - actual
unforntunately i do not understand the table as much as i should so i will just wait till monday to explain more clearly what is going on in the table. sry guys
by the way the next scribe is CRYSTAL
February 14, 2007
After that Mr. K gave us three integrals to antidifferentiate. When solving these three questions you must consider these two derivative rules:
After that Mr. K talked about "Trapezoid Sums" and "Midpoint Sums". I'll sum it all up in four simple diagrams.
For Trapezoid Sums when the curve is concave down, the sum would be an under estimate. When the curve is concave up, the sum would be an over estimate.
For Midpoint Sums, if you rotate it and make it a line tangent to the curve you'll see that when the curve is concave down, the sum would be an over estimate. When the curve is concave up, the sum would be an under estimate.
Well that's all I was able to muster up for this class, as you all know Charlene was the next scribe, her post is already up. Enjoy!
I also have a very important announcement to make. Even though I was convinced to stay in the course by Mr. K himself I truly believe now that it's best if I don't continue on. I experienced something in our last class that couldn't be a better reason for my leaving. I was having trouble with completing a chart, not only because I was unaware of the fact I was doing my task improperly, but also because there was something in there that I SHOULD'VE KNOWN from past pre-calculus courses that I completely forgotten. That explains to me that I haven't done my past math courses to the best of my ability, therefore hindering my learning experiences in my present math courses. I've already begun trying to fix up my past by taking my PRECAL 40S course again. This decision hopefully will allow me to take Calculus again in University, and this time understand things completely with the mind set of thinking I know material I should know and that'll make things easy for me. One last thing, I know Mr. K you tried to keep me in your class and you succeed. However, this time I've decided and no matter how disappointed you may be, I feel this is the right thing for me to do in my life and my soon to be career in education. Please I'm not asking you to agree with me on my decision, but I'd like to know you're always there for me when I need you. I'm terribly sorry Mr. K, but I'm not going to ask for your forgiveness on my decision.
February 07, 2007
Here is the draft of whose in what group and whose in what game:
-Mark (a maybe)
4)Game: Painted Faces **Has yet to be decided on**
At the end of Class Mr. K talked about the rules of the Calculus Exam.
**keep in mind that nobody is NOT allowed to say a word about the exam EVER
**remember in the exam it's better to skip and not answer a multiple choice question than to just totally guess on a multiple choice question. ***you will NOT lose any marks if you skip a multiple choice question***...but if you totaly guess on a m.c.q. and you get it wrong, marks WILL be deducted.***
-starting next class we will be doing timed multiple choice quizes at the beginning of class to prepare us for the exam! DO NOT BE LATE FOR CLASS b/c we're going to be timed!
Homework: read the next exercise: 7.6 (it will help you for the next class)
NEXT SCRIBE: DANNY (for friday's class)
February 05, 2007
With this definite integral, we can find its antiderivative and find the area.
Let's take a look at the graph...
Solving problems like this are not always this easy as this one. We can't always find the exact value of the definite integral.
Take a look at this problem and its graph.
We could use Riemann sums to find the area under the curve.
These are the differentiation and integration formulas. We use these to help us determine the problems below.
Use substitution. (I FORGOT TO PUT dx INTO THE PICTURE)
u= x 3
1/3du= x2 dx
If your wondering why u = x3 its because
(x3)2 = x6. So u is u2. We are trying to make the original function look like d/dx Arctan(x) integration formula.
You don't need " + C " in the intermediate steps because the C are not the same while your substituting.
(There's supposed to be an equal (=) sign in between the pictures)
If the 8 was a 9 in the polynomial we could complete the square.
So we do this,
6x-x2 -8 Factor out the negative.
-[(x2 -6x+8 +1) -1]
-[(x2 -6x+9) -1]
1 - (x-3)2 We can replace this into the integral.
Next we use substitution.
Let u = x+3
du = dx
And we're done
Homework: finish up arctrig exercises (7.5)
Wednesday's scribe will be Katrin.
February 04, 2007
Wrong Way Joe!
Joe Physics was driving down the highway on this past Memorial Day weekend. Joe's car ride was getting boring so he decided to calculate his total distance from his house from t=1 to t=3. Normally he could have used his odometer, but he drove backwards on part of the trip. Joe figured that his velocity as a function of time could be given by v(t)=3t^4cos(t). If at t=1 his displacement is 5 what is the total distance traveled by Joe from t=1 to t=3 using integration by parts?
This is what we came up with as a class. XD
On Friday we went to the University of Manitoba for a workshop on "Personal Learning Environments: A Better Learning Landscape?" It was quite fun. =) Driving down the highway with the Calculus class. I thought it was a day well spent. Bravo to Mr.K for his awesome improvisational skills.
The related rates photo is due on, I believe, Tuesday.
I hope everyone did well on their exams. BACK TO CLASS TOMORROW! Next scribe is ASHLYNN!