Integration by parts

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The next scribe will be...... Robert. I have to do a lot of digging up just to find who hasn't scribe for the sixth cycle. Our homework is exercise 7.4 all odds and number 22. Have a good night everyone!

BOB

I personally think that this sixth unit was very short but compact with tons of information. I think that i will have very little trouble with this test (hopefully) . There was really nothing too difficult in this unit other than the fact that i might mess up on using the proper notations, so i need to work on that tonight after i study for chemistry. Thats it for my last BOB for this year. Good night and study well.

BOB

Wow, this unit happened really fast. I personally think that this unit was a tough one. There are just too much stuff in this short unit. I really had a difficult time understanding the concept related into solving optimization problems, because it ties into all of our previous math courses. Hopefully, with the help of continued practice and A LOT of studying through various calculus sites, I will somehow pass this test. Thats it for my 5th BOB, good night everyone!

Optimization Problems Workshop

Hello everyone, Its m@rk here for Friday's scribe. Last Friday in class, Mr. K separated us into 3 groups just like what we always do during pretest or workshop classes. During the class we practiced solving more optimization problems so we can get used to it. Solving these kinds of questions involves using all our knowledge from previous math classes. The first question involves using diagrams and engraving information into it, in order to solve it easier. The second question requires the use of variables in the problems in order to build 3 sets of equation. The third question requires the use of similar triangles. Notice how our previous math classes are so important in calculus. The questions and the detailed solution can just be viewed in the slides. Thats it for my scribe. Our homework is Exercise 5.4. The next scribe will be Tim_MATH_y.

BOB

Well, this unit about differentiation rules has been a really tough one for me. This unit started off fairly easy, but it got a little challenging when the chain rule and related rates problems came up. I must admit that I'm still having troubles with those two concepts, but hopefully with the help of more studying, i can still get a decent mark. Thats it for my BOB, good luck everyone!

Tangent Approximation and Newton`s Method

Hi everyone! Its Mark here for today`s scribe.

We started today`s class by finding an approximate value of the square root of 37, without using our calculators to get the answer. The burning question is , how are we going to do that? We all know that its some where between the range of 6-6.1 since the square root of 36 is 6, but that is not a good approximation. Here comes the tangent approximation to the rescue. Our plan is to use the tangent line approximation of f(x)=square root of 36 @ x=36. First we need to graph the function, to see whether our approximate will be over or under the real value. If the function is concave up at that given point then our approximation will be over.If the function is concave down at that given point then our approximation will be under the real value. The graph of the function can be seen on the first slide. Second, find the derivative of the function @ x=36. After that create the tangent line at that given point. Then substitute x=37. We can see that our answer is approximately 0.006 larger than the actual value. The operations involve can be seen on the second slide.

We can therefore make a general rule that is true to every function to approximate any given value. The rule can be seen on the third slide.

The second thing we talked about in class is Newton's Method. We started our discussion about this method by going on a side story about Newton and Leibniz feud. After that, we went on describe Newton's method. His method is used for approximating zeros or roots of differentiable functions. Well, the big question is, how does it work? First, it starts with an initial guess, which is reasonably close to the true root, then the function is approximated by using the tangent line , and one computes the x-intercept of this tangent line . This x-intercept will typically be a better approximation to the function's root than the original guess. This process can be done over and over again until the root is found, this process is called iteration. In general, we can apply the formula that can be seen on the fifth slide to find a better approximation.

The third thing we talked about are the scenarios where Newton's method may fail. His method may fail if it produces a derivative of 0 or undefined. Also, it may fail if the steps just continues with the same set of values. Lastly, it wont work if the results diverge away from the root that you are interested in.

If you are feeling a little confused by this new concept, you can check this animation out.

That's it for today's scribe. Our home work for tonight is Exercise 4.7 (all the odd numbered questions). The next scribe will be Van.

BOB

Well, this unit about definite integrals was an easy one for me. I felt like I understand all the concepts and do all the homework without stressing too much or even asking my sister. I am really confident that this test shouldn't be too hard, but as for precautionary measures, I will just make sure that i study well for tomorrows test. I hope that everyone will get a good mark on tomorrow's test. Study well and good luck.

BOB

This unit about derivatives was really tough for me. It started off relatively easily until the concepts just kept coming and coming, but as the days go by and with the continued help from my peers, i am starting to understand the concepts a little bit more. I just hope that my knowledge will be sufficient for the test tomorrow. Well, thats it. Good luck to everyone and study hard!

Scribe version 3: Continuing our discussion of continuity

Well, we started off today's class by forming different groups and working on two problem solving questions about derivatives and continuity.

For the first part of the first question G`(t) represents the exact definition of a derivative so in this case we can see that G(x+h)=log((9+Δx)+1). Therefore G(x)=log (x+1). It can be seen on the second slide.

For the second part of the question since x=9 ,therefore a=9 just by looking at how g`(t) represent the exact definition of a derivative.

Now, for the second question it will be a nonremovable discontinuity. WHY? It is a nonremovable discontinuity because a limit has to exist in order for this to be removable. It can be seen on the third slide.

For the second part of the question the answer would have to be NO, because if you buy more than 3 bags of grain from this company the price jumps up. As it can be seen on the graph of the piece wise function that after exceeding three bags of grain the behavior of the function changes for being linear into being quadratic.

Then after doing those two questions we spent the rest of our time, taking a quiz at visual calculus.

Thats it for today. Our homework for tonight is to start doing the supplementary questions for chapter 2. The next scribe will be......(drum rolls)...CRAIG. (sorry but i have to pick someone)

Graphs, graphs, graphs.....

I don't know why the text are really small after i convert the notebook files on pdf but please just bare with it and enjoy.^__^

Derivative of a function at a point

Oh yeah, i forgot to mention that the next scribe will be DINO.

Exponential Functions Lab Solutions

Question 1

Similarities:

• Range is infinite with the exception of horizontal lines
• Domain is infinite
• both have y-intercepts
• can both be transformed
• both can be used to model real life situations

Differences:

• linear function has common difference ,exponential has common ratio
• linear function has constant slope
• we need the use of a tangent line when finding the slope of a exponential graph

Question 2

The y-intercept is at 3 because of you let x=0 you will get a answer of 3. The growth factor is 4.

Question 3

The value of a is 6,because anything with the base of 1 is just itself.

Question 4

It means that a is less than 1 because the graph is decreasing.

Question 5

f(x)=2*5^x

The 2 signifies the y-intercept and the 5 is the growth factor.

Question 6

It is equivalent to 16 fold because in one growth period it doubles, so growth periods it would double 4 times or in other words 2^4 which is equal to 16.

Group: Graeme, Mark, Ethan (I'm apologizing if i misspelled someone's name)

First quiz of the Year

Well, we started off today's class with a little quiz about functions. Then after that we spent some time on answering some of the questions from lesson 1.6 , which is all about inverse functions. We were supposed to do the odd numbered question from lesson 1.5 but we already finished that yesterday. The main concept in lesson 1.6 is that in order to get the inverse of a function you only need to switch the inputs and outputs. Only one-to-one functions have inverses. One to one function means there are different outputs for every different inputs. Thats all we need to remember in lesson 1.6. Do not forget that we will be having another quiz tomorrow about functions.

The next scribe will be .......... Van or is it Phan? I can't spell so whatever you asked for it. Good night everyone and study .

EDITED: On exercise 1.6 we have to do questions number 1,5,9,10,18,28,30 and 31. That is for the people who missed today's class.