The last BOB

Well, this unit was one of the small units that we had done although I still say the smaller the units the harder they are. Especially for this unit, I am still a little nervous going into this test as I still cannot wrap my head around Euler's method, it is very confusing to understand although I am still studying it. Another spot I am having a little bit of trouble is with the anti differentiating first-order and second-order equations although that also is a working progress which I should have figured out by tomorrow. Well good luck everybody for tomorrow's test and good luck.

Scribe #8

Hello back again for another scribe, Tuesday's class was not very difficult although very important. As we began the day with a simple question, which was "Find the volume generated between the x-axis and the graph f(x)=x^2, from x=0 to x=2. Here is the graph of x^2.

http://fooplot.com/x^2

So we begin to solve this problem by setting up the equation (0,2) ∫ πr^2 dx

= (0,2) ∫ Π(x^2)^2 dx
= (0,2) ∫ Πx^4 dx
= (0,2) ∫ Π(x^5/5) dx
= Π(2^5/5) - Π(0^5/5)
= 32Π/5
Solved.

Then Mr.K began showing us a new form in which solids of revolution can be formed, which was by rotating the function about the y-axis. The equation that represents a function about the y-axis is: V= ∫2 Πrf(x) dx Here is a problem that was worked on in class using this method.


Region S is bonded between two functions, f(x) & g(x), find the volume of the solid generating around the y-axis.

F(x)= 0.5x^2-2x+4 G(x)= 4+4x-x^2

Using the volume equation of a solid of revolution around the y-axis we get, just solve algebraically, do not solve completely:

= 2 Π (0,4)∫ x [g(x)-f(x)] dx

All in all this class was very important as the equation for a solid of revolution about the y-axis was given and explained. Remember all the rest of 8.3 was for homework. The next scribe is VAN.

Scribe #7 (I think)

Hello back again for another scribe post. Sorry in advance to everyone who was waiting for the Friday's post, i was not able to get it up because of my internet connection to the blog was timing out. Well lets get into some math. Friday's class was all about rotating functions around the X-axis. Although this functions were not lines, they were functions like the area between two parabolas, or a function that had never been seen before. It started with a simple solids of revolution function, Find the volume of a solid revolution obtained by rotating the function x^2, bounded by the lines x=2 and x=1 around the x-axis. So here is a picture of that function revolved around the x-axis.
http://fooplot.com/index.php?q0=x%5e2

So between 1 and 2 we want to find the volume of the solid of revolution. So we find that once we make a cut and pull out a piece from the solid of revolution it looks like a circle with a hole in the middle. Where x^2 is the radius of the circle So the area of the circle is A(x)=Πr^2= Π(X^2)^2 =
Π(x^4).

So the volume of the solid of revolution is V=1,2 ∫Π(x^4) dx = Π [x^5/5] 1,2 = [32Π/5] - [1Π/5] = 31Π/5.

After this question we took up questions in our homework from the previous night that we were not able to complete. The next scribe is going to be Dino once again.

scribe^(-1)[scribe(6)]

Well back once again for another scribe. Well class started off very cold, the classroom itself was freezing. Although class started as usually and we began with a couple inverse trig functions. They started off being on the easier side, like sin^-1(sin(pi/3)) in which case sin was in the perfect quadrant, so arcsin was the original value of sin, (pi/3).

Then they became a little harder like, arcsin(sin(5pi/4)) in which case 5pi/4 is -2^(.5)/2. So we know that Sin lives in quadrants 1&4, so arcsin(-2^(.5)/2) would be in quadrants 3&4, so quadrant four will be picked in which case the arcsin(-2^(.5)/2) in quadrant 4 is (7pi/4 ).

Next we went in the opposite direction, which was by taking a length and finding its arc value, then in which case the trig value wanting to be determined will be found. An example is : cos(arctan 1), in which case we work backwards. We first find the arctan of 1, which is pi/4. Then we want to find the trig value, cos of pi/4. The value of cos(pi/4) is 2^(.5)/2. So the answer is 2^(.5)/2.

Next we went to something completely new, and that was to find a length that was not common and find its arc value. Here is an example: cot(arcsin(2/3)), the first thing we do is solve as a function by using x. So it becomes sin(arcsin(2/3)) = sin x ----> (2/3) = sin x. So after we find this we solve using Pythagoras's theorem by using the length and the trig function given. opp=2, hyp=3, and by solving adj= 5^(.5). So now bring in the other trig value of the equation, cot which is adj/opp it solves as 5^(.5)/2, which is the final answer to that question.

Just before the end of class Mr.K asked the class to draw the graphs of arccos(x), arcsin(x), and arctan(x). So just something to remember, the graph of arc trig functions are its Cap trig graphs reflected over the function y=x. due to the fact that if you do not use the Cap trig graphs reflected over the function y=x, the arc trig function would not be a function as it would fail the vertical line test. So remember that piece of advice, as Mr. K said in class today with the use of Star Trek.

Mr.K did not assign any homework today, so we have the night off. Tomorrow's scribe is going to be Tim_Math_y.

Bob VI

Well time to bob except this time I almost lost the mark. Thanks, to Ms. K for letting me use her computer to bob. Well this unit was very quick, a little bit too quick if you ask me. Well the part of the unit that had me a bit in a stir was assigned area questions, as i had become very confused over them and am still not sure if i am able to do them all. Most of the other units in these applications of the integrals unit were not very difficult. Well all in all it is time for me to get back to physics. Good luck everyone, hope everyone had studied hard.

Scribe #5

Well back again for another scribe, sorry that this scribe is late I had other homework. Well we began class with the derivative problem from Friday which we were not able to finish. A good thing to remember when doing a square root derivative problem is that it is always nice to break it into pieces. So for the first question which is asking for the derivative between x and x^2 it is easier to find the value of 0-x and then the value of x-x^2. Due to the fact that it is easier to find a derivative of 0-x. So 0-x first must be reverted to x-0 so that we are able to find the derivative which means that that part of the equation is going to be negative. As you can see on the slides Mr. K had posted, it is much easier to differentiate with 0-x then x-x^2. So you solve by placing x in for t and then find the derivative of x^2 which is simply 2x.

For the next question we are trying to find the area underneath the graph to the x-axis. It is not as hard as it seems. To begin take the equation that is given and find its roots. (To find the roots it is much easier to use synthetic division which is shown on the slide). Once the roots are found the next step can then be followed. As the next step is shown in two different ways, here is the first way explained. When finding area it is not signed area instead is is all the are including both positive and negative. Which means that when evaluating the equations to solve for area you must break up the positive and negative areas in two equations. Then take the positive equation and subtract it by the negative equation so that the negative equation changes sign to positive and the areas may be added. The second method is by evaluating the who equation over the whole interval and then taking the absolute value to receive both the positive and negative area.

For two graphs you follow the exact same procedure, except you are also subtracting the lower graph from the upper graph when finding the area in an set interval, before subtracting the total negative area from the total positive area.

That was basically the whole class. Tonight's homework was 6.4 (all the odd #) and remember to bob before Thursdays test, so that also means pre-test Wednesday.

Tomorrow's Scribe is TIM!

Well back again for another Bob posting. well this unit was not extremely difficult except for one part of the unit. That part was the optimization problems. As there are so many ways optimization problems could be done although only one way of solving them is correct. I find that I am getting better at these questions when doing the homework and exercises, although I find these problems will give me the most grief if they are on the test. Other then these optimization problems I found the first derivative and second derivative rules to be understandable and also the mean value theorem to be very straight forward. Well I should get back to the studying. Good luck to everyone tomorrow and make sure to study.

Optimization Problems

Well i am finally back to scribe, it has been a pretty long time since i had last scribed so here it is. Today in class, Mr. K introduced a new topic Optimization problems. Optimization problems are problems that deal with either finding the maximum or minimum of the function in the problem. although very rarely is the function given so you as a calculus student must use all your knowledge you've learnt up to now and create a formula/function that can be used to answer the question that is being asked. With that said Mr. K gave the class the 6 rules to live by when doing optimization problems, although do not really live by them, just follow them loosely.

6 Rules to Optimization Problems




1. Find what you are trying to either maximize or minimize.

2. Write an equation for it, make sure to use descriptive variables so you won't get confused when doing the problem.

3. Try to get the equation into a two variable form, where you are finding one of the variables, because you want to find the derivative.

4. Write the derivative of the equation in the two variable form.

5. Set the derivative to zero, and solve for the remaining values. (First Derivative Test or Second derivative Test)

6. If needed depending on what the question asks, Plug the value of that variable into the one or (two) equations to find all dimensions.

Now here is the explanation of the question done in class.

Well first we identified it was a maximum problems, because it asks for the box with dimensions with the largest volume.

Next the equation is made by using V=h*w*L, so h=x, w=16-2x, and L=21-2x. So the equation looks something like this, x(16-2x)(21-2x).

Then we multiply it out. 336x - 74x^2 + 4x^3

Then we take the derivative. Which is 336 - 148x + 12x^2

then by using the quadratic equation we find that x = 3 and 9.3, so we do a line graph and find it is positive before 3 and negative after 3 meaning that by the use of first derivative test, 3 is the maximum x can be.

Well, that was it for today's class. Tomorrow's Scribe will be M@rk.

BOB 4

Well hello again doing my fourth bob for AP calculus. Well this unit was very challenging for me especially when we had gotten to the related rates problems. The chain rule gave me a little bit of problems in the very beginning although I was able to correct my mistakes and set myself on the right course. All in all I believe this test is going to be very difficult, in terms of the difficulty of the questions, especially the related rate problem that is probably going to be on the test. Well study hard and good luck to everyone for tomorrow's test.

Scribe Post # 3

Hello again, I am today's scribe. Today was the first lesson of the unit "The Definite Integral". So there was not a lot of information that was covered in class today except the basics of the unit The Definite Integral. If you take a look at the second slide that Mr. K posted just below this scribe post you will see a question like; Give a lower and upper estimate of the distance traveled in 5 seconds.

In order to find the lower estimated distance you must add all the values on the table excluding the last value of velocity at 5 seconds which is 5.7 ft/s. The reason why you must exclude the 5.7 ft/s is because the lower estimate should not have the highest velocity which was recorded. As even with the high estimated distance you do not count the lowest value on the table, because you want to find the highest estimated distance, in this case 1.4 ft/s is not used. So the lower estimated value is 1.4 + 2.7 + 3.5 + 4.5 + 5 = 17.1 the reason why we are adding is because the values in change of time is 1 so 1.4 divided by 1 is still 1.4. The upper estimated value is 2.7 + 3.5 + 4.5 + 5.0 + 5.7 = 21.4.

If you move to the third slide it asks for the actual distance travel. Well you are never going to find the actual distance, only the approximate distance travel. So you begin by adding the upper and lower estimated values and divide by two ( like when trying to get the average of two numbers). Now in this case the value is 19.25 Although there must be an error because more than likely the actual distance is not the average. So we say that there is an error of + or - 2.15.
As there is a distance of 2.15 from 19.25 to either 17.1 or 21.4.

If you move to the fourth slide it asks for representation of the lower estimate, well it is the blue graph, as it is lower then the red graph, which means it is the lower estimate.

If you move to the fifth graph it asks to fill out the rest of the table and pick values for the 0.5 second interval ( which cannot be the average of the whole seconds) so intern once done picking values find the lower and upper estimates in my case I had gotten lower: 18.25 upper: 20.4 . (When finding the estimates you have to divide by 2 because of the 0.5 second intervals) Then find the actual distance in this case after adding and dividing by 2 and you received 19.325. In this case also we received the same error which is + or - 1.075. So in other words the shorter the interval of time, the smaller the error for actual distance.

So that was the class, not much information, but a lot of knowledge on Definite Integrals.
Next class scribe will be Mark.

BoB 1

Well, this unit was a unit that I had struggled with. It was all alright until the day Mr.K introduced the graphing of a function to find its derivative and second derivative. After doing all the homework and discussing it with other students in the class I felt more reassured, although many times I do still get confused on the second derivative of the parent function. All in all i believed i had studied enough and get the understanding of derivative functions. So good luck to everyone on today's derivative test.

Derivative Functions Homework

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Scribe Post 2

Well the class was suppose to start with a quiz, but due to a part of the class not being there at the time of the quiz it was postponed until Monday. So instead we began class by programming a program into our calculators called intercept. It is on page 122 in the calculus text book that Mr. K gave out to all of us if any one has still not programmed their calculators. After programming the intercept program, Mr.K split us into groups where we did 2 problems which are just below this post on Mr.K class slides. That was basically Thursday's class, and for homework Mr.K wants us to create a problem like that on the first slide that Mr.K posted. Make four functions on one graph in which three functions are related by f(x), f'(x), and f"(x). and the fourth function to be unrelated from the other 3 functions and post it to the blog. The next scribe will be Mark.

The Derivative Assignment

Sorry for the late scribe post, but I had a lot of homework last night. Well yesterdays class began as usual with Grey-M posting the word of the day on the board and everyone awaiting the explanation. Except today was different, we had a fire drill, so we had to leave class and return a few minutes later. Then class had begun, with Mr. K reviewing a little bit on derivatives, and had showed us that in order to find out the equation of a derivative, you sometimes have to do some algebraic massage. An example on how it may be used are in the 3 slides Mr. K published yesterday, which explains the multiple steps needed.

Mr. K then gave out an assignment on the many different derivative questions which was done in groups for the rest of the class. The assignment explains how derivatives can be explained by a graph, a table, or an equation. This derivative assignment is due Wednesday. Also Mr. K assigned section 2.3 ( question #: All odd and #12) due for Wednesday as well. Tomorrow's scribe will be Sandy, and do not give the scribe duty to Grey-M as he has no Internet right now.

Scribe Post

that by blogging you Hello once again, this is Dino and I was today's scribe. well i have to say that today's class was not as extravagant as the other class that we have been having throughout the past weeks. Mr. K started of the class with a podcast recording that referenced Mr.k and all the class blogger's receive a world wide audience, which is so very true especially after hearing that podcast. Mr. K then began explaining to the class that he was able to get each member of the class smart board programs which he was going to give us through the means of the blog, which to me sounded pretty cool. Then it was time to get into the inverses workshop he had planned for us today. He began with this problem:

Problem: Given the function h(x) = (3x)/(x+5)

Explain how you know that h has an inverse? The function passes the horizontal line test.

Find a formula for h^-1(x). h(h^-1(x)) = x To get the inverse put the inverse of h into the (x) value of the function h(x). So intern the function h, (inverse of h) = x.

h(h^-1(x)) = 3(h^-1(x)) / (h^-1(x))+5

x = 3(h^-1(x)) / (h^-1(x))+5

x ((h^-1(x))+5) = 3(h^-1(x))

(h^-1(x))x + 5x = 3(h^-1(x))

(h^-1(x))x - 3(h^-1(x)) = -5x

(h^-1(x))[ x - 3] = -5x

(h^-1(x)) = -5x / (x - 3) This is the inverse of the function h.

Suppose f & g are inverses of each other, what is true about their composition?

f(g(x)) = x
g(f(x)) = x Both functions are equal to x.

That was it for class today, although remember to do questions 1,3,5,7,11,13 in section 1.9 for homework. Tomorrow's scribe is going to be......Craig.