Mr. O'Brien's 2009-10 Red Pre-AP Calculus: April 2010

Friday, April 30, 2010

Scribe Post April 30, 2010

We started class with a homework quiz. Mr. O'Brien said that homework quizzes are like a doctor's thermometer, and that we shouldn't really stress about them, but they can tell you how much you need to focus on the homework.

We also talked about the final a little bit, which we'll take with the seniors.

And then we took some notes...

Vector Intro
A vector is a directed line segment. There is a vertical and horizontal segment to a vector.

- The length of the vector is the magnitude
- Theta is the direction angle

Homework

29. This problem was trying to find the bearings of ships. We labeled the most of the angles, and used the law of cosines to solve for the triangle. We got C by doing the cos inverse. After finding C we used the law of sines to help solve the rest of the problem.

9. We used the law of cosines, with a^2 being on the left, and then used the law of sines.

21. There was a domain error because it was a bit too short to make a triangle, so the answer is that there is no triangle.

35. Drawing a diagram is really helpful, and then you can use the law of cosines.

The homework for next class is p. 456/3-11 odd, 57-61 odd.

Wednesday, April 28, 2010

April 28, 2010: Math Class

Today, we started class off with a warmup, which utilized the Sine Rule that we practiced for homework last night. This warmup was tricky, for it gave:
A=26 (degrees)
a=5
b=8
Which from our reading last night, was the tricky A.S.S. problem. This was interesting because there were more than one triangle that is possible from these three pieces of information.
We realized that you have to be careful using the Sine Rule! We learned that you could solve for the other triangle by finding:

and

. Once you find these values, you can solve just like we did in the homework.

We also talked about how we may be given values that do not form any triangles!
From the warmup, we learned about the dangers of the SSA triangles, for there are two triangles you have to watch out for, as long as b (or the side opposite the given angle is greater than the height, but less than the other given side.

After solving the problem, we were challenged to make an accurate representation of it on Geogebra, which was a little harder than many thought it would be.

Next, the class looked at several applets that helped us understand more about the Sine Rule, by playing around with the a, b and c values.

We learned several rules, which can be seen at this applet: http://home.comcast.net/~paulbirdsall/class/ambig.html

O triangles means that:
b < h

1 triangle means that
a. b = h
b. b > h and b > c

2 triangles means that
h < b < c

After this, we looked at several problems from the homework, mainly the word problems:
These were all hard, because there was usually a trick we had to discover in order to get the ball rolling on finding A,a,B,b and C,c. Once the trick was solved however, it was easy to follow the Sine Rule, and solve them.

Here were the tricks!

37. Find the 48 degree angle from the right triangle, then you have the crucial angle, and can easily solve for theta using the Sine Rule.

39. The parallel lines are key here. you have to look at the parallel lines, to find the alternate interior angles! Then you can fine the triangle with the added angles, which again, is the crucial answer. Easy to solve from there!

41. In this problem, you again need to look for alternate interior angles. If you look at the diagram that Mr. O'Brien drew, it is easy to see the triangle that can be used to solve b. From there you can solve for c and d easily.

43. This again, is just a matter of looking at the geometry of the problem, you simply need to find the triangel to solve for x, then you can find d!

This was most of the homework we went over, and we turned next to looking at the Cosine Rule. We derived this in a similar way to the Sine Rule. I don't think anyone was missing, so it shouldn't be a problem of getting notes from this!
It does seem a little more complicated than the Sine Rule, but we'll be spending more time on it in the coming classes, so I'm sure that with homework, and classwork, we will understand it fine.

Overall, this class was productive because we covered word problems extensively, and learned about that tricky A.S.S. problem! The homework will enforce the Cosine Rule, and any doubts will be answered in class on Friday.

Check the iCal for homework!
There's a quiz on friday over all of the past homework!

Hannah will be the next scribe!

Remember:
The big unit test is two weeks from Wednesday.

Links for Wednesday, April 28th

http://www.geogebra.org/en/upload/files/english/dtravis/ssa_ambiguous.html

http://home.comcast.net/~paulbirdsall/class/ambig.html

Monday, April 26, 2010

April 26, 2010 - the greatest day in the history of mankind

Happy birthday to Annieeee, happy birthday to me!

For a warmup today we had about 25 minutes to...

1. Graph $y=\frac{(2tanx)}{(1-tan^2x)}$ . We wanted to be able to do this problem on a non-calculator section. This problem required the use of the double angle identity for tangent. The graph was basically that function compressed by a factor of 2.

2. Solve the equation $sin2x-cosx=0$ on the interval $[0,2\pi)$ . We also tried to be able to do this one without a calculator. This one required the use of the double angle identity for sine. Then you can factor out the cosx and solve for the cosx and sinx values - then find the spots on the unit circle where those values are true.

3. Find the exact value of cos150˚.
We first thought of doing this by splitting up the cos150 into parts for a sum identity- 60˚ and 45˚. Maci pointed out that you could also transfer it into radians if that helps you. Mr. O'Brien reminded us that we can also use the difference identity for cosine or a half angle identity. We decided to try the half angle identity. So we split cos150˚ into $cos\frac{1}{2}(210)$ and then used the half angle identity to progress through the problem. Our answer comes out to be $-\sqrt\frac{2-\sqrt3}{4}$ . This is an exact value. Then we compared this answer with the answer the sum identity people got, which was $\frac{\sqrt2-\sqrt6}{4}$ . Turns out, the two values are the same! Mr. O'B challenged us to show why these two values are the same thing - anyone can try this for an IOU.

4. Finally, Mr. O'Brien gave us a simple graph with a line coming from the origin that went through the point (1,2). The angle created by this line was called u. We had to use this graph to find the exact value of cos2u. Again, this one required the use of the double angle identity for cosine. We found the answer to be $\frac{1}{5}-\frac{4}{5}$ or $-\frac{3}{5}$ . We then tried to do it with technology using arcsin and arccos, which confirmed our answer. Easy peasy pumpkin squeezy.

Today we discussed what's coming up. All we have to do for the next couple of weeks is keep up with the homework - we have homework quizzes on Friday and next Thursday. The big unit test is two weeks from Wednesday. So make sure to keep up with homework, and remember to only use the answers from the book as a checker - if you just copy them down, you won't learn the processes.

We also talked about the homework quiz from last class. Mr. O'Brien handed them back... There seemed to be a general consensus that it was difficult and that we wished we had had more time. The best way to be prepared for these is to just be thorough with your homework and make sure you ask questions and check your answers. Mr. O'B reminded us to go back and check all our quiz questions that we got wrong so we can learn from them. All that good stuff.

Notes from the quiz:
1. Remember to sub back in answers and include the +2πk.
2. Remember to always take the plus and minus of a square root. You don't have to list all four solutions. We could have just used a calculator and then figured out what the answers are, but that can take more time.
4. If you get stuck on the algebra, just solve it with technology and move on. Then come back later. But this one required the use of tangent and secant identities. 0 was an extraneous solution.
5. Using the quadratic equation resulted in $u=\frac{11\pm1}{30}$ . Then you had to remember that u really meant sinx, and you could solve using arcsin. Then you had to subtract both those from π to find the other two answers. This one was tricky tricky.
6... was quite fun. Mr. O'B suggested making a table with x values of 1, 2, 3... 12. Wherever s is larger than 100, those numbers represent the months of January, November and December.
7. Remember when you have a denominator of 12, you can break it up into 3π/12 and 4π/12. Then you can solve using a difference identity.

Next we went over our homework questions from the p.420 assignment - specific questions were #41, #43, #77, and the tan part of #53. We went over each question individually and worked them out together. If you were absent today and need help on those, you can ask someone for their notes.

41. $cos^2x+sinx=1$

43. $2sin2x-\sqrt2$

53. $tan(\frac{25\pi}{12}) = tan(\frac{11\pi}{6}+\frac{\pi}{4})$

77. Prove: $sin4x=8cos^3xsinx-4cosxsinx$

FINALLY... we went back to "The Geometry Days", as Mr. O'B fondly refers to them. The next 2 days are going to be spent on the Sine Rule and Cosine Rule. We learned how to derive the Sine Rule in class today. If you need notes, I'd borrow them from Lange. She does pretty colors and everything. Or Nick. He goes everywhere she goes. For the homework tonight you'll need to use the Sine Rule plus the formula for the area of a triangle (half base times height).

Next scribe is Sir Nick!

Tuesday, April 13, 2010

Tuesday, April 13th

We began class by working on the Sum, Difference, Double, and Half Angle Identities packet, which we received last class.

We reviewed the distance formula; when finding distance between two points, you subtract the distance between the x-coordinates and the distance between the y-coordinates, square them each and add the two values.

Think the theorem of pythagoras!

"Mother of All Identities"
"It's from God..." -Mr. O'Brien.

When looking at the relationship between sin and cos, we wrote sin(a - b) in terms of cos:

This gives us the difference identity for sine. In class, we completely derived both the difference and sum equations. In addition to the identities for sin and cos, we also derived the identities for tangent (use FFOO for these guys!).

*Remember, you can reference these identities by looking in the front cover of your textbook.*

Next, we went over the following homework questions the class had: *19, 29, 37, 47, 53, 71

We then began to go over the following Double Angle Identities:

Homework for Next Class:
* p. 415/3, 9, 11, 13, 19, 21, 23, 25, 49
* Prepare for HW quiz next class. You will be given the sum and difference identities for use on the quiz. Happy studying!

And the next scribe shall be...Julia! :)))

Friday, April 9, 2010

April 9th Class

Today we started off class with the warm up. It said, Solve on [0,2 $\pi$ ) a) $2cos^{2}3x=1$ b) $sec^{2}x+tanx-3=o$ . We found that there were several answers for A, which were $\frac{\pi}{12},\frac{3\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{9\pi}{12},\frac{11\pi}{12}, \frac{13\pi}{12}, \frac{15\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}, \frac{21\pi}{12}$ . So the short answer is $x=\frac{\pi}{12}+\frac{\pi}{6}k$ . A common mistake on A is not including the plus or minus when the right side is squared. For b, the four answers are $x=\frac{\pi}{4},\frac{5\pi}{4}, tan^{-1}(-2)+2\pi, tan^{-1}(-2)+\pi$

Then we went over the homework. The first problem we went over was #33. On this problem we used the Squaring both sides method from page 393. This method squares both sides which allow you convert it to a quadratic type, which is easier to solve. Other problems we went over were numbers 55, 73, and 75.

Then we looked at a new set of identities. These are called the sum, difference, double and half angle identities. THESE IDENTITIES DO NOT NEED TO BE MEMORIZED. We will have access to them during quizzes and tests.
The six identities are:
Cos(a-b)=Cos(a)Cos(b)+Sin(a)Sin(b)
Cos(a+b)=Cos(a)Cos(b)-Sin(a)Sin(b)
Sin(a-b)=Sin(a)Cos(b)-Cos(a)Sin(b)
Sin(a+b)=Sin(a)Cos(b)+Cos(a)Sin(b)
Tan(a-b)= $x=\frac{\pi}{4},\frac{5\pi}{4}, tan^{-1}(-2)+2\pi, tan^{-1}(-2)+\pi$
Tan(a+b)= $\frac{Tan(a)+Tan(b)}{1-Tan(a)+Tan(b)}$
These identities are also located on the front cover of your book

Towards the end we went over and corrected the quiz.

Homework p. 404/5, 19, 29, 33, 37, 47, 53, 63, 65, 71 [these questions should be answered without the use of technology, but be sure to use our new identities!]

Everyone has scribed 3 times now so someone will have to step up next class

Keegan

Wednesday, April 7, 2010

4/7/2010

Today we started off with a little quiz on 5.1 and 5.2. The back page was a little tricky.

Then we got to look at all the Quarter 3 projects from R3 and W2. Mr. OB was right, Andy's was very entertaining.

We got to go over some problems from last class's homework on page 396: 5, 21, 33, 35, 56, 68, & 75.

The goal of last night's homework was to get us to practice solving them only with technology. Then we will practice solving them entirely without technology.

$[0,2\pi )$

33) csc(x)+cot(x)=1 (enter into calc as $\frac{1}{sin(x)} + \frac{cos(x)}{sin(x)}$ and set scale to pi/12, domain )

Be careful when entering cotangent into the calculator, it should be entered as cos(x)/sin(x), not 1/tan(x) which is undefined to the calculator.

Answer: $\frac{\pi}{2}+2\pi k$ , $k\epsilon Z$ (funny looking Z that wasn't on the equation editor)

75) $h(t)=53+50sin(\frac{\pi}{16}t-\frac{\pi}{2})$ Factor....

$h(t)=53+50sin(\frac{\pi}{16}(t-8))$

a) So, at 8 seconds and 24 seconds it is 53 ft above the ground.

b) During 160 seconds, the person will be at the top after 16, 48, 80, 112, and 114 seconds.

We can then confirm this on a calculator and set limits/scale.

Solving with Algebraic Methods:

5) Use the unit circle when solving algebraically, it's quite helpful to check your work.

If you have to solve this... begin by factoring:

$2sin^{2}x-sinx-1=0$

Let sinx=u

$2u^{2}-u-1=0$

$u = -\frac{1}{2}, 1$

This is when you would use the Unit Circle:

$sinx=-\frac{1}{2}$ at $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$

at $\frac{\pi}{2}$

$x= \frac{\pi}{2}+2\pi k$

$\frac{7\pi}{6}+2\pi k$

$\frac{11\pi}{6}+2\pi k$ , $k\epsilon Z$

21) $cos^{3}x=cosx$

$cos^{3}x-cosx=0$

Factor out a cosx...

$cosx(cos^{2}-1)=0$

and

$cos^{2}x-1=0$

$cos^{2}x=1$

$cosx=\pm 1$

You can than look at the unit circle to see where cos is equal to 0, 1, and -1.

Or, if you prefer a graphical analysis, you can look at the graph of a cosine wave and see where the graph is at 0, 1, and -1.

$x=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$

**28) This was not on the homework but Mr. OB thought it would be a good one to go over in class.

Separate the sec and the tan on separate sides and square...

Factor...

$sec^{2}x=1-2tanx+tan^{2}x$

and since we know that $sec^{2}x=1+tan^{2}x$ through the pythagorean identity, they cancel on both sides to get...

and tan equals 0 at...

$x=0, \pi$

However, if we evaluate this on a calculator and check the values on the table, we can see that pi does not work - it is an EXTRANEOUS SOLUTION, that comes from squaring both sides in the beginning of the process. We must always remember to check our answers before moving on. As Mr. OB says, "It's like brushing your teeth and flossing. You can't get away without it"...

35) $cos2x=\frac{1}{2}$ This is considered a multiple angle function because the 2 shrinks the period of the function to pi, meaning there are more revolutions in a 2pi segment. We know that cos is equal to 1/2 at pi/3 and 5pi/3, so: