Workshop Period

Ohhh my goodness. I nearly forgot I was scribe today because of the basketball game XD. (Go lakers!)

So the title above describes todays class. Mr. K was away again but no worries, he'll be back tomorrow...i think. Anyways, we were given a worksheet that we worked on in our little groups for the whole class. I thought the sheet had some challenging question like 1 and 2. I hope we go over those tomorrow. Yeeaah it was a pretty uneventful period other than that.

So I'll get to the point. Next scriibe is NELSA. "Beautiful" right Nelsa?

Bob for Conics

*Sigh* What a nice time to have a long weekend. Now I can bob early.

Anyways, this unit was really fun! Not to mention it was also short and simple. What really helped make me understand this unit more was the folding exercises with the ellipse, parabola, and hyperbola. Yes it really did help I'm not just making my bob extra fluffy this time XD. Folding was great for homework assignments too! I usually have trouble with anything that deals with graphing but not in this case. I found it really simple and *coughs* I enjoyed doing them. But what I'm worried about is mixing up all the standard formulas for the ellipse and hyperbolas because they're so similar. Other than that there wasn't anything that troubled me surprisingly and I'm really confident with this unit! Hopefully I'll still feel the same way about this unit after we get through word problems.

Parabolas and More...

What we did in todays class:

1) learned formula of a horizontal and vertical parabola

2) graphing the focus, directrix, vertex and I almost forgot the parabola

3) completing the square

How we get (x-h)^2=4p(y-k)

From that piece of paper where we did all those folds, we realized that the distance of PF was equal to the distance of PD. So by putting the coordinates of PF and PD into the distance formula we can get the formula for vertical parabolas. As you can see from the slide we first expanded (y-(k+p))^2 and (y-(k-p))^2. Also (x-x)^2 = 0 if you were wondering where that went. After expanding, we see that most of the like terms cancel out except the py's and the kp's. Then after isolating (x-h)^2 and factoring out the 4p from the right side, we're left with (x-h)^2 = 4p(y-k). The equation for the vertical parabola. The same process is done with finding the horizontal parabola's equation except the coordinates are different.

y=a(x-h)^2+k vs. (x-h)^2=4p(y-k)

-the binomial (x-h) is squared on both

-by moving k to the left side of first equation we'll see that 1/a is equal to 4p

-if a>1 we use to say that the parabola would be narrow, however in our new equation when 4p >1 then it will be wide

From looking at the equation we know:

-vertical parabola since equation starts of as (x-2)^2

-vertex coordinates are (2,-1)

-direction of parabolas opening, which is up

-wider than a standard parabola

To find the focus and directrix, we must find the value of p. p is the distance of the focus to the vertex and the distance between the vertex and the directrix. From the equation we can see that 4p=4 and we get p =1. Now we know the focus is 1 unit above the vertex and the directrix is down 1 unit of the vertex. Then you can find the roots by making y = 0 and voila there's our forgotten parabola.

(y+1)^2 = -8(x-3)...equation tells us:

-horizontal parabola since it's (y+1)^2

-4p is negative so opens to the left-parabola going to be very wide because 4p > 1

-v=(3,-1)

*note* kinda tricky because x and h aren't in the same position as last question

Just like last question, we should find p first. p = -2 since we know the -8 = 4p. We'll ignore the negative sign since distance is always positive. So now we know the the directrix is 2 units to the right of the vertex and the focus is 2 units to the left of the vertex.

Completing the Square!

Feels like grade 11 all over again. Anyways, Mr. K said that most of the time equations will be given to us in general form. General form is when all the terms are on one side of the equation. So to get it back into standard form we complete the square. First we get all the like terms to one side. We usually put the variable that is squared along with its like terms to the left so it'll match the equations for horizontal/vertical parabolas. So to complete the square we take the second terms coefficient and divide it by 2 then square it. Now what we add to the left side of the equation we must do to the right. Then we factor the left side and factor out the -8 from the right and we're back to standard form.

Well there is my short and early scribe post. You better be reading this Nelsa *squints*. Anyways, next scribe is.....who to chooose?! Wait there's only Kristina left psha i want more choices jk. Yeah so kristina your next.

Bob for Combinatorics

This is the most difficult unit out of them all for me. I have a long list of problems in this unit. What I am having trouble with are the word problems. I have trouble figuring out where to start first of all like since there's so many ways to solve these problems. But todays class really helped me. I have a better understanding of how to tackle these word problems after seeing how Mr. K solves them by breaking down the question into parts. I also realize I am making a lot of mistakes with questions because I use pick when I should really be using choose. There was also that binomial theorem stuff that was really a pain but I get that now fortunately. It was just so confusing to stare at all those numbers and letters for the first time. Oh and how could I forget those circle seating questions?! I get half the work right for these questions and mess it up at the end by like multiplying numbers I shouldn't be but I think I've learned my lesson...

There wasn't much I was good at in this unit but I liked doing the easy things of course like the simplifying of those factorial questions. I'm relieved this unit is almost over...I can't wait actually. Knowing those numbers in Pascal's triangle are everywhere will haunt me forever though XD.

Logarithms and Exponent Bob

I almost forgot to bob! How could I forget such a thing? Hmmm shall I start with the good news or bad news first? Didn't Mr. K say it was always better to start off with the good news?

Anyways, I find this unit pretty straight forward. It was just in the beginning where it was sort of confusing because of all the "logarithms are exponents" saying. After finally understanding and getting the concept of treating logs as exponents, things became waay easier. I really hope there's a lot of those power, quotient, and product law questions on the test because I really like solving them. However, on that quiz we got where we had to solve for K I messed it up really bad. I forgot to divide everything so I could get the base e by itself an just ended up going straight to the step where you put log on both sides. Well I guess that quiz helped me see what I was doing wrong. I find myself not making that mistake anymore because of that quiz now. Oh yeah, am I the only one who likes using log better than ln?!

The worst part of this unit for me would probably the graphing. All the inverse talk confuses me but I'm starting to get it now....slowly. What I was doing wrong was I didn't start off trying to graph the original but went straight to the ln or log so I was just making it harder for myself.

I hope I do well on the test tomorrow :D. Sooo I better start reviewing the slides.

Logarithms and Their Laws

Okie please bare with me, I can't write the subscripts so the questions might look funny XD. The first part is somewhat a review, so just scroll down to the 3 laws for the new stuff. Enjoy!

Class started off with a discussion over that flickr project that was due on friday at midnight! But back to math, the first slide was yet another reminder from Mr.K about a logarithm being an exponent. He still insists we're going to forget but we'll prove him wrong riiight?

The anatomy of a power is really important. If you know all the terms and what each part is called it'll help when you're solving these log questions..for real.

The first slide were some warm up questions that we've learned. This is a logarithm question so we know what logarithms do right? Say it with me. A logarithm is an exponent! So how I think when I do questions like the ones on the bottom slide is what exponent will give me that power. For example the log2(2), so we know the base is 2 and the power is 1. What to the exponent of 2 will result in a power of 2? That is how we get 1. In math we are always asked to go backwards and forwards soo there isn't an exception in this unit.

Rewriting a logarithm to it's exponential expression isn't too hard. First, we take the base and make the exponent in the logarithm it's exponent. An eponential expression turns an exponent to a power so the expression will equal the power. For example we use the question log2(2)=1 and rewrite it to it's exponential expression. The base is 2, the exponent is 1, and the power is 2 so our expression is 2^1=2.

To solve the question 2.5^x=4 we turn it into a logarithm. As you can see from the slide above, we had 3 ways the question was solved but only one was correct. The mistake made was multiplying 2 and 5. You can't arbitrarely multiply the numbers. The correct way to do it was the 3rd choice. What must be done first is to divide everything by 2 so the 5^x is by itself. Then we turn it into a logarithm log5 (2) = x. If the question asks for the exact value of x that is how you would write it since log5 (2) is not a whole number.

On Friday's lesson we also learned how to write our answers in an elegant way again. For example we'll use the question y =Loga 3x and solve for x. Once again we write it in its exponential expression. 3x = a^y then isolate x. To make our work elegant we would write the answer as (1/3)(a^y) instead of (a^y/3).

On the slide above, we graphed one of the logarithms. First we shall ignore the absolute value signs and save it for later. Then we turn y = log4 x into its exponential expression and we get 4^y=x. After this step we use our past knowledge of inverses, which we get by switching x and y. So the inverse of 4^y =x is 4^x=y and we know how to graph this now. Unfortunately that isn't the graph we want. So after graphing 4^x=y we take its x and y points and switch em to get the graph we do want! Now it is time to deal with those absolute value signs. We learned in the past that whenever we have an absolute value to just take the negative values and flip them up. You may be wondering what the word cusp up on the slide is. A cusp is different from a corner. It is a point where 2 points meet and makes a very sharp curve. As for where it is increasing the answer is (0, infinite) because that's where the x values start to increase. (0,1) is where the x values start to decrease from what we see on the graph. Notice that we did not put square brackets around the 0 on both because it cannot be increasing and decreasing at the same time. A helpful fact we learned on Friday was if a questions asks "where" it wants x values. If it asks what it is for the y values.

Now it is time for what I thought was the most important part of the lesson, the laws! Let's start with the product law. It is the same concept as (3^5)(3^7)=3^12. When we have powers with the same base we are able to add the exponent. So we can use that same rule for the question log2 (8.16). Since it is multiplying 2 powers with the same base, we know that it must be adding exponents (logs are exponents remember). So the question should look like this now: log2 8 + log2 16. The log2 8 = 3 and the log2 16 = 4. Those are the exponents and we add them to get 7. What if we get a question like express loga 2 + loga 3 as a single log? It is exactly what we did up there but going backwards this time. They both have the same base of a and when we add that means the the powers (2 and 3) are being multiplied. The work should look like this: loga (2.3) and can be further simplified to be loga 6.

Another law we learned was the quotient law. It is also the same concept as (3^7)/(3^5). When we are dividing the same bases we can subtract the exponents. Again it's the same concept when we do a question like log2 (32/128). Since it is dividing the powers it must mean it is subtracting exponents. So we can write this question like this now: log2 32 - log2 128 because logs are exponents. Guess what we'll do next?! If you guessed work backwards...then you're right. We'll use the question loga 5 - loga 4. In this question it's subtracting logs, which is like subtracting 2 exponents. When are we able to subtract exponents? Only when we are dividing powers with the same base. So with that we can write loga 5 - loga 4 as loga (5/4) and that's the answer.

Finally, the last law, which is the power law. Just like the two other laws, it has the same concept as saying (5^3)^7. When there is a power to the exponent of something we can multiply the exponents. So for that question it'll be (5^21). Now using that concept, we can apply it to a question like log2 8^5. Since the power, which is 8 is to the exponent of 5, we know that it must be multiply exponents. Just like multiplying 3 by 7 we multiply the log2 8 by 5 because log is an exponent. We can write the question like this now: 5(log2 8) which is 15. Once again we'll do this process backwards. Let's simplify 3loga 2. The answer is loga 2^3. This is because the exponent, which is the log is being multiplied by a number. Since we are multiply, we know that it must be multiplying the power with an exponent.

That was a really really long scribe post. It was really hard wording the power law. But the concept of a log being an exponent is really helpful if you understand that part. Yes, this is long enough. Too bad, I didn't even get to talk about Star Trek.

Duck duck duck...goose! *points at kristina* You're next scribe.

Flicker Project

Well I finally finished my project with 30 minutes to spare! I did think at the beginning that it would be kind of rare to find a really nice wave, unit circle, or any other trig related topic in the real world since I didn't remember seeing any before. I was sort of wrong, you just have to know where to look sometimes. Since doing this project, I've been looking at everything so differently. I started noticing the little things out there that had to do with the things we learned in math. Just tonight while driving home, I thought I saw some kind of wave made from the street lights in the distance.

Anyways...here's my new picture captured this afternoon. Who would've thought you'd see a wave under the blinds?

http://www.flickr.com/photos/25355345@N05/2407051864/

OH YEAH and i'll publish my scribe post laater it's late.

BOB for Identities

Wow it's already time for another bob?!

Out of all the units, I like this the best for some reason. That doesn't mean I didn't struggle with it though. The first time I saw those proof questions was not pretty. It didn't help that I couldn't remember the formulas or derive them either. I always thought I'd never be able to solve those kinds of questions. But with practice and finally figuring out how to derive, I slowly started getting the answers right. I also found that what works for solving identities is just experimenting at first when you don't see the answer yet. With that said, I just have to work on solving these questions faster because there's only an hour or so to do a test and to find the clever idea in solving them.

Now what I really liked in this unit was the Sine dance. At first I was like "How's this going to help me?" but it did. The sine and cose formula will never leave my mind and I'll always remember it when I need it. That's why I find that I'm really confident when doing the sum and difference questions.

Yeah so the test is on Friday. Just have to get it over with and then it'll be Spring break!

More Pythagorean Identities

K hi guys. My turn to blog for the day or should i say...night.

So we started the morning class off by watching that article 13 video. After that it was straight to math. Today's math class was a workshop since many of us enjoy doing this activity.







Anyways, the lesson for today had a great deal of emphasis on...elegance in showing our work. By elegance, I mean the quickest and most direct way of solving or proofing an identity. The formulas above help a great deal in achieving this elegance. To make things elegant we can change everything into sine, cosine, or any of the other identies to solve faster. Like the question on slide 3. Work did not need to be shown because it is already proven that tan^2a is = sec^2a-1. Showing work is not bad but we are trying to be elegant. Have I overused that word yet? Where was I...oh yeah. Elegance however does not mean skipping steps! Work will need to be shown if the formulas above aren't present in the question. Another thing about the picture above, it is not needed to memorize all those formulas. They are all derived from sin^2x+cos^2x=1. That is the only important one to remember.





This lesson also focused on do's and dont's when solving an identity. The first thing we learned was the "Great Wall of China", which is the line we put under the equal sign. This line can't be passed! This is because it is not an equation. We can't do things like multiplying or adding to balance both sides. However we can "algaebraically massage" the seperate sides into something we can work with. Another don't is putting equal signs, instead we just show the work going down. Putting an equal sign is a no no because we don't know till the end of the work if one side is really equal to the other side. We were also shown some strategies to solving identities, which are above.

Then we were given questions like these, where we could apply those strategies. For the first question, we multiplied top and bottom by 1+sint because we can see 1-sint is a factor of a difference of squares. We multiply it by its other factor to make things easier. Now the top is (1+sint)cos and the bottom (1-sin^2t). The bottom can be changed to a simpler form which is cos^2t! We know that sin^2t + cos^2t=1 without proving it. So after moving sin^2t to the other side we'll get cos^2t=1-sin^2t. That is elegance folks! Then we just continue simplifying it as much as possible. When we're all done we put Q.E.D to indicate we are finished. Q.E.D is some latin word, which I can't remember. So don't ask XD.

The rest of the questions we're mostly the same but using different strategies like factoring and such. Doing these questions need a great deal of practice to get better! Alrightey, there's my scribe post. I hope I included all the important things so yeah. I shall see everyone tmr. Bye bye.

Whoops forgot to name the next scribe. You know who you are! AnhThi. Also, why won't my paragraphs seperate up there...oh well.

Trig Assignment team "Jabawakeez"?

So these are the solutions for the question:

Tidal forces are greatest when Earth, the sun, and the moon are in line. When this occurs at the Annapolis Tidal Generating Station, the water has a maximum depth of 9.6 m at 4:30 am and a minimum depth of 0.4m 6.2 hours later.

Here's how our graph looks like. In the question we were given the max and min values of the wave. The max is at 4:30 and we changed the time to its number value by taking the minutes and dividing it by 60 so its new value was 4.5 on the graph. Now it is easier to add 6.2 hours to it, so the min will be at 10.7 on the x axis. With the max and min we can figure out 1/2 of the period. Since the min and max are 6.2 hours apart we use that. 6.2 * 2 = 12.4. But here's the tricky part. If you do 12.4*(1/4) it will not equal 4.5. So we can't do the 1,2,3,4 ticks and automatically write in the values. What we did was 4.5+12.4 to make it easier to graph. The value in the 4th tick on the x axis is now 16.9. The values on the x axis are now 1/4, 1/2, 3/4 of 16.9. Then you'll just put the points on at the max and min. Then we know the patterns of the curve (after the max) will go to the avg value, then min, avg value again, and then the max.

a)Here's an equation in cose. To find D we found the average of the min and max which was 5. We can now find parameter a by subtracting 5 from the max 9.6. B=2pi/period so in our equation it'll be 2pi/12. In our graph here it does not start at the max where cosine starts so it must've moved by 4.5 our parameter c.

b) First we changed 2:46pm to its number value. Since it has passed into night time we can add the 12 hours from the morning. Then all we do is add the 2 hours and 46 minutes. But we divide 46/60 first. So its number value should be 14.7667. We take this value and plug it into t. This can be done on the calculator...and will get an answer of h=7.1641m.

c)In this question we just plugged in 2 into the equation and solved for t.

BOB for Transformations

It's time for another test. It was a short unit but I liked a lot of the things we learned. I feel really comfortable with those asymptote questions. At first that whole biggering smallering thing confused me but I get that now. I've been enjoying those inverse questions too. It really helped that we were showed how to understand them graphically, algebraicllay, and that baby play concept. What I liked best from those three was the algebraic one. It was so simple to get since all we had to do was replace the x and y places. But graphically helped me understand what was happening as a result of the x and y place changes.

What I had some trouble with in the beginning was some transformations. After getting it through my head that -f(x) meant to take the opposite of the y values and that f(-x) meant to take the opposite of the x values it became easy. The whole reciprocal thing confused me too in the beginning. I didn't understand that all we were doing was undoing everything that was done until like near the end of class. One last thing, I'm also really sloppy when it comes to questions like -3f(1/2x-2)+1. I always forget to factor out the 1/2 and to use its reciprocal.

K there's my Bob for this unit. I'm kinda looking forward to this test too. Hope I'll do better on it than the last one where I like blanked out. Anywhooo that's all lol.

bob for Circular functions

Since our test is approaching for this unit I thought it was a good time to bob!

I was able to understand most things in this unit but my "muddiest point" would be the graphing part. I need still need some more practice with graphing those kinds of questions where all the DABC changes are there. I'm really comfortable with the A,B,C,D transformations but just not all together.

The part of this unit I really liked was probably those kinds of questions where you just have to find the exact values and simplify. At first that was kind of hard since we've all been accustomed to using degrees. I remember converting all the radians to degrees for each question, which was really time consuming. As I learned the exact values on the unit circle, I've found those kinds of questions way easier and didn't need to put it to degrees.

There's my bob for this unit. Good luck to everyone on the test this week?!

Trigonometric Equations

Alrightey I finally got home. Sorry for the late post had some technical difficulties with the computer also. Anyways here's the recap of what we did on todays slides.

Slide 2:

Here we see the reciprocal trig functions we learned about yesterday. Just for review, the new trig functions we learned were cosecant, secant, and cotangent. Mr. K emphasized that most kids go wrong when they think that cosecant is with cos and secant is with sin. So make sure to not do that! Anyways as it says on the slide, we get cosecant by 1/sin(theta), secant by 1/cos(theta), and cotangent by 1/tan(theta). Or we can just change the numerator with the denominator to save time. As I learned today, don't switch any signs. If it's negative in the denominator it will still be negative when it becomes your numerator when you flip the fraction.

Slide 3:

Ok the question here was a bit of a review from the questions we've been doing for the past few days. First we must find the hypotenuse of the big triangle. We do that by plotting the point (9,-40), which are your x and y values. Then we must find the cos and sin values. We do this by finding the angles adjacent to theta over the hypotenuse (for the cos), and for sin we put the angle opposite of theta over the hypotenuse. The cos and sin value you get will be the coordinates for theta! As for the second question on the slide, just list the cos, sin, tan, secant, cosecant, and cotangent values. Remember for the reciprocal trig functions, just flip the fraction.

Slide 5:

We've been practicing how to find the exact values with questions like these on this slide. Doing these questions are a lot easier if you know the values around the unit circle. K i'll make this brief, all you have to do is substitue the exact values in and simplify!

Slide 6:

Oh yes I remember this slide from this morning. Mr. K challenged our mathematical minds to do these questions. But like he said there's a method to his madness or something like that. I think we all know how to do these kinds of questions. We just isolate x or factor.

Slide 7:

We now see what Mr. K was trying to do by starting us off easy. We do exactly what we did on the last slide. We simply isolate sinx and get the value of 1/2. Now what on the unit circle do we know has a y value of 1/2? pi/6, and 5pi/6 of course.

Slide 8:

1+2cosx = 5cosx

The question gets a bit tougher but we still do the same things. Isolate cosx and it will equal 1/3. Now we take our trusty calculators and press 2nd fnc cos then 1/3. Make sure to be on radians. We then get the value 1.2309. But there is another value because cos is also positive in quadrant 4. How do we get the value in quadrant 4? Just put in your calculator 2pi - 1.2309 to get the other value of x, which is 5.0522 the related angle. K a quick summary, if looking for related angle in q2 you take the reference angle and subtract it from pi. If looking for related angle in q3 you take the reference angle and add it to pi. And that's where we ended!

Hope you guys understood my rambling. Oh yeah next scriiibe is...........................JAMIE since I took your turn today. KK goodniight everyone *faints*.