Precal Blog

On Friday during class we were in the cafeteria because the math room was having work done in it. In the cafeteria we worked with our peers around us to help us get set for Monday. There was 2 problems. The first problem was

You decide to begin contributing to a savings/retirement fund. You find an annuity with projected annual earnings of 6% compunded quarterly. (In other words, 1.5% every 3 months.) You will make a quarterly payments of $600.

All we had to do was figure out and write down the pattern. The pattern my group came up with is as follows.

A1=600

A2= 600+(1.015(600))

A3=600+(1.015(600)+(1.015(600)squared)

A4=600+(1.015(600)+(1.015(600)squared)+(1.015(600)cubed)

The pattern works and we will discuss in detail what goes on during Mondays class.

The second question is very similar to the first problem. The second problem is below

Suppose you take 500 mg of acetaminophen (Tylenol) and continue taking 500 mg every 4 hours. About 59% if the drug is eliminated by your body every 4 hours (so 41% remains). Assume that each dose immediately enters your system. Below is the pattern we found.

A1=500

A2=500+(.41(500))

A3=500+(.41(500))+(.41(500)squared)

A4=500+(.41(500))+(.41(500)squared)+(.41(500)cubed)

On monday we will discuss what these patterns mean and how this is relating to what we will be learning the next few days.

Today in class we just went over the logarithms test so I’ll just bring up a few things about sequences.

1. A sequence is an ordered list of numbers

2. For our purposes they are either arithmetic(adding the same value), geometric(multiplying by the same value), or neither

3. They can be written as recursive or explicit functions(As addressed in Emily’s post)

4. Remember that when your doing problems that the first term might be t(1) or t(0) so it’s important to pay attention to that

5. If you want to find the nth term just plug n into either your explicit or recursive methods

6. If you want to find out what term a given value is set it equal to your equation(solving will require, everyone’s favorite, logarithms for geometric functions)

If you understand these basic things about sequences you should be fine. On the bright side our class can’t do any worse then we did on the logarithms test. Good luck with series tommorrow everyone.

Scotty P.

(Just a side note .9 repetant is equal to 1)

If a sequence is neither arithmetic or geometric then a matrix can be used to write an equation for it.  I will admit, I barely remember how to use a matrix considering we learned them freshman year.  The good news is that they aren’t hard to pick up on again!

Let’s use the following example because it is neither arithmetic or geometric.  The sequence goes like this:

4, 10, 18, 28, 40, 54, 70…

Notice that the differences between each of these numbers aren’t constant. 6, 8, 10, 12, 14, 16… But, the differences of the differences is constant, 2.  This means that the equation for the sequence has a squared value.   We can use the quadratic formula, which is tn=an²+ bn+c (where n is the term number and tn is the value in the sequence).  Since there are three unknowns (a, b, and c), we need to three equations.

4=a(1²)+b(1)+c
10=a(2²)+b(2)+c
18=a(3²)+b(3)+c

Now, this is where the matrix come in.  We use the general equation Ax=b (where A is the coefficient, x is the variable, and b is the solution).

Notice how the first row corresponds with the first quadratic equation, the second row with the second equation, and the third row with the third equation.  Now, in order to get the x column all by it self we would have to divide b by A, which is the same as multiplying b by the inverse of A.  This can be done by putting the matrix in your calculator.  First, go to second inverse, which is the matrix key.  Scroll over to edit.  Make sure A is 3×3, then fill in the numbers into the matrix.  Now edit b, make sure this is 3×1, then fill in the numbers.  Go to the home screen.  To get the solution, hit the matrix key then select A.  Now, hit the inverse key to get the inverse of A.  Go back to the matrix menu and select b.  Hit enter, and you should get 1, 3, and 0.  You plug these numbers in for a, b, and c in the quadratic equation.  

tn=1n²+ 3n+0 (n is the term number)

We now have the equation for the sequence!

Today in class we had to claim problems from section 14.2. The claimable problems were numbers 2,4,9,13,14, and 20. Since no one claimed 20 and because I wasn’t paying attention when 9 was presented, I will go over 2,4,13 and 14. My apologies to Jake for not paying attention to his presentation but I’m sure it was great. To save you the time of scrolling down past all of this just to see who the next scribe is, it will be Scoot (yes I spelled it that way on purpose)

For problems 2 and 4 you had to

a. tell whether the sequence is arithmetic, geometric, or neither

b. write the next two terms

c. find t100

d. Find the term number of the term after the first elipsis marks

2.  27, 31, 35, . . . , 783, . . .

a. because the sequence increases by a consant (4), it is arithmetic

b. to find the next two terms just keep adding four, (39, 43) 

c. you can create an explicit formula for the sequence, (tn=27+4*n) and just plug 100 in for n to find the 100th term, 427

d. use the explicit formula to find n when 783=27+4n. n=189

4. 100, 90, 81, . . . , 3.0903… , . . .

a. because each term is formed by multiplying the pervious term by a constant (.9), it is geometric

b. multiply the previous term by .9, (72.9, 65.61)

c. the recursive formula is tn=100*.9^n, (.00265…)

d. 3.0903=100*.9^n, you solve for n and you get the equation .030903=.9^n, to solve this you need to use your knowledge of logarithims YA! You know that log(.030903)=n*log(.9), solve for n and n=33

13. An office building is worth $1,300,000 and it’s value depreciates every year.

a. The building depriciates by $32,500 every year. Find the first few terms. What kind of sequence is it? How much would it be worth after 30 years? When will it be worth nothing?

Just subtract 32,500 from 1,300,000 three times (1296750, 1235000, 1170000). It decreases by an constant so it’s arithmetic. The explicit formula is tn=1300000-32500*n, so when n=30, tn=325000. 0=1300000-32500*n, n=40.

b. The building depreciates at 10% a year. Find the first few terms. How much will the building’s value decrease the first, second, and third year? When will the value decrease by less than $32,500?

Creating a chart of years, value, and how much the value decreases will help you solve this problem. The first few terms are (1170000, 1053000, 947700). The values it decreases by are (130000, 117000,105300).  Because the amount the building’s value decreases by 10% of the previous term, the equation to find when the value decreases by less than 32500 would be 325000=1300000*.1*.9^n. using logarithms, you solve for n and get that n=15.

14. You decide to save money by putting $5 into a piggy bank the first week, $7 the second week, $9 the third week, and so forth.

a. What kind of sequence is this? How much will you deposit on the 10th week? When will you deposit $99?

It increases by a constant so it’s arithmetic. The explicit formula is tn=3+2n so when n=10, tn=33. 99=3+2n, n=43.

b. What is the total amount you saved at the end of the 10th week? Show how you can find this by averaging the first and tenth values and multiplying this average by the total number of weeks.

 5+7+9+11+13+15+17+19+21+23=140. ((5+23)/2)*10=140

c. How much would you have saved in a year?

There are 53 weeks in a year so you would save $2912

So that was today’s class. I hope this helped and if your looking for the next scribe, you obviouly missed it in the first paragraph. Good job being observant!

All we did in class today was continue to work on the problems from 14-2, which I’m pretty sure will be claimable tomorrow. And since I was told to write something that is actually of importance, I’ve decided I’m gonna blog about recursive and explicit formulas, since I don’t think anyone has written about these yet.

To start off, I’m going to give a problem for which one can write both an explicit and recursive formula for, and explain these two in further detail as I go on. For those reading from a certain math-blogging honors pre-calc class (which I doubt many are, since honestly; nobody really reads these), you’ll most likely recognize the example problem from the notes we took one day in class.

Example:

1.)Suppose you have 75 mg of a pain-reducing drug in your system when you take 500 mg more. Your body eliminates 30% of the drug each hour. What can you determine about the amount of drug in your system over time?

The first thing one is advised to do with a problem like this is to make a simple t-chart of values, in this case with the variable “n” representing the numble of hours and “y” representing the mg of drug left in your system. Since I can’t exactly make one here (I probably could if I tried, but I’m not feeling particularily ambitious today) I’ll just list them off:

N: Y:
1 402.5
2 281.75
3 197.225
4 138.0575
5 96.64025
10 16.24232682

Now to create an equation. Or equations, in this case. Looking at the data table above, you find that as you add one to the n value, your y value is multiplied by some unknown value. But is this value really unknown? Look again at your y values. More specifically, at the pattern between these y values. If you multiply your previous y term by .7, you get your next y value (this .7 relates to the 30% reduction per hour of the drug if anyone is wondering)

402.5 * .7 = 281.75

This little tidpit of information gives you the material for the recursive formula. This formula would be y(n) = y(n-1) * .7. Now just so you know, the items in parenthesis are little subscripts symbolizing your n value from above. The purpose for this equation is to find the next y value by using your last y value. For example, say you were searching for y for n=2. To do this, for y(n-1), you would plug in your value for y for which you received when n=1. The y(2-1) would give you y(1), which would ask you to plug in the previous value. From there on, you’d just solve it out. Simple.

The only downside of the recursive function is that it’s only useful if you know the previous answer. For example, say you wanted to know how many mg of drug was left after 10 hours. To do this, you’d need to know what y was when n=9. Therefore, you would have to use this formula over and over again just to get there. This is where the explicit formula comes in handy.

The explicit formula is the type of formula one usually would normally write right away to solve this type of problem. The equation would be 575 * .7^n = y, since there was 575 mg of drug from the start 70% of the drug would be left if for every hour, it went down by 30%. This formula now makes it completely easy to find out how much of the drug will be left after ten hours. Simply plug it in, and you’ve got 16.24 mg left, a value that can also be located in the table I provided above.

Now that I got the recursive and explicit out of the way, I will explain how you can tell what kind of sequence and function these values are. First off, this sequence can be one of two. It will either be an arithmetic sequence or a geometric sequence. To tell, you take a look at your y values. More specifically, at the pattern of the y values.

Do you add a constant to the previous term? Or do you multiply a constant? In this case, there is multiplication. Because of this fact, this would make this problem a geometric sequence. To find out what type of function, just think about it. The equation for an exponential function is f(x) = a * b^n and the equation for a linear function is f(x) = mx + b. Think about the explicit formula written above. Which of the two does it resemble the most?

Yes, this problem is exponential. And just so you know, an arithmetic sequence will always be a linear function, while a geometric sequence will always be an exponential function.

Well, I believe that pretty much covers the basics of recursive and explicit formulas. Job well done to me.

Next scribe will be… STEPHANIE Q.

Today was another work day from section 14.2 problems. Here is an example:

11) Grain of Rice Problem: 1 , 2 , 4 , 8 , 16 , . . .  1000.

a) The sequence first off, is an addition of the term number and multiplication situation, so it is Geometric.

b) The explicit formula is (1), starting number * (2) the growth rate, to the x term power. = tx = 1 * 2^x

c) To find the 1000th grain of rice, utilize the log of both sides, so 2^x = 1000 is log(2^x = log(1000 divided out is

x log(1000) /  log(2) and x = 9.9657

This might help: http://www.youtube.com/watch?v=IGFQXInm-co

The next scribe is AJ

good luck!

Today was a work day, and we worked on problems from section 14.2.  Claims for these problems will be on Wednesday. The problems deal with sequences.  There isn’t really much else to say about today…

The next scribe will be Jared.

Today, as in Friday, we learned more about logarithms  and the properties used with them. The three we learned were

log(a^n)=nlog(a)                      log(2^6)=6log(2)

log(a)+log(b)=log(a X b) log(2)+log(3)=log(6)

log(a)-log(b)=log(a/b)            log(10)-log(5)=log(2)

We derived these using different examples like the ones shown above.

Then we talked about using theses properties in equasions like

log(x)-log2(x-3)=3  —>  log((x-1)(x-3)) —> 2^3=(x-1)(x-3) —> x=5

Lastly, we reviewed teh logs we already know and practiced using them to fill in a number line.

log(4) = log(2) + log(2) = .602

log(5)= log(10/2) = log(10) – log(2) = .699

log(6) = log(3*2) = log(3) + log(2) = .778

etc….

Thanks(:

Today in class we reviewed what we learned yesterday and added some new log equations that can help find answers easier. Also, we looked at figuring out logarithms for 4 – 10 using the equations.

The logarithm equations (?) that we now have are:

log(a^n) = n log(a)       ex. log(2^5) = 5 log(2)

log(a) + log(b) = log (a*b)     ex. log(10) + log(2) = log(20)

log(a) – log(b) = log (a/b)      ex. log(30) – log(5) = log(6)

You can use the powers 10 scale to figure out the following problems or solve them using the equations above as I have them shown below:

log(4) = log(2) + log(2) = .602

log(5)= log(10/2) = log(10) – log(2) = .699

log(6) = log(3*2) = log(3) + log(2) = .778

log(8) = log(4*2) = log(4) + log(2) = .903

log(9)= log(3*3) = log(3)  + log(3) = .954

You can also use the equations to figure out more complicated logarithms:

log(8/3) = log(8) – log(3) = log(4) + log(2) – log(3) = .426

Hope this helped! The next scribe will be Steph Q.