Wednesday, May 23, 2007

My friends... The derivative

Tuesday, May 22, 2007

Friday's Chapter 15 Test Topics

Here’s a list of topics for Friday’s test. Can you believe it’s our last one?

Precalculus Chapter 15 Test Topics:
Identify the degree, number of real and complex zeros and the leading coefficient of a polynomial from its graph
Sketch the graph of a rational function.
Identify transformations of functions.
Classify discontinuities.
Simplify rational functions.
Determine the zeros, the sum of the zeros, the product of the zeros, the sum of the pairwise product of the zeros, and a possible equation from the graph of a cubic function.
Sketch the graph of a given polynomial.
Identify the zeros of the polynomial.
Factor the polynomial.
Prove that a quadratic has no real zeros.
Show that a value is a zero of a polynomial.
Find zeros of a polynomial.
Discuss the implications of nonreal zeros.
Determine an average rate of change.
Provide a formula for an average rate of change.
Determine an instantaneous rate of change.

13 questions on the non-calculator portion, 11 on the calculator portion. That’s it! I’ll be around after school on Thursday afternoon, and in early on Friday.

Mathematicians are like Frenchmen: whatever you say to them, they translate it into their own language, and forthwith it means something entirely different.
- Johann Wolfgang von Goethe

Is this me?

Here’s a problem to consider:
Driving problem: Hezzy Tate drives through an intersection. At time t = 2 sec she crosses the stripe at the beginning of the intersection. She slows down a bit, but does not stop, and then speeds up again. Hezzy is good at mathematics, and she figures that her displacement, , in feet, from the first stripe is given by
Use synthetic substitution to show that is a zero of d(t).
Use the results of the synthetic substitution and the quadratic formula to find the other two zeros of d(t).
How do the zeros of confirm the fact that Hezzy does not stop and go back across the stripe?
What is Hezzy’s average velocity from t = 3 to 3.01 sec?
Write the equation for the rational algebraic function equal to Hezzy’s average velocity from 3 sec to t sec.
By appropriate simplification of the fraction in Problem 16, calculate Hezzy’s instantaneous velocity at time t = 3 .

Monday, May 21, 2007

15-4: Discontinuities, Limits, and Partial Fractions

KAORI! You're next, you lucky ducky.

Wednesday, May 16, 2007

15-2 Graphs and Zeros of Polynomial Functions

Alright....APs are over so now it's time to buckle down and focus on Chapter 15!

This section shows how to recognize the degree of a polynomial function and how to find the zeros.

Stuff you should know:
Zero of a Function: A zero of a function f is an x-value c for which f(c)=0.
Example: f(x)=x^2 -6x+5 = (x-5)(x-1) *by factoring*
So...the zeros equal 5 and 1.

The degree of a one-variable polynomial is the same as the greatest exponent of the variable.

Synthetic Substitution:
Synthetic substitution is a quick way to evaluate a polynomial function. (It's easier than long division of polynomials!!)

Synthetic substitution can be used to find out if the zeros are accurate.
The steps:
~Put the zero that we want to try into a little box.
~Write the coefficients of the equation next to the box.
~Write a line under the coefficients, leaving enough room for another number.
~Drop the first coefficient below the line.
~Multiply the number in the box with the number you just put below the line.
~Put the product below the second coefficient.
~Add the coefficient and the product and put the sum below the line.
~Multiply the sum with the number in the box and repeat the steps.
If the last number behind the line is a zero, then the number in the box is a zero of the function.

The remaining numbers behind the line are the coefficients of the other factor of the equationThings to know for Synthetic Substitution:
*If the last number under the line is not 0, then the x-value is not a zero for the function. The number that you get in the last spot is called the remainder.
*The remaining numbers below the line are the coefficients for the other factor of the function. The factor is always one degree less than the degree of the function you started with. So, if you start with a cubic function, the factor you get with synthetic division is quadratic.
*The Remainder Theorem: If p(x) is a polynomial, then p(c) equals the remainder when p(x) is divided by the quantity (x-c).
*The Factor Theorem: (x-c) is a factor of polynomial p(x) if and only if p(x)=0.

The Fundamental Theorem of Algebra and Its Corollaries: A polynomial has at least one zero in teh set of complex numbers.
Corallary: An nth-degree polynomial function has exactly n zeros in the set of complex numbers, counting multiple zeros.

Corollary: If a polynomial only has only real coefficients, then any nonreal complex zeros appear in conjugate pairs.

Translation: You always have the same number of zeros as the degree of the polynomial (if you count both complex and real numbers). If you have an imaginary zero, it will have a pair.

A cubic function will always have three zeros, but two of them could be complex.

Sums and Products of Zeros:
Sums of the Products of the Zeros of a cubic function:
If p(x)=ax^3+bx^2+cx+d has zeros z1, z2, and z3, then--
Sum of the zeros: z1+z2+z3=-b/a
Sum of the pariwise products of the zeros: z1(z2)+z1(z3)+z2(z3)=c/a
Product of the zeros: z1*z2*z3=-d/a

The sum of the zeros usually fills the B spot of a cubic function, the sum of the pairwise usually fills the C spot, and the product usually fills the D spot of a cubic function.

The values of the coefficients for the cubic functions are opposite for the sum and product of the zeros.

Example: 1)Find the sum of the zeros, the sum of the pairwise products, and the products of all three zeros if the z1=.6, z2=2 and z3=4.
2)Find the particular equation of the cubic function with values you find. 3)Prove the sum, sum of pairwise products, and products are correct by using the sum and products property of the zeros of a cubic function.

1)Sum: .6+2+4=6.6=33/5
Sum of Pairwise: (.6)(2)+(.6)(4)+(2)(4)=11.6=58/5
Product: (.6)(2)(4)=4.8=24/5
2) Because you know what places the values take in the cubic function, you can figure out that the particular equation of one of the possible functions is y=x^3-(33/5)x^2+(58/5)x-(24/5).
Since the coefficients of the values are all over 5, we can change the coefficients to be integers.
So: f(x)=5(x^3)- 33(x^2)+58x-24
3) To prove the values we got in number 1 are correct, we can use the properties of this section:
Sum: (-b/a)=(33/5) ~it checks out
Sum of Pairwise: (c/a)=(58/5) ~it checks out
Product: (-d/a)=(24/5) ~it checks out

(thank Leo for this part)
~Low point = critical point = extreme point = vertex
~Number of extreme points = the power – 1
Example: Cubic functions have 2 extreme points
~Odd powers go in two different directions
~In even number powers, the -/+ on the first term determines the up and the down
~At the Point of inflection, the concavity changes.
Definition of concavity:
Concave down = tangent lines are above
Concave up = tangent lines are below
That's pretty much it!

For extra help:
**'re up next!

Everyone should be excited for two really cool movies that are coming out in the next two weeks:
Shrek the Third:And:

Pirates of the Caribbean: At World's End

Thursday's Quiz Topics

Now that we’re through the worst of the AP season, we can finally get back on track (after the Junior retreat on Friday, anyway)! Here’s a list of topics for Thursday’s quiz:

As a reminder, your Weekly Challenge will be due on Monday, and the test for Chapter 15 will be next Friday.

Precalculus Quiz 15.1-3 Topics
Given a graph of a function:
Determine the degree
Determine the leading coefficient
Determine the number of real and complex (nonreal) zeros
Identify an extreme point and a point of inflection.
Given a cubic function:
Determine the product of the zeros
Determine the sum of the pairwise product of the zeros
Determine the sum of the zeros
Perform synthetic substitution.
Interpret the results of your synthetic substitution.
Given the factored form of a cubic, determine the zeros of the function.
Given data for a cubic function:
Prove the constant third difference property
Determine the equation for the function algebraically (matrices)
Verify the equation with regression techniques
Analyze the equation
Given partial information about the zeros of a cubic function:
Determine the remaining zeros
Determine the product of the zeros
Determine the sum of the pairwise product of the zeros
Determine the sum of the zeros
Determine an equation for the function
Analyze a function (word problem!)

That’s it! The format’s the same as always – ½ Non-calculator, ½ Calculator. There are 20 questions on the Non-calculator portion and 20 questions on the Calculator section (all worth 1 point each). If you know your stuff, you should be able to finish quickly. I’ll be around after school on Wednesday (today), and on campus by 7:00 AM on Thursday.

"I never did very well in math - I could never seem to persuade the teacher that I hadn't meant my answers literally."
- Calvin Trillin

Trashing math and country music - ouch!

Wednesday, May 02, 2007

Friday's Quest Topics

Here’s a list of topics for Friday’s test:

Precalculus Chapter 14 Test Topics:
Find terms – arithmetic sequence (determine pattern)
Find terms – geometric sequence (determine pattern)
Definitions of sequences and series
Find terms – arithmetic series (use information)
Determine type of sequence and justify
Determine terms of a sequence – unknown type
Determine terms of a series – unknown type
Explain your method of determination/pattern of series
Determine terms of a sequence (determine pattern)
Determine terms of a series (given pattern)
Determine value of n for given term in a series/sequence
Determine term(s) of a binomial expansion
Word problem – analysis of arithmetic and geometric sequences
Arithmetic series – determine formula, apply formula
Geometric series – determine formula, apply formula
Partial sums of arithmetic and geometric series
Infinite sums of geometric series
Sigma notation

11 questions on the non-calculator portion, 14 on the calculator portion. Since this chapter was so short, each question will be worth 3 points instead of 5, making this a “quest” instead of a test. That’s it! I’ll be around after school on Thursday afternoon, and in early on Friday.

"Knowledge is of no value unless you put it into practice."
- Anton Chekhov

This strategy won't work on me:

Tuesday, May 01, 2007

14-3: Series and Partial Sums

This chapter is all about formulas. The sequence formulas are from 14-2 (they appear in Katie's post), but I will put them here so you can see them in relation to the series formulas.

First of all, a series is the sum of the values of a sequence.
A partial sum is the sum of n terms in the series.

Things to keep in mind:
  • n is the term number
  • tn is the term value
  • Sn is the nth partial sum
  • "lim" stands for limit. This means that in a geometric series, if the constant ratio is less than 1, the partial sum of that series will approach a limit. In the geometric partial sum equation, as n gets larger and larger, r^n gets smaller and smaller, closer and closer to 0. Therefore, r^n becomes more insignificant and is "eliminated" from the lim equation. This situation is known as a convergent series because the series converges to a particular value as a limit.
  • If in a geometric series |r| > 1, it is called a divergent series because the terms do not go to zero and thus the series diverges.
Q: Find the 10th partial sum of the series 1, 1/2, 1/4, 1/8...
To what limit does the series converge?

r = 1/2 (a common difference), so this is a geometric series
Use the formula for the partial sum of a geometric series.

Then use the limit formula.

Binomial Series Formulas

A binomial series (binomial expansion) is of the form (a+b)^n.
2 Ways to get the coefficients of the expanded series:
  1. Pascal's Triangle
  2. Binomial Theorem

Things to keep in mind (in a binomial series) - these can come in helpful when checking your work:
  • There are (n+1) terms
  • Each term has degree n.
  • Powers of a start at a^n and decrease by 1. Power of b start at b^0 and increase by 1.
  • Sum of exponents in each term is n.
  • Coefficients symmetrical w/ respect to series' ends.
Q: Given (2a-3b)^7, find the term with b^4.

  1. Write down (-3b)^4, because you know this is your b^4 term.
  2. Multiply that by (2a)^3, because 2a is your "a" term and the exponents of the entire term must add up to 7. --> 7 - 4 = 3.
  3. Multiply everything by 12C7, the coefficient of the term.
  4. 35 * 8 * 81(a^3)(b^4)
  5. 22680(a^3)(b^4)
Harder Example:
Q: In the binomial series (a-b)^17, find the 8th term.

A & Solution: This requires one more step in your thought process. You must figure out what b term the "8th term" will be. You know the first b term is b^0, the second is b^1, the third is b^2, etc. So go forward and the 8th b term is b^7.
  1. Write down (-b)^7.
  2. Multiply by a^10 because the exponents of the term must add up to 17.
  3. Multiply by 17C7, the coefficient of the term.
  4. -19448(a^10)(b^7)
Extra help and practice with binomial expansion:
Formulas, Worked Examples, Pascal's Triangle

This is how I feel after finishing that blog

**Christina, you're up next!**

Monday, April 30, 2007

14-2 Arithmetic, Geometric, and other Sequences

14.2 Arithmetic, Geometric, and other Sequences

Sequence - is a function, with a term of pattern (n term - integer) and value (a number)

Recursion formula- specifies tn as a function of the previous term (tn-1)

Explicit formula - specifies tn as a function of n.

Arithmetic Sequence - Add or Subtract the same number to each preceding term
Common Difference - the constant number being added or subtracted in an arithmetic sequence

Arithmetic Linear Function -
Explicit: Tn = To + D(n-1)
Recursive: Tn = Tn-1 + D and To = x
where Tn is a value in the sequence, To is the first value in the sequence, D is the common difference and Tn-1 is the previous term.

Geometric Sequence - Multiply or Divide the same number to each preceding term
Common Ratio - the constant number being multiplied or divided in a geometric sequence

Geometric Exponential function -
Explicit: Tn = To * R(^n-1)
Recursive: Tn = Tn-1*R and To = x
where Tn is a value in the sequence, To is the initial value in the sequence, Tn-1 is the previous term and R is the common ratio.

Sequence Mode on the Calculator -

  1. MODE

  2. SEQ

  3. Y=

  4. nMin = (this is the starting term number..normally 1)

  5. u (n) = (your equation goes here: u(n-1) + 5 ("u" can be found by 2ND 7 and "n" is the x, T, Theta, n button)

  6. u (nMin) = (this is your starting term value)

  7. in order to find a certain term in a sequence go to TBLSET (2ND, WINDOW ) and make the TblStart whatever term you are looking for then go to TABLE (2ND GRAPH ) and the value will be in the table under u(n).

Example Problem:

a. Tell Whether the sequence is arithemetic, geometric

b. Write the next 3 terms

c. Find t76

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

1. 26, 41.5, 57... ,6071.

2. 843, 140.5, 23.417... ,0.1087.


a. First you must look for a common difference or ratio by subtraction terms into each other like 41.5-26 = 15.5 and 57 - 41.5 = 15.5 so the common difference is add 15.5. If there is no common difference then you must divide terms into each other like 843 / 140.5 = 6 and 140.5 / 23.417 = about 6 because of rounding. So the common ratio is divide by 6. The first sequence is arithmetic because there is a common difference whereas the second sequence is geometric because there is a common ratio.

b. Use the seq mode TABLE with the TblStart = 1 in the TBLSET to see the next 3 terms.

c. Use the seq mode TABLE with the TbleStart = 76 in the TBLSET.

d. You can search for this value in the TABLE but it is much easier to solve for n in the equation for the sequence. So, since the first sequence is arithmetic use Tn = To + D(n-1) to solve for n: 6071 = 26 + 15.5 (n-1). So subtract 26 from both sides and then divide by 15.5 to both sides and 390 = n-1, then add one to both sides and n=391 meaning 6070 is the 391st term in the sequence. For geometric sequences, using the explicit formula to solve for n, you must use log to get the n out of the exponent. So Tn = To * R(^n-1) : 0.1087 = 843 / 6^(n-1). Divide 843 by 0.1087 (=7755.289). Then take the log of both sides which moves n-1 infront of log6. Then divide both sides by log 6. So log 7755.289 / log 6 = 4.999 = 5 = n-1. Then add one to both sides to get n= 6. So 0.1087 is the 6th term.


1. a. Arithmetic

b. 72.5, 88, 103.5

c. 1188.5 is the 76th term.

d. 6071 is the 391st term.

2. a. Geometric

b. 3.9028, 0.65046, 0.10841

c. 4 * 10 ^(-56)

d. 0.1087 is the 6th term.

Additional Information can be found here:

Quiz yourself at this neat website:

Madison you are up next!!!

Trey Kozacik, Xander Berry and I did a science fair project and we won 2nd place in our division on Saturday in the LA County Science fair. This means that we are going to STATE!!! Who would've thought considering we did this project purely for fun and then were convinced by teachers to turn it into a science fair project!! Our project is called Shooting For Distance (I know, lame right..well I wanted something a little fun..I was going for SHOOTING FOR THE STARS....) We built a 10 ft air cannon that shoots weighted water bottles up to about 900 ft.

Monday, April 23, 2007

Thursday's Test Topics

Here’s a list of topics for Thursday’s test:

Precalculus Chapter 13 Test Topics:
Relationships between polar and Cartesian coordinates
Plot polar points.
Convert polar equation to Cartesian
Transform a polar graph/equation
Parametric equations – linear motion (word problem)
Determine a point on a curve given polar data
Describe the formation/creation of a polar curve
Parametric equations – nonlinear motion (word problems)

I’ve posted the solutions to the review handout on the class website .
See you in class! I’ll be around until 3:00 pm on Wednesday afternoon, and in early on Thursday.

If you have built castles in the air, that is where they should be; now put foundations under them.
- Henry David Thoreau

13.5 - Parametric Equations for Moving Objects

The main idea of this section is to separate what's going on with the x-axis and what is going on with the y-axis

The best way of describing this is using an example. Let's pretend that a firetruck is able to navigate around at 40 mph eastward and 60 mph northward to reach the next city. The firetruck is currently at the point (50, 20)-- 50 miles east and 20 miles north of the origin (we'll say the town center).

First, we'll find a parametric equation for the path of the firetruck, using t for the hours.

To do this, we simply take our coordinates and our velocities to get:

x = 50 + 40t
y = 60 + 20t

When we graph this, we get a line that can help us answer the following:

Predict the time that the firetruck will be 80 miles north of its current location.

To do this, let y = 80:
80 = 60 + 20t

It will take it about an hour to get 80 miles north of its current location.

Here's an image to help visualize this:

The next idea that we learned is: Parametric Equations of a Cycloid

The reason that we learn this is to calculate where a point is on a rolling wheel along the x-axis at any point in time.

In class, we learned how to derive all of this, but here are the equations:

x = a(t - sin t )
y = a(1 - cos t )
where 't' is the number of radians the wheel has rolled since the point was at the origin.

Here's a really really good explanation on how to derive this, if you weren't in class today:

Another Example

Let's take a look at another example of parametrics in action:

Rocky the flying squirrel:

is flying at 1,000 m/hr west and 800 m/hr vertically. At t=0, he is at point (100, 50).

a) Write parametric equations for his flight, using t hours as the parameter.
x = 100 - 1000t
y = 50 + 800t

b) When will Rocky reach the top of the eiffel tower, which is 270 m higher than his current location?
270 = 50 + 800t
220 = 800t
t = .275 hours -- which is 16.5 minutes.

Here's two helpful websites:
is a summary which has this entire explanation reworded on a single page.


which has the flash animation I posted above as well as a couple others. And something called LiveMath which I don't have, but you may.

In celebration of the recent talent show that we had, here is a clip from it:

Wednesday, April 18, 2007

Section 13.2: polar equations of conics and other curves


This section gives you the basic information about polar equations and how to relate polar coordinates (r,θ) with cartesian coordinates (x,y).

Polar coordinates are a different way of identifying a point in space. With polar coordinates we use a radius (r) and an angle (θ) to specify a point. The coordinate plane looks like this:

The pole is the origin. The polar axis is the same as the positive x-axis on a cartesian graph.

To relate r, θ, x, and y, we can use the following trigonometric equations.

x^2 + y^2 = r^2
sin θ = y/r

From these equations we can derive values for x,y, and θ.

y= r*sin θ

x=r*cos θ


Now, we need to know how to plot polar graphs on our calculators. First, we need to change the mode to polar. When you go to the y= menu it will now say r=. The standard equation for a polar graph is r=a+bsinθ or r=a+bcosθ. These equations will result in a limaςon or cardioid.



The last bit of information in this section is the general polar equations for conic sections.

These equations are r= k / (a+bsinθ) or r = k / (a+bcosθ). With these equations you can make a parabola, ellipse, or hyperbola.

Depending on the relationship of a and b you will get a different conic shape.

abs(a) = abs(b) -> parabola

abs(a) is greather than abs(b) -> ellipse

abs(a) is less than abs(b) -> hyperbola

Note: the focii of a parabola will be at the origin as well as one of the focii of a ellipse or hyperbola.

Thats it for now!

Next up is EDWARD


Thursday's Quiz

Here’s a list of topics for Thursday’s quiz:

Precalculus Quiz 13.1-3 Topics
Plotting polar coordinates
Multiple representations of polar coordinates
Plotting a curve given a table of polar coordinates
Interpreting/Reading a polar graph
Determining the equation of a polar graph
True/False intersections of polar graphs
Converting a polar equation to Cartesian

That’s it! The format’s the same as always – ½ Non-calculator, ½ Calculator. There are 7 questions on the Non-calculator portion and 9 questions on the Calculator section. I’ll be around after school on Wednesday, and on campus by 8:00 AM on Thursday.

People are like stained-glass windows. They sparkle and shine when the sun is out, but when the darkness sets in, their true beauty is revealed only if there is a light from within.

My job:

Tuesday, April 17, 2007

Chapter 13-3: Intersections of Polar Curves

Ok, it's a nice and easy 2 page lesson. A break for all you lucky juniors and seniors in time for prom.

This lesson is discusses the basics on finding the intersections points when given two polar curves.
Basically to find the number of intersections, and the point they intersect, we insert the equation into the regular function mode. After that, we just do the regular calc -> intersect stuff...

For those that do not understand what I am saying...
I'll point out the different steps
1) Get out your calculator
2) Press Mode

3) Select Func if it is not already selected. (If you realized that Func stands for function, then you are really smart. =])
4) Press Y=
5) Insert equations!!!

6) Go to Window and set Xmin = 0 and Xmax =360 (this is because the polar coordinate system is 360 degrees.) The Ymin and max vary depending on the graph
7) Now, press graph!
8) OMG, a graph. Now we can not only find the number of times the two graphs intersect each other, but we can also find out WHERE they intersect each other.
9) I'm not going to explain the intersect stuff...we did it a million times already.
Ok, now that we have the intersection point...

The X-Coordinate is the degree, and the y is the r...
so the format will be (r, theta) to put all of this information into use...

We are given 2 equations, and we are required to find the intersection points between the two

Now, we graph them on our calculator. In polar mode, the two equations should look like this...

Ok...Now that we see the equations in Polar mode, now we switch to function mode. (For those who don't know how to do this...scroll up.)

Alright, now insert the equations the EXACT same way into y=

The resulting graph should look something like this.

Awesome...right? Now we use the Calc-> intersect powers of our calculator and find the intersections between the two graphs.

The resulting answers should be...
(117.531, -.387) (193.264, -1.919) (235.510, -.699) (287.063, 1.880)

Now, we have to change them into polar coordinates. Which is NOT that difficult guys.
All we have to do is swap them around and add a degree symbol to the x-coordinate...
(-.387, 117.531
°) (-1.919, 193.264°) (-.699, 235.510°) (1.880, 287.063°)
Hurray! We solved the problem!!!

Since there I can't find a website truly discussing the intersections of Polar Curves...I'll post one up about some of the things we already learned for last lesson. Besides, this lesson was the easiest one we have had in a long time.

********** Katie You're Up Next. ***********

Of course everybody knows that the Lakers are IN the playoffs right? Well, I hope you all remember last year, when we were up 3-1 against the Suns and LOST. Tomorrow night, they are up against the Sacramento Kings (remember the rivalry?) and if they beat them...they clinch the seventh seed. Yes, and if they clinch the seventh seed, they are once again against the Suns. Except this time...they'll win. So tomorrow, when they're up against Sacramento, I hope that all of you will be rooting for the Lakers...and Kobe Bryant.

Monday, April 09, 2007

Thursday's Test Topics

Here’s a list of topics for Thursday’s test:

Precalculus Chapter 10 Test Topics:
Unit vectors
Planes – Standard Equation vs. normal vectors and points
Points on (or not on) a plane
Lines in Space – points and unit vectors
Direction Angles
Direction Cosines – properties
Vectors between two points
Angles between vectors
Scalar and vector projections
Cross products

I’ve posted a sample test and review on the class website which I’ll distribute in class tomorrow. Review 10-d will not be as helpful as 10-b, 10-c and 10-e.

See you in class!

Here’s a little something that shows the beauty of mathematics:

Math Art: The Beauty and Symmetry of Mathematics.

1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321

1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 +10= 1111111111

9 x 9 + 7 = 88
98! x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888

Brilliant, isn't it? And finally, take a look at this symmetry:

1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111 = 12345678987654321

A unique approach to homework:

Thursday, March 29, 2007

10 - 8 : Vector Equations of Lines in Space

10-8 : Vector Equations of Lines in Space

First of all let me explain the purpose of this lesson. It is very simple. If you have a line, and random line through space, and one point on that line, and a unit vector, you can find the equation of the position vectors to any point on that line. You must choose a distance, or "d" to work with. "d" is just a distance or magnitude, which represents the distance to the next point on the line you want to find.

Ok so i find that it is easier to explain this verbally rather than graphically, however I know some people would like to know precisely where everything im talking about is in a diagram so, I decided to make a diagram which u can refer to whenever your heart so desires!

ok, i am not a very good "graph maker," so I color coded it so that I can try to make up for that with my explanation.

  • the red A is referred to as P0 in the book, and is the given point on the line.

  • the pink vector is referred to as vector P0, and is the vector to the given point.

  • the blue line is the line, and the unit vector that is used, is the unit vecotr of this line.

  • the yellow portion of the line is "d" or the distance from point A to point B.

  • the red B is the point which the position vector we are finding the equation for.

  • the orange vector is the actual position vector we are finding the equation of.

In essence, we are just adding vectors. Or in the case of the diagram, we are adding the pink vector to the yellow vector, in order to get the orange vector. Mathematically we would say, we are adding vector Po and vector PoP.

*note : Vector du = Vector PoP*

The method we use to figure out the equation of the position vector is :

  • Vector v = vector Po + d(unit vector)

This is when we are given the unit vector.

  • Vector v = (xo + c1*d) + (yo + c2*d) + (zo + c3*d)

This is when we must work with the cosines instead of being given the equation.

Sample problem when given the unit vector.

Problem : The point is (5,11,13) and the unit vector is 3/7 i + 6/7 j + 2/7 k. D = 5. Find vector PoP (v).

Work :
Vector v = (5 i +11 j +13 k) + 5(3/7 i + 6/7 j + 2/7 k)
Vector v = (5 i +11 j +13 k) + (15/7 i + 30/7 j + 10/7 k)
Vector v = 7.143 i + 15.286 j + 14.429 k

Answer : Vector v = 10.714 i + 47.143 j + 18.57 k

Now I will show you a problem where we are given the cosines

Problem : The point is (5,3,-1) and the cosines are : c1 = 6/11, c2 = -2/11, and c3 = 9/11. D=5.

Work :
vector v = (5 + (6/11)(5)) i + (3 + (-2/11)(5)) j + (-1 + (9/11)(5)) k
vector v = (5 + (30/11)) i + (3 + (-10/11)) j + (-1 + 45/11) k
vector v = 7.727 i + 2.091 j + 3.091 k

Answer : vector v = 7.727 i + 2.091 j + 3.091 k

Both of these methods are purely plug and chug. They are a sinch and aren't ridiculously long either. All in all they are pretty darn awesome.

Challenge Problem!!!

Problem : Timmy is looking at the great wall of china. He knows that the exact point he is staring at is located at (5,10,20). He knows that the unit vector for the wall is
(9/17) i + (12/17) j + (8/17) k. He wants to know the position vector from him to the point 10 feet to the right of the one he is looking at currently. Help Timmy find the position vector.

Work :
first set up the problem
vector v = (5 i +10 j +20 k) + 10 ((9/17) i + (12/17) j + (8/17) k)
distribute the distance
vector v = (5 i +10 j +20 k) + (90/17) i + (120/17) j + (80/17) k
group according to like terms
vector v = (5 i + (90/17) i) + (10 j + (120/17) j) + (20 + (80/17) k)
add like terms
vector v = 10.294 i + 17.059 j + 24.706

Answer : vector v = 10.294 i + 17.059 j + 24.706

This concludes MY lesson. But if you really want to indulge yourself in a very vector-like bliss, i suggest you check out

This site literally has everything on vectors. It has formulas, but also explains them. It also has pretty good pictures and diagrams to go with them. So everyone, even Alvin and the chipmunks, should go check this out!

Here is a little bit of me : This is the cover the the Silversun Pickups debut cd, Carnavas. It is absolutely phenominal in every way. So I insist you go buy it now, or have me burn you a copy, because they are just that kool. They kill 2 stones with one bird, just like chuck norris.


Wednesday, March 28, 2007

Snakes, Direction Angles, Planes, and Direction Cosines

10.6: Vector Products of Two Vectors

We now have that long-awaited SECOND way to multiply vectors, so brace yourself. This multiplication process is called the "cross product," and it results in a vector answer, rather than a scalar one.

So basically, Vector C = Vector A X Vector B (the X represents "cross", so you'd read this as "vector C" is equal to Vector A cross Vector B). This process is NOT COMMUNITATIVE (authoritative use of capital letters, no?), so don't you dare assume that Vector B X Vector A will equal the same thing as Vector A X Vector B. If you do, by God, I will fight you. BUT, Vector A Cross Vector B WILL equal - Vector B Cross Vector A. Good to know.

Resultant Vector C will lie perpendicular to the plane that Vectors A and B reside on. What we must decipher is the direction in which Vector C is going; is it going toward us from the plane or away from us? We determine this by using the Right Hand Rule, and the steps for said rule are as follows

1) Cut a hole in the box

2) Put your ju.....oh wait, wrong steps, my bad.

1) For "Vector A Cross Vector B," Place your right-hand fingers along Vector A, and curl them towards Vector B

2) The direction in which your thumb is pointing represents the direction in which Vector C will travel.

3) For "Vector B Cross Vector A," simply do the opposite; place your fingers along Vector B, and curl them towards Vector A.

4) A good thing to remember is that if Vectors A and B are parallel, the product will be zero and the right hand rule consequently will fail miserably.

Other stuff to know:

Vector A Cross (Vector B + Vector C) = Vector A Cross Vector B + Vector A X Vector C

Here are the cross products of the unit coordinate vectors:

i x i=0 K x K=0 j x i=-k
k x i=j I x k=-j i x j=k

j x j=0 j x k=i k x j=-i

Example of Cross Products:

Find Vector C for Vector A Cross Vector B

Vector A= 8i+3j+4k

Vector B=9i+7j+2k

Vector A Cross Vector B= 72 i x i + 56 i x j + 16 i x k+ 27 j x i+ 21j x j + 6 j x k+ 36k x i + 28k x j + 8 k x k

Vector A Cross Vector B Reduced: 56k - 16j - 27k +6i +36j -28 i

Vector C = -22i +20j + 29k

Find Vector C for Vector A Cross Vector B Using Determinants:

Here is the basic rubric for know it. It's important. I swear.

(i j k)

(Ax Ay Az)

(Bx By Bz)

Vector A Cross Vector B = (AyBz -AzBy)i - (AxBz - AzBx)j + (AxBy - AyBx)k

So then, let's apply it that rubric to determine Vector C (Using the same Vectors A and B)

(i j k)

(8 3 4)

(9 7 2)

Vector C = (6-28)i - (16 - 36)j + (56-27)k

Vector C = -22i + 20j +29k!

Alright, we're ALMOST done. Other stuff to know:

The Area of the parallelogram formed by Vectors A and B equals the absolute value of Vector A times the absolute value of Vector B times the Sine of the angle the two vectos form.

The Area of the triangle formed by Vectors A and B equals 1/2 times the absolute value of (Vector A times Vector B.)

And...THAT'S IT! Thank god.


My personalization.......IS LAAAAARRRYYYY!

Leo, you're next. Peace out.

Monday, March 26, 2007

10-5 Planes in Space

We already know what a vector in a three dimensional setting looks like:

A plane in space is the set of all points that satisfy the equation: Ax + By + Cz = D
It looks like:

To find a vector ,normal or perpendicular, to the plane Ax + By +Cz = D, just plug the coefficients A, B, and C into the vector formula:


  • given 3x - 8y + 5z = 24

  • plug 3, -8, and 5 into the vector formula

  • your answer should be

You can also determine a plane by using a normal vector and a point, as shown below:

courtesy of Mr. French

To find a plane using a normal vector and a point P:

  • Place the vector's tail at the point P =

  • Pick another point and draw another vector to that point

  • Find the displacement vector (head - tail) =

  • *the vectors are perpendicular so their sum is 0

  • Use algebra so that

  • Then, and that's your answer!


given and P = (2,5,-1)

You will find that essentially, you only need the equation

These are pictures from a JACK'S MANNEQUIN concert I went to in February. They're my favorite band and everyone should go listen to their music. Their rooted from SOMETHING CORPORATE, which is my other favorite band, because the lead singer was originally from SoCo. Andrew McMahon, the lead singer, fought and beat cancer, which i think is amazing and he's amazing so everyone should go buy his music NOW!