Category Archives: Math

This is Calculus 3++…

We reviewed Vectors today, how vectors can be used to describe a line in 2D and 3D space. <1,3> + t<1,1> or (1+t)i + (3+t)j and in 3D 2si + (2+2s)j Why the s and the t? They are just variables describing the scaling factor of your place in the system.

For describing planes, this type of generalization works as well. There are two direction vectors and one initial vector pointing to the zero-zero location of the plane and two direction vectors describing the tilt of the plane.

For example, the Equation of three points, as a plane is:
POINTS: (1,2,3) , (5,5,5) , (-1,6,9)
Form a line with a starting zero zero location of <1,2,3> plus a scaling in t direction and s direction.
<1,2,3> + t<4,3,2> + s<-2,4,6>

Vector Rules
u · u > 0 unless u = 0
u · v = v · u
(u + v) ·w = (u ·w) + (v ·w)
r(u · v) = (ru)  · v = u ·(rv)
*****r is a scalar

u · v = |u| |v| cos θ
u · v = u1v1+u2v2+…unvn

Why does the dot product work this way?? Because of the Law of Cosines, which states:

c2 = a2 + b2 – 2ab cos θ

and if we work out a triangle, eventually we will see
a = u
b = v
c = u – v
|u – v|2 = |u|2 + |v|2 – 2|u||v| cos θ

If we want to find the angle between two vectors, we can use the dot product.
arccos (a∙b / (|a|*|b|) )
That is regular multiplication of the length of a and b.

Now we can project b onto A by using |b| cos θ

And we can describe a plane, by saying (x – p) ∙n = 0
So we are describing the plane with a normal n and a point (is that p?)

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Introduction to Multi-Var Calc and Lin Alg

Previously in calculus we did things like y=mx+b with m being the slope and for x not varying much.

The goal of this class is to understand things like z = x2+y2, a three dimensional shape, similar to a bowl. One dimensional curves in space and vector fields, like the flow of water in a river.

The best way to do this is through Vectors and linear transformations (Matrices).

The right hand coordinate system, has x emerging from the paper, y traveling to the right and z to the top of the page. The left hand coordinate system swaps y and x.

A vector has length and direction. For this class we will write in notation R3 meaning the real number system in three dimensions. We can describe locations in Rn by position vectors. Points are different than vectors. You can’t add points, but it vectors can be added or subtracted.

For the purposes of this class [#,#] is a vector Square Brackets, (0,0) is a point.

There is a special vector, the zero vector [0,0]. The length of vector v is the absolute value of v. I should write little hats on top of my vectors, but I can’t easily type those in.

Describe a Line in a plane
At some time t, the position of the vector passing through (0,0) of the plane and with a slope of one, the line can be described as t[1,1]

For some other line with a slope of one-half and passing through the point (0,2) this line can be described as [ 2t , t+2 ]

Line = starting point + t(direction)
Line = u +tv
and v are vectors. t is not. u is the zero vector for this point.

To describe a plane, use two vectors
[1,0,0] + t[-1,1,0]+s[-1,0,1]

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Looking At Functions

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Multiplication of Matrices

Some basic terms:
Ith row and jth column
m row by n column matrix

When multiplying matrices multiply row by column.

Using a third variable p to describe the row or column in a pair of matrices to be multiplied

A = m x p
B = p x n
A(B) = m x n

So what I just said is that m rows by p columns when multiplied by a matrix which has an equal number of rows p as the columns of the original matrix, will give an answer of a matrix which has as many rows as matrix A and as many columns as matrix B.

So For example, we have two matrices. A and B

A has 3 rows and 2 columns
B has 2 rows and 4 columns.
We can multiply!
And the answer should have 2 rows and 4 columns. We can multiply if p is equal in both matrix A and B.

What happens if p is not equal in both matrices?
Then AB is undefined

Practice Problems from HotMath

Problem 19: Is the matrix product of the given matrices defined? If so, find it. If not, give reasons.
Observing the matrices shows p to be equal, giving us an m x n matrix of 2×1
[2 x 3][3×1] = [2×1]
So what exactly did I do?
First step was to lay the B matrix on it’s side to equal the number of rows in matrix A that is what gives me the 2 rows part. Since I have only one column to work with in B I only lay that column down, but if I had two columns I would have laid it down twice per row.

Then I drop the rows from matrix A down to match up.

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Log questions continued…

Question #4

Question #5

Question #6

Question #7

Question #8 – Find the domain

Question #6a

Question #6b

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Some more log problems

Example #5 Evaluate

Example #6 Evaluate

Example #7 Evaluate

Quiz questions! No outside help allowed
Question 1a – Evaluate exponentially

Question 1b – Evaluate exponentially

Question 2 – Evaluate logarithmically

Question 3 – Evaluate Logarithmically

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Square roots, negative exponents & fractional exponents

Describing the relationship between square roots, negative exponents & fractional exponents.

Describing how to turn negative exponents positive.

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Logs are another way to write exponents.
This shows ‘The common log’. When a base is left out, it is assumed to be base ‘10’

My curiousity was sparked by a Chain Rule problem, which included ln in the answer. I did not know how to handle ln. ln is ‘the natural log’. It involves a relationship with an unknown variable ‘e’, which is its base.

Some practice problems

Example #1 evaluate exponentially

Example #2 evaluate exponentially

Example#3 Evaluate Logarithmically

Example #4 Evaluate Logarithmically

Ran out of time. Using problems from West Texas A&M University

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The Chain Rule

Practicing problems from UCDavis.

Problem 1

Problem 2
Note to Self: Square roots. To find the derivative of a square root, change the square root into a fraction and proceed, using the exponent rule. Ie. The derivative of the square root of x is 1/2x-1/2

Problem 3
According to the shown solution this answer is wrong. I should have included (-4+35x4) inside of the exponent “29”. I don’t know why I should do this. Is it because a simple version of this might look like 3(x5)2(x4)? I believe with that sort of problem I could write out 3(x5)(x4)2 and if I solve either one of those two I would get the same answer.

Lets try that:

Let x = 2


That’s not right either.
I’m going to have to do some research on this. And in the mean time make a mental note, to move the values bracketing the (u) formula to the end! Solved; See Problem 6

Problem #4
OK… So I answered this one wrong as well. (The correct answer is shown though..) This time I should not have pulled g’(x) inside of the exponent, like I needed to in problem #3. So what is going on here? How to do I handle exponents? Solved; See Problem 6
As a side note. I’ve learned how to “Align along = sign” How nice!

Problem #5
Ok. This one immediately threw me for a loop with the division AND the exponent. First things first, I was able to reduce u, leaving the inner function division free but with negative exponents. After that it was smooth flowing, except by the end of my problem, I still have three pesky negative exponents which I need to simplify out of existence.

Problem #6
I got this one right! And it was EASY!
Dxy Sin(x) = Cos(x)
I believe I’ve also solved the exponent trouble I was having. The trick is to simplify first when u is still a variable, and then once simplified completely, replace u

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A Review of the Limit Proof

To review my knowledge of limits, I need to think back. A Limit expresses a relationship between two changing but related variables, Epsilon eand Theta strong>d.To review my understanding of limits, the limit is a variable L, which has a relationship to a secondary variable a. a is the change along the x-axis as we approach the limit, which is represented by the variable L. To think back on what I understand, the variable L is bordered on the y-axis by Epsilone. As far as limits are concerned, we are in absolute space. We do not have negative limits. So to start out, I know the limit must be greater than Zero. Both variables, Epsilon e and Theta d must be larger than Zero, or else the limit does not exist. So I can write e > 0.

The purpose of this exercise is to describe the relationship between Epsilon and Theta. So what can I write about Theta? Theta d represents the border of variable a along the x-axis. Since both Epsilon and Theta represent both sides of the border, there are absolute values.So what exactly is the relationship between our variables? L, a,e and d, as well as our line, y = f(x)

L is the limit. It is reached by x – a. Variable a is found when you look at a limit and there is an equivalency x ->a. Epsilon is a real number, greater than 0. The limit is a point on the line f(x). Epsilon deals with the y vertice, which is described as a function of f(x). So the variable x describes the location along the x-axis, and f(x) describes the location along the y-axis. Lets break this problem off into x and y axis sections. To the x axis, we will give the variables, x,a and d To the y-axis we give the variables f(x), L and Epsilon. x – a should be less than d and f(x) – L should be less than e. The reason our x and y coordinates are less than variables Epsilon and Theta, is because, the Limit is contained inside of the variables Epsilon and Theta, because they form the boundrays. Because the limit is not negative, we need to use absolute values for both of the expressed changes, x-a and f(x)-L so we would say |x-a| < d and |f(x)-L| < e .

There is only one more thing I am leaving out, which is that the limit is in positive space, so the whole thing has to be greater than 0. Back at the beginning of this essay, I stated e > 0 I could say the same thing by writing 0 < e . We can include this note about the limit being in positive space, by sticking that Zero Less Than in front of the entire expression to write. 0< |x-a| < d and |f(x)-L| < e Can I use this relationship to find Theta with a known Epsilon.


The first step is to solve the f(x)-L side, because we know what f(x) is, we know what L is and we know what Epsilon is. Using our known variables we will be able to find the values of x.


So now we have found out an exact value of variable x, it is between 2.00083 and 1.99916. x – a should give us a value for Theta

*Incase you haven’t noticed, I’m having some trouble typing out the Greek letter variables. I’ll figure that out some other time.

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