the blue and red slopes are not at the same angle, the bottom figure has just a little hole in the middle left because of this.

>>7648
Oh and the green and orange ones as well

leave it to 9gag to be tricked by an 8 year old meme.

I almost feel we could make a calendar out of /math/ based on how regularly this gets posted.

>>7671

library.nu is no longer with us :(

>>7664
AW FUCK

>Is this generally considered a difficult portion of Calc 1 courses, or am I just going full retard here?

Completely depends on your ability to interpret questions and solve problems. I find them terribly easy but I know a lot of people that don't.

if you understand derivatives well, and are a good problem solver, then they should come pretty easily to you (at least with some practice).

I recommend doing as many of them as possible. Not really recommending quantity over quality here; work through each problem, understand the necessary steps (relate the variables, find necessary equations, solve for needed derivatives, etc.) but working through more problems helps you recognize patterns within different problems etc. i dont need to harp on dis' shit ya'll.

>>7654
I certainly agree with practice.

I'll post a few examples of the level of difficulty my instructor expects of us.
The problems in the text are far to easy. Math for me isn't like chemistry or biology where once I understand the concepts I can quite well.

Mathematics takes that extra step of understanding...and lots of practice. At least for me.

related rates were the hardest part of calc 1 for me. instead of slamming your head against a wall, just look up solutions to lots of problems and see the method. then just do a few problems until you get it.

I found related rates to be my saving grace. Until then I could do the math, and I understood it could solve some problems but it didn't seem like I was solving real world problems. When I saw a problem involving like the fuel level of a rocket while its traveling, or the acceleration ( x mass) to get the force required to propel something, I started to realize it really is pretty useful. For me, the physics connection really makes it click, but I love physics so...

>>7617
It is the rate of change with respect to the variable x in that case. An easy way to think about it is that the derivative gives the slope of the function. So f'(x) tells you the value of the slope on a graph y=f(x).

Think of the graph of f(x) = x^2. (let's consider the positive-x side for simplicity). The y-values increase as you go along x, but the rate of the increase grows as well. From the rules you probably already know, the derivative is f '(x)=2x. That's the function that represents the rate of change of f(x) = x^2.

Now imagine the graph of f(x)=2x. It's a straight line. It increases, but the rate at which it increases is constant. The derivative of that would be f '(x)=2.

Now think of the graph f(x)=2. It's a flat line. It's not increasing. It's derivative is 0.

Like the above guy said an easy way to think about it is: the derivative gives the slope of the function you're deriving. It's like slope in function form. You can have a general formula f '(x), and you can use it to plug in a point to find the slope (rate of change) at that particular point.

So what's the rate of change of f(x) = 5x^3 + 2x^2 + x @ x=3? Just calculate that f '(x) = 15x^2 + 4x + 1, and plug in the x-value you want to know about.

You can even go on to higher-order derivatives, which is like "the rate of change of the rate of change". Or f ''(x), f ''(x).

Has your prof. confused you with the whole dy/dx notation yet? It's all related.

>>7618
Geometrically speaking a derivative is the slope of a line tangent to the function at any point along the function. Mathematically speaking it is the rate of change of some value with respect to the variable you are deriving by, and in terms of how it is taught it is the slope between two points infinitesimally close together both on the same function.

It's crystal clear now, thanks guys.

Make f(x) the miles your car has traveled after x hours
f'(x) is the speedometer, or better, a function to find where the speedometer would be at any place x.

For starters lets take f(x) = 60x. So back in the analogy we're gaining sixty miles for every hour. The derivative would be f'(x) = 60 because the speedometer would always be at sixty.

Alright, let's forget the analogy now. Let's use the normal math terms.
What about f(x) = x^2 ? A quick way to check derivatives is to find the slope between two points. How do we do this?

f(x,2) - f(x,1)
---------------
x,2 - x,1

Where x,2 is a place after x,1, get them as close as you can, but know there's always closer (a infinitive limit)

Apply this to our problem of x^2 (1.1^2 -1^1)/(1.1 - 1) = 2.1 (2.1^2 -2^2)/(2.1 - 2) = 4.1

Comment too long. Click here to view the full text.

http://en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization

hmm is there a way to do this without eigenvectors or eigenvalues

Well, I can't remember any other ways to do it and looking here: http://en.wikipedia.org/wiki/Matrix_decomposition
Eigen decomposition is the only one with the form you want so I think you're stuck

>>7494

First, think about storing the number 5857458489 using the CRT. This can be done by simply storing the residues modulo the primes below 27 (or whatever upper limit you need, this will depend on the size if the number to store).

>>7516

this should give you something like:
1, 0, 4, 1, 3, 12, 12, 17, 5, 6

Now do the same for the second number. Then find the difference between each corresponding residue. And finally use the differences to reconstruct the new value (the answer to the difference question).

man, legit, this professor is whack. expects us all to be geniuses to know how to solve half of what's on here...

for each observation / bucket:
# of that one seen / total number

so if i have a bunch of measurements { 1, 2, 3, 4, 1, 1, 3, 1, 3, 2}

1s: 0.4
2s: 0.2
3s: 0.3
4s: 0.1

x
x _ x
x x x
x x x x
1 2 34

>I remember someone posting a link to their professors page with notes and practice problems and stuff, that would be nice.

This guy has great notes. They definitely helped me get dat A in Calc 2. I wish I found it earlier
http://tutorial.math.lamar.edu/

Khan Academy is useless. Math is about doing, not watching. Download Sullivans algebra books and work through em.

>>7567
It's by no means useless. You won't master any material by watching the videos alone, but if you watch them before and/or after a lecture covering the same material and do the problems in your book you should be doing just fine.

>>7558
Khan Academy is best after attending and paying attention in class. The guy who does the videos does a great job at slowing and breaking the whole thing down to it's most basic parts, something most professors don't really do.

It's great for figuring out why you have to manipulate that variable that way or why you have to use this equation to solve that problem.

Read letters to a young mathematician, by ian stewart. Look up math topics on wikipedia and see what interests you.

>>7555
Then google it, dumbass

I want to fondle those tits.

DICKS EVERYWHERE

What is your reasoning for believing that life is "about" a shitty American electronic DJ , some random graphing utility, a South Asian plant, and a single glass of H20? You better pony up a proof and some definitions if you expect this bullshit to fly.

The meaning of life, or what we do on a basic level, is simply to survive and reproduce. Although all the trappings of human society are nice, but we cannot justify how any part of it is the purpose of or necessary for our existence.

>>7587
You sir are awesome

>>7587

>he thinks the meaning of life isn't math

sorry, but you only get a cookie if you say the right answer

Your answer is correct. I'm not exactly sure what form of factoring or manipulating they did to get that answer but I'm guessing they must be equivalent.

(x-5)^3 *[(x-5)+4x)]
(x-5)^3 *[(5x-5)]
(x-5)^3*5(x-1)
5(x-5)^3(x-1)

>>7568
Thanks to you both. It was late and I was getting frustrated.

Something one of my cal profs told me not to worry too much about the answers provided in the book as the questions and answers are all written by grad students so they like to provide needlessly obtuse answers.

The product rule and the chain rule.

y = x(x -5)^4

Affectionately memorized: the derivative of the first times the second, plus the first times the derivative of the second

y' = (1)(x - 5)^4 + x(4)(x - 5)^3(1)
y' = (x -5)^4 + 4x(x - 5)^3 factor out the (x - 5)^3
y' = (x - 5)^3 ((x - 5) + 4x)
y' =(x -5)^3(5x - 5) you can factor the 5 out of here
y' = (x -5)^3 * 5(x - 1)

or, rewritten in the form they have,

y' = 5(x - 5)^4(x-1)

couldn't you square root both sides and be left with 7x=5. Then simplify that down to x=5/7.

>>7592
I meant + or - 5/7

• 5/7, + 5/7