that's dumb - there's no paradox there.

civilization was pregnant with calculus. you'll notice that almost all monumental breakthroughs in science and technology are arrived at by multiple independent people.

1/14/2010 9:50:50 PM

nerd argument is imminent

1/14/2010 9:52:52 PM

it is fundamentally not a paradox.

if anything trapezius' thought experiment implicitly asserts that calculus will be invented no matter what

[Edited on January 14, 2010 at 9:54 PM. Reason : s]

1/14/2010 9:53:38 PM

Can we get back the movie argument and what is art and not art? That was more fun.

1/14/2010 9:56:13 PM

how about fractals?

math and art together!

1/14/2010 9:57:12 PM

I like how Trapezius glosses over his massive pwnage ITT

1/14/2010 9:57:53 PM

First quiz today. Topics include: Shifting and reflecting a function, composing and decomposing functions, and finding the inverse of a function.

1/21/2010 1:45:48 PM

DID YOU [fail] IT YET OR NOT???

1/26/2010 3:07:54 PM

Quote :

"Unrelatedly, my MA141 professor wore the same thing every day of the semester. I dunno if he had a bunch of the same clothes or just wore the same set over and over."

Was a little hunched over african, w/ a HEAVY accent, that was an absolutely terrible prof.?

If so, the clothing (2000ish) looked to be a decade old and not well matched.

[Edited on January 26, 2010 at 3:25 PM. Reason : m]

1/26/2010 3:24:37 PM

Ummm

I currently have a 100% average on 8 out of 8 webassigns and a 9/10 on the first quiz.

So, I think I'm doing pretty well so far.

1/26/2010 3:26:18 PM

I owned calculus I when I was like 16 with an A+

calculus II was a bitch though

1/26/2010 3:28:26 PM

calc 2 = bad word

1/26/2010 3:30:44 PM

I told my teacher that I'm taking it next. He, in his usual Dr. House style, told me that statistically it was the most failed class in the history of college.

1/26/2010 3:31:35 PM

I can't understand why. If someone is good at calc 1, calc 2 isn't that much of a stretch.

calc 3 on the other hand, is a nasty bitch.

1/27/2010 2:28:29 AM

^I remember 3 being much easier than 2

1/27/2010 2:29:39 AM

26 and taking calc webassigns

I wonder if they have ap english webassigns, that would've helped my GPA get that 6.0 I was looking for senior year

1/27/2010 2:29:41 AM

I have a test in one week. FUCK.

1/28/2010 3:06:49 PM

I have a test today. The professor said that we can make thirty notecards with example problems from the material on each subject for extra points. I've spent all day at work (in between actual work) doing it. I haven't had much time to study. So, hopefully I pull one out on this test.

I have a 100% total average so far in the class. Additionally, if I do bad on this test, I can swap my grade on the final exam with my lowest test grade.

2/2/2010 1:25:32 PM

I ended up getting a 94 on my test. It was actually an 87, but I got 5 points extra credit and a 2 point curve. Apparently I was one of three people who got above a C on the test.

This one fucking question, though:

lim x-->infinity of (x-2)/(sqrt(9x^2 +1)

I knew the answer from webassign, because I had a question similar (except the bottom was 4x^2). So, I knew the answer was 1/3. However, I had solved the webassign problem with my calculator by plugging in values, so I had no idea how to prove that I knew the answer.

I wrote down some bs work and basically wrote "therefore 1/3" at the end. The guy gave me 0 out of 8 points for the question . I would have gotten a 102 on the test.

I'm going to go back on Tuesday and see if I can at least get a few points back for knowing the right answer.

2/5/2010 3:12:04 PM

Two Webassigns due tomorrow night.

SHIT.

SHIT.

2/8/2010 8:35:41 PM

If the tangent line to y = f(x) at (3,4) passes through the point (0,3), find f'(3).

2/9/2010 10:48:22 AM

4

2/9/2010 10:58:09 AM

congrats, you don't know high school level math.

2/9/2010 11:04:01 AM

This one fucking question, though:

lim x-->infinity of (x-2)/(sqrt(9x^2 +1)

As x->infin the equation reduces to x/sqrt(9X^2)

and x/(sqrt(9)*sqrt(x^2))

and x/((3)*(x))

and cancel the x's out and you get 1/3

2/9/2010 11:05:31 AM

^ no, step 2 i dont think it would reduce to that.

2/9/2010 11:13:41 AM

Quote :

"If the tangent line to y = f(x) at (3,4) passes through the point (0,3), find f'(3)."

answer is 1/3 i think

f'(3) is just the slope of the tangent line at x=3

you're given two points on that line

slope=rise/run=(4-3)/(3-0)=1/3

[Edited on February 9, 2010 at 11:25 AM. Reason : sd]

[Edited on February 9, 2010 at 11:28 AM. Reason : and (3x)(3x)=9x^2, so sqrt(9x^2)=3x]

2/9/2010 11:17:19 AM

graphing anything should be easy, just plug in numbers for one variable and get a result of the other variable

ex: Y=X^3 plug in several nums for x and get a y result

x=1 y=1
x=2 y=8
x=3 y=27

then graph those points on a graph
i won gold at the international math championships in 2006 so i know what im doing

[Edited on February 9, 2010 at 11:29 AM. Reason : ok i just realized this was three pages and this has been covered extensively, but im not deleting]

2/9/2010 11:28:52 AM

^^Problem says the tangent line ot f(x) passes through (3,4) and (0,3). The tangent line is the derivative of f(x), so f'(x)=1/3X+3.

So f'(3)=4.


Or if you're on top of your shiz, the problem says the tangent line (AKA f'(x)) passes through (3,4), so f'(3)=4 without even having to figure out f'(x)

2/9/2010 11:38:15 AM

oh shit i guess that's right

2/9/2010 11:47:24 AM

Quote :

"congrats, you don't know high school level math."

at least he's trying to learn

2/9/2010 11:56:33 AM

Quote :

"^^Problem says the tangent line ot f(x) passes through (3,4) and (0,3). The tangent line is the derivative of f(x), so f'(x)=1/3X+3. So f'(3)=4. Or if you're on top of your shiz, the problem says the tangent line (AKA f'(x)) passes through (3,4), so f'(3)=4 without even having to figure out f'(x)"

sorry, that's wrong. (both bolded things)

river rat was correct.

f(3) = 4, not f'(3) = 4.

f'(3) = slope of tangent line. (1/3)

also, there is no way to find out f'(x). [(1/3)x + 3 is the equation of the tangent line... that doesn't mean it is the derivative of the function]

2/9/2010 12:04:49 PM

I somehow got through 3 calculus classes without ever knowing what the hell I was doing. And then I decided "fuck this shit, I don't want to even know what I'm doing anyway. I'm switching majors."

2/9/2010 12:08:13 PM

I somehow got through 3 calculus classes without ever knowing what the hell I was doing. And then I decided "fuck this shit, I don't want to even know what I'm doing anyway. I'm switching majors."

2/9/2010 12:10:16 PM

If the tangent line to y=f(x) passes through (3,4), then f(3)=4 and f'(3)=4.

2/9/2010 12:11:10 PM

yes, f(3) = 4.

the derivative is equal to the SLOPE of the tangent line. that's the definition of derivative in the first place.

so, f'(3) = slope of tangent line = 1/3 [(4-3)/(3-0)]

2/9/2010 12:13:46 PM

FYI, I just plugged these into Webassign. I had already tried 4, it was incorrect. 1/3 is correct.

Thanks for the explanation though.

2/9/2010 12:16:40 PM

The way I read the problem was that f(x) passed through (3,4), and f'(x) passed through (3,4) and (0,3).

Or to re-state the problem, f'(x) passes through (3,4) and (0,3), find f'(3).

The derivative of the tangent line of f(x) (f'(x)), would be f''(x), which would be 1/3 at all points.


ninjaETA: I'd have complained about the problem statement.

2/9/2010 12:32:12 PM

^ no dude... that's a classic calculus problem, and that's how it is phrased in every book in the world.

you need to brush up.

Quote :

"f'(x) passed through (3,4) and (0,3)."

it says that no where. the tangent line passes through those 2 points, not f'(x). if you graph f'(x) against x, it will not pass through those 2 points.

2/9/2010 12:40:26 PM

^^The problem was literally copy/pasted from Webassign so... whelp

2/9/2010 12:45:28 PM

^^Apparently the phrasing has changed since I took MA141. When I took Calc, a line tangent to the curve defined by f(x) is the derivative, f'(x).

I understand that the answer that's correct per WebAssign is 1/3, which is the slope of the tangent line, but to me that's f''(x), which is true for all values of x for that line. Which makes the problem kinda stupid.

[Edited on February 9, 2010 at 1:29 PM. Reason : []

2/9/2010 1:29:03 PM

http://lmgtfy.com/?q=May+15%2C+2007+cotter548&l=1

2/9/2010 1:31:34 PM

Quote :

"When I took Calc, a line tangent to the curve defined by f(x) is the derivative, f'(x). I understand that the answer that's correct per WebAssign is 1/3, which is the slope of the tangent line, but to me that's f''(x), which is true for all values of x for that line. Which makes the problem kinda stupid."

Sorry, again you are mistaken. No one taught you that way, you just remember wrong. I will say this for the last time:

Derivative of f(x) at a certain point = SLOPE of tangent line to graph of f(x) at that point. [This is the definition of derivative, and if you are contesting this, you don't know calculus.]

2nd derivative of f(x) at a certain point = SLOPE of tangent line to graph of f'(x) at that point.

2/9/2010 1:35:13 PM

^yup

Quote :

""When I took Calc, a line tangent to the curve defined by f(x) is the derivative, f'(x). "

Think about this for a second. Take y=x. You're suggesting that the derivative is the tangent, which is also y=x.

2/9/2010 1:37:39 PM

I've worked through this problem at least 4 times, and I keep getting the same answer. I'm stuck.

If G(x) = x/(1 + 2x), find G'(a).

Obviously this is a problem that requires a lot of "busywork," so it's possible I'm making a small error somewhere.

So, using G'(a) = lim h->0 (G(a + h) - G(a)) / h

I would write this as:

top: ((a + h) / (1 + 2(a+h))) - (a / (1 + 2a))
bottom: h

Now, working just with the top, I can find a common denominator of (1 + 2a + 2h)(1 + 2a)

This means that I would combine both fractions as:

[(a + h)(1 + 2a) - a(1 + 2a + 2h)] / [(1 + 2a + 2h)(1 + 2a)]

and then when I multiply everything out, I would have:

(a + 2a^2 + h + 2ah - a - 2a^2 - 2ah) / (1 + 2a + 2a + 4a^2 + 2h + 4ah)

I can cancel stuff out, leaving me with

h / (1 + 4a + 4a^2 + 2h + 4ah)

with all of this over h.

I can then multiply by the inverse of (1/h) on both the top and bottom, cancelling out the h. This leaves me with:

1 + 4a + 4a^2 + 2h + 4ah.

Since this is a limit as h --> 0, the 2h and 4ah cancels out, leaving me with:

1 + 4a + 4a^2.

However, this is wrong according to Webassign.

2/9/2010 1:40:25 PM

Quote :

"1/(1 + 4a + 4a^2 + 2h + 4ah) Since this is a limit as h --> 0, the 2h and 4ah cancels out, leaving me with: 1/(1 + 4a + 4a^2)"

There is your correction.

Also can be written as 1/(1 + 2a)^2

2/9/2010 1:45:28 PM

Ohh, h/h equals 1, duh........ Thanks.

2/9/2010 1:46:04 PM

Quote :

" Sorry, again you are mistaken. No one taught you that way, you just remember wrong."

Yeah, that post got me straightened back out. It's been a while.

2/9/2010 1:55:15 PM

Quote :

"FYI, I just plugged these into Webassign. I had already tried 4, it was incorrect. 1/3 is correct."

2/9/2010 2:22:38 PM

My 100% Webassign average is in jeopardy. I have two Assignments due at 10pm. One is 64.1% done; the other is 87.5% done.

2/9/2010 2:55:27 PM

start postin dem questions

2/9/2010 2:56:27 PM