Recent content by Noober

Ok, thanks for the help everyone. Well, actually, just Gib Z, but whatever :).

Ok, with that, I got k=-1/2, and a=-3. Right...?

I did that (see above pics), and it looks right. But part C doesn't work. If I set -2ak=1, and 2a²k/3=1, a comes out to be -1/3 and k=3/2. That doesn't work for the third term...

2. I did. I made an Excel thing to find the sums, and it converged to -0.617209760262333. The actual value of cos(63º) is 0.4539905. I can't figure out what's wrong... Edit: I was doing it in degrees. I got it now. 3. Ah, thanks. Can anyone help me with Part II B and Part IV?

Homework Statement http://img87.imageshack.us/img87/6784/dsc07215xt5.jpg http://img72.imageshack.us/img72/9435/dsc07216aa0.jpg Homework Equations None. The Attempt at a Solution Answers for part I are circled, here's parts II-IV. http://img485.imageshack.us/img485/8642/dsc07217uq2.jpg...

Let y be the distance from pulley to the angle opposite 15. Then you have x^2 + y^2 = 225, 2xdx + 2ydy = 0. Then plug in? Will that work?

Well, that's the way it's done in my Calc class. You don't really need a table of values, just a numberline: <---|---> x where x is where the second derivative is 0. Just pick a random value greater than x and see if the second derivative ends up positive or negative, and do the same for...

It's a triangle: (not to scale) Pulley |\ 4| \15 |__\Truck 1| X Box With dx/dt=0.5

Surface tension.

It's basic algebra - substitute -2 for x in f(x), then substitute that for x in g(x).

If you differentiate y=\frac{x}{x-1}, you get \frac{dy}{dx}=\frac{(x-1)-x}{(x-1)^2}. Divide by \frac{x-1}{x-1}: \frac{dy}{dx}=\frac{1-\frac{x}{x-1}}{x-1} Substitute y=\frac{x}{x-1}: \frac{dy}{dx}=\frac{1-y}{x-1}

Alright, thanks. The series stuff in my textbook is confusing and unorganized, so it doesn't help that much :/.

Bad algebra, sorry.

I have an AP Calc BC exam this wednesday, and I know pretty much nothing about series (specifically Taylor/Maclaurin, errors, and some Lagrange thingy). If anyone has taken this exam, can you help me? What do we have to know about those things for the test? Which formulas do we have to know...

x^5+y^5=5 5x^4dx+5y^4dy=0 5x^4dx=-5y^4dy -\frac{5x^4}{5y^4}=\frac{dy}{dx} \frac{5y^4(-20x^3)+5x^4(20y^3)}{25y^8}=\frac{d^2y}{dx^2} \frac{d^2y}{dx^2}=\frac{100x^4y^3-100x^3y^4}{25y^8}