Recent content by Gibybo

Best I could do was a MS Paint recreation: http://www.warcraftcheese.com/math/polar%20area.jpg Note that the loop on the right should be pedal shaped, does not dip below the x-axis on the interval [0,pi/2], and is 6sin(2t). The question says inside both, so the idea is to calculate the inner...

[SOLVED] Area inside Polar Curves Homework Statement I have spent several hours beating myself up over this and I just can't seem to solve it. It's the only problem I haven't gotten correct and it is particularly frustrating. Can you math gods here save me? :) Find the area of the region...

Ah hah! Thanks :) [hits self on head]

I agree, but why is it wrong when I do? I think we can agree that there are no algebraic mistakes there. Why do I HAVE to pull them out for #2?

I'm not sure what you're saying. I shouldn't pull out constants? Why is it wrong to pull them out then? And how do you correctly do #2 without pulling out constants first?

Right, I did both of them both ways. If you do not pull out the constants, you get the correct answer for #1. However, #2 is not (1/2)*ln(2x), it is ln(x)/2. Summary: Pull out constants: #1 wrong, #2 right Do NOT pull out constants: #1 right, #2 wrong

Oh sorry, just an error while converting to [tex] images I think. -ln(x-2)/4+c is incorrect though, which is my problem.

Ah hah, that's how you make it pretty! Anyway yeah, I have \frac{-1}{4}\int\frac{1}{x-2} which is equivalent and still valid, right?

Yeah, by integrate(4x,x) I would mean 4x*dx. It's standard notation when typing integrals for calculators/CAS's etc, I did it that way because I was thinking the forums might automatically convert it to the proper mathematical symbols.

[SOLVED] Integration, u substitution, 1/u -- +C at the end of the integral solutions, I can't seem to add it in the LaTeX thing -- Homework Statement #1 \int\frac{1}{8-4x}dx #2 \int\frac{1}{2x}dx The Attempt at a Solution #1 Rewrite algebraically: \int\frac{1}{x-2}*\frac{-1}{4}dx Pull out...

That works great if air resistance was negligible, but this is a ping pong ball. It flies slightly better than a balloon. Air resistance is the real problem here, not gravity. I'm very competitive with this sort of thing and have thought of at least 20 different designs from gravity powered...

I doubt he had any plans for us to solve the trajectory at all, it's a fairly basic physics class and this type of air resistance calculation is way beyond what we've ever done. I've looked for computer programs to help solve this (I'm sure they exist) but haven't had any luck finding one. The...

I'm in a high school AP physics class but I suspect this question is considerably above the course level because I don't think it was intended to be a part of the project, so I'm not sure which forum to put it in. I would like to find the initial velocity required to launch a ping pong ball 25...