Recent content by formula107

Okay, here's my next problem. Assuming the slope is 2/3, which I sure hope it is, being that that is at -2, the y-intercept would be 4/3, right? So then I would have y = (2/3)x + 4/3? How do I convert this to a polar equation? Would I put -2 in for x, and 0 in for y, since that is the...

Even the book has the answer as 2/3. I had: 2cos(T)-6sin(t)cos(T) / -2sin(T) -3(cos^2(T)-sin^2(T)), with T = pi I get -2/-3 = 2/3.

Sorry man, but I'm still confused on the first part. So if I have a slope of 2/3, would I just make the equation r=2-3sin(2/3)? The original was r=2-3sin(T) Thanks again

Few stupid questions about polar curves and stuff... Okay, here's the first dumb question. I have to find the tangent line where r=2-3sin(T) at the polar point (2,pi). To find the slope, you just take dy/dx, and I come up with 2/3. I know that part is right. T=theta (I know it's spelled...

Yep, I finally got it. I end up with u^1/2, and with the numbers plugged in, I got (ln(2))^(1/2). Seems right to me! Just one more question: Is there any way to solve this problem without solving for dx before integrating. I came up with dx=xdu, since du=1/x, but we haven't done this in...

I tried bringing the bottom part to the top, so it makes \int_2^{16} \dx (2 x (\ln x)^{1/2})^{-1} With dx on the end, but I can't get that to show. Then from there I just get completely lost. Is that a good first step?

Thanks for the advice. Yep, that was the integral I was trying to describe. I have attempted it, but I still get confused on what to do with dx on the top. One more question: Can I use the same text you are by using the [tex] code? Thanks

Here's the problem: find the integral of dx/2x(lnx)^1/2 In other words, dx over 2x times the square root of lnx. It has to be evaluated from 16 and 2. I don't know how to say it, but the integral sign with 16 at the top, and 2 at the bottom. If you can find the answerm and show me...