Recent content by sporff

ahh. thank you HallsofIvy. it makes sense to me now.

thanks guys. very strange that my book would do this. he doesn't go over homogeneous for another couple of sections but he goes and gives me a problem on it. at least now that I know what it is I can go about figuring it out. but anyway... what is the reasoning behind the substitution of...

I'm almost finished my calculus book (I'm self-teaching) and in the last 2 chapters it's giving a brief intro to differential equations. the second section is for "separable" and I'm stuck on this one halfway through the exercises. It doesn't seem to be separable by any means I can see unless...

Ahh I got it now. I guess I wasn't thinking straight when you were talking about expressing cosh in terms of it's natural log equivalent. I forgot about that identity. Your help is much appreciated

Thanks for the help. You are the man. BTW, What did you mean by do you mean for taking the integral of \frac{1-\cosh 2x}{2} and expressing it as (\frac{1}{2}x\sqrt{x^2-9} - \frac{9}{2}ln[\sqrt{x^2-9}+x]) ? Excuse my newbness.

Ah I will try that out. Offhand, do you know if it yields the answer in my previous post?

For this integral... \frac{2}{3} * \int{\sqrt{x^2-9}} dx Here is this answer that my TI-89 and Wolfram's Integrator give: \frac{2}{3} * (\frac{1}{2}x\sqrt{x^2-9} - \frac{9}{2}ln[\sqrt{x^2-9}+x])

Hmm I've tried integration by parts a few different ways like tan^2 * sec (which ends up being too hard or circular and cancelling out), and also rearranging it in different ways first. And as for the cosh substitution. It requires taking the integral of sinh^2 though. I haven't been taught...

This seems like it should be easy but I can't seem to wrap my brain around it right now. I'm integrating to find the area of a hyperbola cut off by the line x=4 (I assume just the nose of the hyperbola cut off by the line on the positive side) hyperbola: \frac{x^2}{9} - \frac{y^2}{4} = 1...

I just looked at it tonight (jan 1st). Doesn't look like much for me either in my 10" dobsonian. not really any tail. It looked almost like andromeda but a more pronounced nucleus and more circular.

Has anyone seen approximate distances, in miles, from Earth that it may pass? I've been following this asteroid a little but haven't seen that info yet. The image is pretty crazy though if it's accurate.

It's cloudy here as usual. I've missed the Jupiter occultation and now this. oh well at least I should get to see comet Machholz at some point.

I've wondered this too. I love math (even though I'm still in early stages of learning it because the only knowledge i have is high school and the rest is self-taught) and someday I'd love to go to school for a degree in mathematics. It's such a beautiful subject to me that I'd love just...

will it be visible to the northern hemisphere?

thanks turbo-1. I think this 'scope will be great for me. A beginner who wants at least a little bit of power. I ended up going with the higher model 10" Dob and got the keypad too. I'm being helped with the cost here so that's why I needed to expedite my purchase a little bit. So the...