Recent content by AATroop

Ah, yep. I was hasty.

Right, and you also know the distance so now would be a good time to find the potential.

I got a different value for k, could you double check it? My answer was a magnitude different.

You need to find k first. What's the spring doing BEFORE the elevator starts moving?

I don't want to throw you off, but I think it might be easier to find the initial speed first then find the radius. It's a geosynchronous orbit, so what do you know about its period?

I would start with the first equation you provided. F_y = ma_y. What's the acceleration (a_y) in this problem? Where you can find a force on the system?

Well, it definitely seems like now there is an error in the answer key. So, yes, I would not follow it.

Awesome! Thanks a bunch for your help.

Yeah, I think I just got it. My new result for d^2y/dx^2 = dy^2/dv^2 * \frac{1}{x^2} - dy/dv * \frac{1}{x^2}. I think that's right because the diff eq worked out pretty well from there.

OK, can you also show how you're getting s(3)? Because, if the equation is s(t) = (2/3)t^2 + 4t^2 + 6t + 2, then at t = 3, the current position is 6 + 36 + 18 + 2. Which is equal to 62. If the equation is s(t) = (2/3)t^2 - 4t^2 + 6t + 2, then at t = 3, the current position is 6 - 36 + 18 + 2...

Can you explain how 2/3 * t^2 = 4*t^2 + 6*t + 2?

Homework Statement Solve x^{2}\times y'' - 4 \times x \times y' + 6 \times y = 0 for y(x) by first using the substitution v = ln(x) to obtain an equation involving y, dy/dv, d^2y/dv^2 and no x. Solve for y(v), then return to y(x). Homework Equations NA The Attempt at a Solution I know how...

OK, I will. Thank you very much for your help, glad to know I'm finally getting this.

Yes, I do. Awesome. Hey, if you don't mind, could you just briefly look over the next question? It's the same format, but y(x) = Cx^3 and I concluded \frac{dy}{dx} = \frac{3\times y}{x} . I just want to make sure that one is correct because the next few questions rely on it. Thanks again.

Thanks for helping me. I was working on it and I reached \frac{dy}{dx} = 3\times y^{2/3}. I found the derivative and basically just substituted.