Recent content by elvishatcher

You are not trying to find \frac{dr}{dt} . Each part of the question wants you to find \frac{dV}{dt} for a given rate at which the radius is changing (think about what represents the rate of change of the radius).

Glad I could help - if you ever figure out some reason why I'm wrong and it should be k^2 not 4k^2, let me know

To find your limits, think about the description of the area A. It is the area enclosed by the function and the region y >= 0. Picture this area on a graph (make a sketch if it helps), and it should be pretty clear that the integral should go from x=-3 to x=3. Using the other method, your...

Well, it makes sense to me that it would be (2k+1)^2. Since you're summing up to 2k+2, going backwards from the final term you would have (2k+2)^2 then (2k+2 - 1)^2 then (2k+2 - 2)^2, etc. which would make the second to last term (2k+1)^2 and not (k+1)^2. What doesn't make sense to me is that it...

Ah, that makes a lot more sense. I'm no expert, but it seems to me that you're correct about the problem being due to ∞ - ∞ not being a defined value. It seems like what you do to get the two different results is solid. Infinity's a kinda weird thing to deal with, and a lot of its complexities...

I don't understand the part where you replace a_i with S_n - S_{n-1} and then put that into the original sum. It seems to me like you're doing something meaningless here as the sum takes values of i from 1 to n and you are plugging in S_n - s_{n-1} That is, you're taking the sum of that...

I think you made a mistake in determining the integrand. What you have for the dy/dx is not what I got when I differentiated the function (you also write the function as containing x(1-x)^2 initially and then when you rewrite it in terms of y you have an x(1-x^2) but I believe I got something...

Ah, I hadn't thought about the periodicity before. Thanks

Very interesting - thanks for the info!

Yep, seems like that oughta work

Oh dear, how embarassing. What a silly mistake. However, now that I think about it, while that may have been a silly arithmetic mistake, it is true that e^{2i\pi} = 1 which leads to essentially the same question I had originally.

Oh dear, how embarassing. What a silly mistake. However, now that I think about it, while that may have been a silly arithmetic mistake, it is true that e^{2i\pi} = 1 which leads to essentially the same question I had originally.

It seems like you could solve this by first manipulating to get y' + \frac{Bx-C}{A}y = 0 at which point you now have an ODE of the form y' + P(x)y = Q(x) and there is a general way to solve such ODEs.

So, I was thinking about Euler's formula, and I noticed something interesting. Based on the fact that e^\frac{i\pi}{2} = 1 , it seems as though \frac{i\pi}{2} = 0. However, this doesn't make any sense. Not only can I not see how this expression could possibly equal 0, but that would imply that...

Ah, I see it now. I was getting a little mixed up with the y's and the t's, and I wasn't thinking about the fact that once it settles into an equilibrium value it's not going to change anymore. Thanks.