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).
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...
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.