Recent content by MacLaddy1

Thanks again.

So it still is wrt "t." Could you use the implicit formula to figure this out. $$-\frac{F_t}{F_y}?$$ I suppose not.

Yes, I believe I can. However, before I try would you mind showing me why that first y is canceled and the second isn't? If you could do it in Leibniz notation I would be even more appreciative. I get mixed up when dealing with Newton's notation. Thanks, Mac ** Just tried anyway. That made...

Hello all, This is probably a simple question, but for some reason no matter how much I fight these figures I can't seem to make it work out correctly. $$y' = -r(1-\frac{y}{T})y$$ $$y' = -ry + \frac{ry^2}{T}$$ $$y'' = -ryy' + \frac{2ryy'}{T}$$ $$0 = -ryy' + \frac{2ryy'}{T}$$ $$ryy' =...

Ahh, yes. $ \frac{dy}{dx} $ was a nearly-midnight-brain-... Well, it was mis-typed.Piece of cake, thank you very much for your help. The book vaguely suggested doing it this way, but I didn't understand the way it was presented. $$ \frac{du}{dt} = \frac{1}{t} $$ $$ \int{du} =...

Hello all, it's been a long time. Hoping I can get some assistance with what is probably a simple substitution problem, yet it's flummoxing me. $$\frac{dy}{dx} = \frac{y+t}{t}$$ I've tried substituting $$ v = y+t $$ $$ y = v - t $$ $$ \frac{dy}{dx} = \frac{dv}{dt} - 1 $$ $$\frac{dv}{dt}-1 =...

Thanks, Sudharaka. This is one of those problems that when finally worked through to the end, I end up kicking myself for making it far more difficult then it really was. It's an off week, I think. Thanks all, Mac

Alright, not sure if anyone is still looking at this problem of mine, but I think I may have figured out the final solution. However, there is a bit of a trick I am doing here that I do not know if it's valid.$\displaystyle \int \int_{S} f(x,y,z)\ dS = 2\ \int \int_{D} (x^{2}+y^{2})\ \sqrt...

I have solved the integral the following way. $$\int_0^{2\pi}\int_0^2 (\sqrt{5})rdrd\theta = 4\pi\sqrt{5}$$ Thanks everyone for their contributions.

Just a quick observation, \(A=\int_{z=0}^4 \int_{\theta=0}^{2\pi} \frac{\sqrt{5}}{2} z d\theta dz \neq 4 \pi \sqrt{5}\)

Actually, I did draw a picture, still didn't help. Thanks anyway. Mac

Actually, no. Now I'm doubly confused. Why does your limit of integration $\rho$ have 0 to $2\pi$ as it's limits, and $d\theta$ have 0 to 2? Isn't $\rho$ just the radius of the paraboloid $0 \leq \rho \leq 2$? And $d\theta$ is the circle $0 \leq \theta \leq 2\pi$? Doesn't the height, $0 \leq z...

Ahh, good. So I was on the right track. It looks like I'm just using r where your using $\rho$. Now I'll have to figure out how to evaluate that integral. Thanks, Mac *EDIT* Your integral has different limits of integration. I'll have to dig into it and see if I can figure out why.

Thanks Captain, I appreciate the assistance. I do have a couple of follow up questions, as I'm still not quite grasping this as well as I should. I never actually took geometry, which is funny considering I'm now in Multivariate Calculus, but I haven't really needed it up to this point. So...

Here is another that I am stuck on. Please doublecheck my work, and let me know if where I am stuck is correct, or if I am on the completely wrong path. Evaluate the surface integral \(\int\int f(x,y,z)dS\) using an explicit representation of the surface. \(f(x,y,z) = x^2 + y^2;\mbox{ S is...