Recent content by kevtimc

Use an integral test: \int 1 / (n*ln|n|) A certain substitution should eventually lead to an antiderivative of: ln(ln(n)) I'll leave the rest to you (remember to look at what lim n->inf: ln(ln(n)) does)

Thanks, but I'm having trouble with the second integral in the problem: 2pg \int (y-1) * \sqrt{9-y^2} For: 2pg\int \sqrt{9-y^2} from 1 - 3 (when integrals are separated) doesn't that evaluate to 2pg \int9*cos^2\theta ? I'm not getting the correct answer when I evaluate this integral.

Homework Statement Find the hydrostatic force on any side of a bottom half-circle with a 6 m diameter with the top 1 m above water level. http://img377.imageshack.us/my.php?image=45095885io3.png" Homework Equations P = 1000 * 9.8 (mass density of water) * displacement A = l * w F = P * AThe...

EDIT: Nevermind, I see the image now, and I missed the T part. It's correct.

I think your theta should be flipped to the bottom. Currently, your amplitude is 150^{o}.

Darn, I thought I was going to be the first to catch that one :approve:

Yeah, first year momentum-impulse problems usually don't account for gravity. For a: I = \Delta p, p = m * v For b: I = F*\Delta t, I = \Delta p, F = \Delta p / \Delta t

ABG and DEH are congruent GBC and HEF are congruent angle ABG = angle DEH angle GBC = angle HEF AGB + GBC = DEH + HEF ABC is congruent with DEF Can you give reasons why this is? (Hint: the first is given, the second has to do with properties of...

It should be \sqrt{-63} in your first equation. You confused the heck out of me for a few minutes :wink:

[SIZE="1"]I don't know that it will be traveling at 10 m/s at the instance of the catch . . . Actually, if it follows a perfectly horizontal path, I guess it is safe to assume.

Mentallic is correct, but I would just like to add that the first boy should be removed from your thought process. You have a ball traveling at you (50 kg person) with a certain momentum (mass * velocity). It doesn't matter how it got its momentum at this point. The momentum when the ball is...

I'd just like to add that you can think of the bullet and the leopard as an inelastic collision. By the way, this is kind of a gruesome physics question :biggrin:

Sorry, I'm just getting used to all the quirks of latex. The question is to find the length using surface area revoultion formula of y = sin (pi * x) (rotated about the x-axis) from 0 - 1. \int2\pi*sin(\pi*x) * \sqrt{1 + (sin(\pi*x)')^2} I think sec^3 is actually correct. You have to...

Homework Statement y = sin \pix Using arc length and surface revoultion on x-axis 0 <= x <= 1 The Attempt at a Solution d/dx sin \pix = \pi cos \pix (\pi cos\pix)^2 = \pi^2 cos^2\pix \int sin pi * x * 2 * pi * \sqrt{1 + pi^2 * cos^2 (pi*x)} u = pi cos (pi * x) du = -pi^2 * sin...

Story of my mathematical life.