Recent content by DavidWi

Ok, right now in my college physics class, we're going over work in a circular motion. I see how gravity does the amount of work that the thing falls in a circular motion. However, my teacher told us that the tension of the string does no work. Gravity makes the ball go down, so gravity does...

wow, finally i see the right answer. Now, I wonder if I can get my calc teacher with that problem. I don't think he'll be able to do it. Maybe he will though.

OHHHHH! I GOT IT! Maybe... Maybe... I forgot to take into account my domain. I'm going from pi / 2 to 3 pi / 2. But when i solved for the arcsin 0, i thought it was a 0 radians, which is out of my domain. arcsin 0 = pi, not 0 sin (c / 2) = 0 c / 2 = arcsin 0 c / 2 = pi c = 2 * pi...

no, i meant sin (3 pi / 4) because it's sin (x / 2). if x = 3 pi / 2, then it would be sin (3 pi / 2 / 2) = sin (3 pi / 4)

ohhhh, you're right! wow! I guess I got too caught up in the integral mean value theorum. But there's the much easier to do differential one. f'(c) = (sin (3 pi / 4) - sin (pi / 4)) / pi f'(c) = (sqrt (2) / 2 - sqrt (2) / 2) / pi f'(c) = 0 / pi = 0 sin c / 2 = 0 c / 2 = arcsin 0 c...

The five answers are (A) 2 pi / 3, (B) 3 pi / 4, (C) 5 pi / 6 (D) pi, (E) 3 pi / 2 I can't figure out a way to simplify arcsin (2 sqrt (2) / pi), so I don't think it's any of these answers.

soooo... what you guys are saying is that I should set (2 * sqrt (2)) / pi equal to my function and solve for x... 2 * sqrt (2) / pi = sin (x / 2) x / 2 = arcsin (2 * sqrt (2) / pi) x = 2 * arcsin (2 * sqrt (2) / pi) But this still isn't any of my answers... I'm still doing something...

As far as I know, the constant of integration only applies to taking the indefinate integral. Since I'm taking the definate integral, the constant of integration cancels out. Expalin the rest though...

I don't know... I've worked through this problem about ten times, but each time, I get the same answer. It's a kind of easy problem too. I don't konw what my problem with this is. If f(x) = sin (x/2), then there exists a number c in the interval pi/2 < x < (3*pi / 2 that satisfies the...

Ok, I was thinking today during my calculus class about taking the integral of a function in a different way. Let's assume for a second that we want to find the area between the function and the y axis, on the interval x = [0, 2] of the function y = 2x. What I was thinking I could do, is...