Recent content by FabioTTT

thanks Doc Al. I did this, dividing the three regions of the velocity graph and calculated the area beneath each of those regions. I came up with 215 meters but the online assignment checker counted it as wrong so I am a little confused.

I can't for the life of me figure this out. The question is, based on this graph, determine the total distance the particle traveled during the first 20 seconds. http://seminario.no-ip.com/graph.gif I graphed the velocity graph just fine based on the acceleration graph. but the position...

wow, nevermind. I'm such a tool. I didnt realize all that under the sqrt came out to be ( 3t^2 + 1 ) ^2

I have to integrate: sqrt( (3t^2)^2 + (1)^2 + (sqrt(6)t)^2 ) at the limits t=1 and t=3 I can't figure out how to simplify the problem so that I can do the integral. So far the farthest I've gotten is the obvious: sqrt( 9t^4 + 6t^2 + 1 ) now I am stuck.

oh ok, thanks

can i ask what programs you guys use to write out your mathematical formulas? i realize they are images.. so you must be making them some where. edit: by the way, thanks chen

question, i understand the way you worked this, however, why would i have to calculate the velocities separately for each axis? wouldnt the position vector be r(t) = acos(wt)i + asin(wt)j + btk ? then i would just do dr/dt to find the velocity vector wouldn't i? and dv/dt to find the...

thanks. I'm aware of how to obtain the velocity, speed and acceleration vectors, however I just didnt know why he would include a,b,w. doesn't make much sense to me.

Thanks for the info... it's funny this problem is nothing like any of teh other problems I've encountered in this section of my math textbook.

http://seminario.no-ip.com/problem2.gif I'm confused about the above problem because I don't know why there would be an a,b, or greek letter "w" in there. Anyone?

yep that's what i got. thanks

im an idiot.. of course its valid.. its just an angle. thanks guys. i guess this thread was pretty pointless ;)

(z^2) / 2 = tan(2theta) if sin(2theta) / cos(2theta) = tan(2theta) is valid... is it?

i know what youre hinting at that i could divide both sides of the equation by 2 and then sub x and y for r cos(theta) and r sin(theta) and then cancel the r^2 on both sides and have sin(2theta) left.. but that would leave me with (z^2 cos(2theta)) / 2 = sin(2theta) is that correct though...

yeah i realized i could use cos(2theta) from the previous problem but I am still stuck since i still have the 4xy on the other side of the equation :(