Recent content by Minusu

:S isn't what I'm doing integration by parts? Or do you mean separate it into two integrals? I don't think you can do that with a sqrt sign

I'd end up with sqrt(2+t^2), but then I don't know how to integrate that from 0 to pi. At first, I tried u-substitution, so u =2+t^2, du=2tdt, du/2=tdt, but there is no tdt outside the sqrt so that I can substitute.

Homework Statement Find the length of the curve traced by the given vector function on the indicated interval. r(t)=<t, tcost, tsint> ; 0<t<pi Homework Equations s= ∫||r'(t)||dt The Attempt at a Solution r'(t) = <1, -tsint + cost, tcost + sint> s= ∫||r'(t)||dt ||r'(t)|| =...

Thanks for all the help. I just solved for the angle. The tension in Tp1 and Tp2 is the same, so Fnety=Tp1sin∅+Tp2sinθ-Fg Fnety=0 0=68sin17+68sinθ-39.2 (39.2+68sin17)/68=sinθ .868=sinθ θ=sin^-1(.868) θ=60.3 degrees.

Would these two forces be the Fg on the boot and the tension of the ropes? So Fnety=-Fgboot+tp1cos∅+tp2cosθ=0 39.2-68cos17/tp2=cosθ And then solve for the x direction so Fnetx=Tp1sin17+tp2sinθ=0 68sin17/sinθ=tp2 Plug into first 39.2-68cos17/68sin17=tanθ θ=52.4

Homework Statement An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of 4.0 kg, and the doctor has decided to hang a m = 6.9kg mass from the rope. The boot is held...

http://gyazo.com/62a49a2ec489bb6662afb35a72f981b9 . So how about r^2+h^2 = c^2? But if that is the case, I'm not sure what I would do with c^2. Or should I rearrange the equation in order to combine r^2+h^2=c^2 and V=1/3*pi*r^2*h?

I've tried these problems but I just can't wrap my head around them. Thanks to anyone that can help me out here. I have a couple more problems similar to these to solve, but if I can get to understand these then I'll have no problem with the others. Here are some diagrams I drew...