Recent content by Superman1271

Ok, Thanks =]

So on the unit circle the halfway point is always along the positive x-axis. And if that is halfway, the y components will always be the negative of each other? Also how did you realize it was exactly half way?

Homework Statement Hi Everyone, I am trying to show why the given sum is zero. I am pretty sure it is zero. Homework Equations sin[8*\pi*n/5]+sin[12*\pi*n/5] n is an integer. The Attempt at a Solution n----sin[8*\pi*n/5]----sin[12*\pi*n/5] 0 ----...

I had this problem before, and got better after using suppositories. Maybe worth asking your doctor about them.

I really like vector, sounds so cool =]

When you push something at rest, what direction does it move with respect to the force you applied?

I will check it out, thanks!

m\int\frac{dv}{dt}dx =m\int\frac{dv}{dx}\frac{dx}{dt}dx =m\intv\frac{dv}{dx}dx \frac{d(1/2v^2)}{dx} =1/2*(2v)*\frac{dv}{dx}=v*\frac{dv} {dx} ∴m\intv\frac{dv}{dx}dx=m\int\frac{d(1/2v^2)}{dx}dx =\frac{1}{2}mv2! Thank you everyone for your help :).

Thanks for this, I can sort of follow it. R(x,z) is some sort of remainder function? I don't have a definition of the differential, which is my problem. The definition of the derivative which I have been using is lim h->0 \frac{f(x+h)-f(x)}{h} Could you recommend some problems to...

I do not think this is a matter style. The derivative is based on a limit, and is given the notation\frac{dy}{dx}. For you to treat them as if they were fractions is quite strange to me. How does that hold by definition? How can you isolate dv like that? It seems intuitively to be correct...

I've looked at that one. In one step they choose to do this: F\frac{dx}{dt}=mv\frac{dv}{dt} F dx=mv dv They are clearly using differentials as fractions.

That is interesting. How does the chain rule allow you to manipulate differentials as fractions.

My equations are 1 dimensional, force is acting in the positive or negative x-axis. Isn't that a scalar equation from Newtons's equation of motion? Work is defined as the sum of all contributions of force in the direction of motion multiplied by their respective distances? I'm sorry I don't see...

Yes, this is how I was taught to derive this theorem. I still hope for something more, since I have not looked at non-standard analysis and have not defined differentials in a rigorous way.

Homework Statement Trying to derive the work-energy theorem, without manipulating differentials.Homework Equations a=\frac{dv}{dt} v=\frac{dx}{dt} W=\int F dx =ΔKE=\frac{1}{2}mvf^{2}-mvi^{2} The Attempt at a Solution F=ma \int F dx=m\inta dx =m\int\frac{dv}{dt}dx <-- I cannot continue...