Recent content by samoth

What kind of weight lifting do you do? Just casual? I really recommend looking into powerlifting. Not so much as to push heavy weights around, but more because PLing really concentrates on your 'core', specifically, your lower back. I hear so many people, young and old, comment about back...

Actually, I am a competitive powerlifter, so I wasn't implying anything 'cheesy' with posing. If I had better upper body symmetry, I would be oiled up on stage myself. You have decent stats... but taking yourself down to 180 will be a lot of hard work if you plan to do so whilst conserving...

What BF are you at now? How tall are you? I am unsure if you were being sarcastic about the bodybuilder part, but if you weren't, 12% won't look very good oiled up.

Ah-ha! That's where things weren't making sense in what I was doing. Ok, I managed to solve for v. The equation v approaches c like it is supposed to. However , how is the equation related to the slope of a line? I believe the professor (also) wants an equation which can be inputed into...

The professor states that there is an exponential involved because the function of the line on the v versus t graph asymptotically approaches c, compared to the linear function (given to us to be a=F/m) which goes unbounded past c. So that means I need a function of...

The more I try to do this, the more confused I get! So, you are saying to integrate \frac{F}{m} dt = d\Big(\frac{v}{\sqrt{1-v^2/c^2}}\Big) and that each side is an integral of one, which should give \int \frac{F}{m} dt = \int a dt = at on the left hand side, and the right hand...

The professor solved for the Newtonian v... however, I don't know why exactly. I am trying to solve for the function that gives the asymptotic line, which is linear and a=F/m classically.

This is an extra-credit problem. The question covers materials some students may not have had prior, hence not being mandatory. Question posed: Force is given by F=dp/dt F=d/dt (mv) F=m dv/dt F=ma and a=F/m is the slope of the line on a v versus t graph. This does not take into...

Please do, as I am quite curious now. Good luck as well with your course!

Yes, A is symmetric when A^T=A. First of all, I was wrong about the eigenvectors of A and A^T being the same. They are not. However, I cannot help as to why, as our text offers only two sentences in this matter. Further, I do not believe I am at a level of knowledge upon which speculation...

Don't know how much help I can be, but since I am studying the same material at the moment, I will help with what I can. The eigenvalues of A are equal to the eigenvalues of A^T because det(A-λI)=det(A^T-λI). The diagonal/trace stays the same here. (I am guessing on the next...