Recent content by hmvince

f''(x) = k(x-1)(x-3) = kx^{2} - 4kx + 3k and f''(x) = 12ax^{2} + 6bx + 2c (differentiating from general formula) ∴12a = k ∴6b = 4k ∴2c = 3k

Either way I do it, if I work back from the general form, I have unknowns on both sides, the k in front of one lot, and the a, b, and c in front of the second derivative of the general form. Please help me Solving for A, b, and c I get: k = 12a k = -1.5b k = (2/3)c

Thankyou, and sorry, I should have explained. a stationary point of inflection is a point of inflection at which the rate of change is zero. Like this: http://mathworld.wolfram.com/images/eps-gif/InflectionPoint_700.gif I've been sitting here trying to sketch a quartic with two of these points...

Hey everyone! Recently got a question in maths which asks: "Use integral calculus to find the equation of the quartic that has stationary points of inflection at (1, 23) and (3, 15) and a y-intercept of 24" This means that the second derivative has the form (as inflection points are...

Thanks for the reply, In (d) the charged particle would be accelerating (would this produce X-rays, or is that only decelerating electrons?). as for (e) I suppose there would have to be a switch, what causes this pulse of radiation when it is switched? As for a, b, and c, would you say they are...

Homework Statement Which of the following will produce Electromagnetic radiation? Explain Your answers. a) Alternating current in a TV transmitter b) The cord of a vacuum cleaner while it is being used c) The circuit of a battery operated CD player d) A positively charged particle...

Thankyou :D

Just to make sure I'm doing things right: Alpha Centauri is located 4 light years away. A spacecraft travels at 0.8c towards this star, as seen from earth. How long will the spacecraft take to reach its destination? The spacecraft sees the distance to Alpha Centauri as 2.4 light...

which once graphed, shows that optimum time is to go as fast as possible.

sweet as guys. Thanks for all your help! So back to the equation at the start and finding minimum time (observers reference frame) I would use the equation (to find minimum time over a distance of say 100km): t'=tγ since t = D/V and distance(moving reference frame) = 100000/γ t' =...

Ok, now we're getting somewhere! :biggrin: I have just had a quick read through this thread: https://www.physicsforums.com/showthread.php?t=599015 and shatteredmerc seems to be thinking pretty much exactly the same thing as me! reading through the last few posts in this thread and the posts in...

precisely. I changed tγ to (distance/velocity)γ as t = (distance/velocity). That is the time it takes the rocket to travel from A to B, which is calculated from the distance IT has to travel / its velocity. Yes. Yes. And is t'=tγ incorrect? I got this equation from...

I used the spaceship's distance/velocity Is that incorrect? Please explain why, I really don't understand.

to get minimum time (according to the rest frame) I used this formula: where γ is the lorentz factor: t' = tγ t' = (distance/velocity)γ graphing that you get a minimum at v = 0.707c

Optimum for getting from point A to point B in the least time possible (from the rest reference frame) By balance I mean getting the least time with t = distance/velocity while keeping the effect of time dilation (as the faster the rocket travels, the more time is going to pass in the rest...