- 12 Oct '09 00:44

Are you really that dense? You can't see that after 4 seconds the thing is going EXACTLY 0.5 C? And you can't figure the dilation with just the simple formula we all know?*Originally posted by voltaire***A satellite is accelerating at a rate of 37474057.25 meters/second. Find the rate of change of the satellite's time dilation when time=4 seconds.**

I need to solve this problem using derivatives and obviously rates of change - 12 Oct '09 01:23

thats not the point queermo. i made up the problem for a calc project and none of them will know*Originally posted by sonhouse***Are you really that dense? You can't see that after 4 seconds the thing is going EXACTLY 0.5 C? And you can't figure the dilation with just the simple formula we all know?** - 12 Oct '09 04:35

well for starters, a function or relatioinship is needed in order to differentiate?*Originally posted by voltaire***A satellite is accelerating at a rate of 37474057.25 meters/second. Find the rate of change of the satellite's time dilation when time=4 seconds.**

I need to solve this problem using derivatives and obviously rates of change

so what is the relationship between time dialation and constant acceleration? what are its variables , what are its constants?ect......

once you have the formula relating the afformentioned quantities you need to find its first derivative, and evaluate it at t=4 (s) - 12 Oct '09 06:31 / 1 edit

I calculusate that when dilation=10cm better have some diapers, a car seat, and be ready for little to no sleep for about 6 weeks.*Originally posted by voltaire***A satellite is accelerating at a rate of 37474057.25 meters/second. Find the rate of change of the satellite's time dilation when time=4 seconds.**

I need to solve this problem using derivatives and obviously rates of change - 12 Oct '09 16:16 / 1 edit

Yeah, sorry got a bit overboard there, was just thinking of the easy solution, I think at 0.5C it turns out to be about a 20% dilation if memory serves, haven't done that calc in years but I know that closer to C for spacecraft, the dilation isn't much help till you get to around 0.99C to help you still be alive after a long interstellar journey. Even at that velocity it's not all a bed of roses, not like you could go on a thousand light year journey, at that velocity it would still only be good for the nearest stars, say those within 20 light years. Better than nothing though! I'll have to find my notes to see exactly what it turns out to be, if I remember right, at 0.9C it's only about 3 times C effectively and I think at 0.99C around ten C, that is to say you could go on a 10 LY trip and spend a year at the star of choice and come back and only about 3 years would go by on your craft but 21 years would go by on Earth.*Originally posted by joe shmo***Easy there sonhouse, noone makes fun of you when you ask questions.**

Of course that ignores the effects of slamming into interstellar H2 at that velocity and the amount of shielding it would take to survive such an onslaught, just talking about the actual velocity effects. - 12 Oct '09 18:00

t'=t0/sqrt(1-(v/c)2)*Originally posted by joe shmo***well for starters, a function or relatioinship is needed in order to differentiate?**

so what is the relationship between time dialation and constant acceleration? what are its variables , what are its constants?ect......

once you have the formula relating the afformentioned quantities you need to find its first derivative, and evaluate it at t=4 (s) - 13 Oct '09 02:50

ok, but I think the equation is*Originally posted by voltaire***t'=t0/sqrt(1-(v/c)2)**

t' = t0/(sqrt(1-(v/c)^2)

you had

t'=t0/sqrt(1-(v/c)2)

the "^" means "to the power of".

anyhow, the above relationship is to velocity, and you have acceleration in your problem.

so by substitution

v = a*(t0)

so

t' = t0/(sqrt(1-((a*t0)/c)^2)

so now time dialation is a function of the initial time.

so the equation looks like

t' = t0/(sqrt(1-[(a/c)^2]*t0^2)

lets let (for the sake of clarity) (a/c)^2 = b

now the equation becomes

t' = t0/(sqrt(1-b*(t0)^2)

now find the rate of change of t' with respect to t0 (ie dt'/dt0)

you'll need to use the product rule, or the quotient rule ( whichever floats your boat)in combination with the chain rule to arrive at

dt'/dt0 = 1/[(1-b*(t0)^2)^(3/2)]

then just evaluate it a t0 = 4(s), as long as everything is in standard units that is the result you desire to obtain.

any questions about the actual differentiation processes, and calculation, ask

later - 14 Oct '09 03:33

yeah i know the answer i was just making sure it was workable. .3849 is what i worked it out to be. this is just for a project where we make up a problem to assign, supposed to be difficult, creative etc.*Originally posted by joe shmo***ok, but I think the equation is**

t' = t0/(sqrt(1-(v/c)^2)

you had

t'=t0/sqrt(1-(v/c)2)

the "^" means "to the power of".

anyhow, the above relationship is to velocity, and you have acceleration in your problem.

so by substitution

v = a*(t0)

so

t' = t0/(sqrt(1-((a*t0)/c)^2)

so now time dialation is a function of the initial time. ...[text shortened]... questions about the actual differentiation processes, and calculation, ask

later