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Re: Units
*To*: mathgroup at smc.vnet.net
*Subject*: [mg23810] Re: [mg23718] Units
*From*: "Richard Finley" <rfinley at medicine.umsmed.edu>
*Date*: Sat, 10 Jun 2000 03:00:19 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
Enrique
Notice that you have your units wrong...since t is time in seconds, your acceleration should be
{3 t Meter/Second^3, t^2 Meter/Second^4}
when you enter the ODE you need to be careful of the units:
DSolve[{x''[t]==3 t Meter/Second^3,y''[t]==t^2 Meter/Second^4,
x[0 Second]==2 Meter,y[0 Second]==5 Meter,x'[1 Second]==2
Meter/Second,y'[1 Second]==3 Meter/Second },{x[t],y[t]},t]
This will give the solution:
{{ x[t] -> Meter (4 Second^3 + Second^2 t + t^3)/(2 Second^3),
y[t]-> Meter (60 Second^4 + 32 Second^3 t + t^4)/{12 Second^4) }}
Which gives the result in Meter as it should. When you plug in a value for t...say
t -> 3 Second you will get the correctly dimensioned answer.
Of course, the easier (and usual) way to do it is to just leave the units out until the end.
regards, RF
>>> "Enrique Cao" <cao at mundo-r.com> 06/04/00 11:09PM >>>
Hi:
I need help in order to solve this problem.
I have the vector aceleration {3 t Meter/Second^2,t^2 Meter/Second^2}.
This is a vector dependent of time (t).
I want to get the position vector with initial conditions {x[0]==2
Meter,y[0]==5 Meter,x'[1]==2 Meter/Second,y'[1]==3
Meter/Second}.
I do this with the follow expression:
<< Miscellaneous`Units`
<< Miscellaneous`SIUnits`
DSolve[{x''[t]==3 t Meter/Second^2,y''[t]==t^2 Meter/Second^2,
x[0]==2 Meter,y[0]==5 Meter,x'[1]==2 =
Meter/Second,y'[1]==3 Meter/Second },{x[t],y[t]},t]
The solution is
{x[t]->(Meter(4 Second^2+ 8 Second t +t^3 ))/(2 Second^2),
y[t]->Mter(3+(6 t)/Second+t^4/(12 Second^2))}
The unit of position is Meter but the solution is not correct.
How can I get the correct solution ??
Thank you very much
Henrique
cao at mundo-r.com
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