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For Physics Gurus Only by adebayo18015(m): 5:58am On Oct 27, 2020
a particle moves along a curve whose parametric equations are.

x=e^-t, y=2cos3t, z=2sin3t {where t is the time }



a. determine its velocity and acceleration at any time



b. find the magnitude of the velocity and acceleration at t=0
Re: For Physics Gurus Only by NDSMELODY(m): 7:19am On Oct 27, 2020
adebayo18015:
a particle moves along a curve whose parametric equations are.

x=e^-t, y=2cos3t, z=2sin3t {where t is the time }



a. determine its velocity and acceleration at any time



b. find the magnitude of the velocity and acceleration at t=0

Re: For Physics Gurus Only by Aluminumquadri(m): 7:22am On Oct 27, 2020
x=e^-t, y=2cos3t, z=2sin3t, where x, y, z are positional vectors dependent on t. x'=-e^-t, y'=-6sin3t, z'=6cos3t where ' denotes d/dt (speed/ velocity). x"=e^-t, y"=-18cos3t, z"=-18sin3t (acceleration). When t=0 these are x"=1, y"=-18, z"=0. The magnitude of the acceleration denoted by " is: √(x"^2+y"^2+z"^2)=√ (1+324)=√325=5√13=18.0278
Re: For Physics Gurus Only by Aluminumquadri(m): 7:24am On Oct 27, 2020
x=e^-t, y=2cos3t, z=2sin3t, where x, y, z are positional vectors dependent on t. x'=-e^-t, y'=-6sin3t, z'=6cos3t where ' denotes d/dt (speed/ velocity). x"=e^-t, y"=-18cos3t, z"=-18sin3t (acceleration). When t=0 these are x"=1, y"=-18, z"=0. The magnitude of the acceleration denoted by " is: √(x"^2+y"^2+z"^2)=√ (1+324)=√325=5√13=18.0278
Re: For Physics Gurus Only by NDSMELODY(m): 7:25am On Oct 27, 2020
Last part

Re: For Physics Gurus Only by adebayo18015(m): 6:52pm On Oct 30, 2020
Aluminumquadri:
x=e^-t, y=2cos3t, z=2sin3t, where x, y, z are positional vectors dependent on t. x'=-e^-t, y'=-6sin3t, z'=6cos3t where ' denotes d/dt (speed/ velocity). x"=e^-t, y"=-18cos3t, z"=-18sin3t (acceleration). When t=0 these are x"=1, y"=-18, z"=0. The magnitude of the acceleration denoted by " is: √(x"^2+y"^2+z"^2)=√ (1+324)=√325=5√13=18.0278
A very big thank you for this
Re: For Physics Gurus Only by adebayo18015(m): 6:52pm On Oct 30, 2020
NDSMELODY:
Last part
Thank you for this
Re: For Physics Gurus Only by adebayo18015(m): 6:53pm On Oct 30, 2020
[quote author=NDSMELODY post=95381356][/quote] Thanks for this

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