Question

A particle moves along a curve whose parameter equations are

x=e−tx=e−t

,

y=2cos3ty=2cos⁡3t

,

z=2sin3tz=2sin⁡3t

. Find the magnitude of the acceleration at t=0

Answer 1

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 approx.

Answer 2

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 approx.