Find the tangential and normal components of the acceleration vector.
r(t) = t i + t2 j + 4t k
aT=
aN=
just use the definitions in your text. Simple derivatives.
To find the tangential and normal components of the acceleration vector, we need to differentiate the position vector twice with respect to time.
First, let's find the velocity vector v(t) by differentiating r(t) with respect to time:
v(t) = dr(t)/dt = d/dt (t i + t^2 j + 4t k)
v(t) = 1 i + 2t j + 4 k
Next, let's differentiate the velocity vector v(t) to find the acceleration vector a(t):
a(t) = dv(t)/dt = d/dt (1 i + 2t j + 4 k)
a(t) = 0 i + 2 j + 0 k
We now have the acceleration vector a(t) = 0 i + 2 j + 0 k.
To find the tangential component of the acceleration (aT), we need to project the acceleration vector onto the direction of the velocity vector.
The tangential component of the acceleration aT is given by:
aT = (a(t) dot v(t)) / ||v(t)||
where "dot" represents the dot product and "|| ||" represents the magnitude.
Taking the dot product, we have:
a(t) dot v(t) = (0)(1) + (2)(2t) + (0)(4)
a(t) dot v(t) = 4t
The magnitude of the velocity vector is:
||v(t)|| = sqrt(1^2 + (2t)^2 + 4^2)
||v(t)|| = sqrt(1 + 4t^2 + 16)
||v(t)|| = sqrt(4t^2 + 17)
Substituting these values into the equation for aT, we have:
aT = (4t) / sqrt(4t^2 + 17)
Now, to find the normal component of the acceleration (aN), we need to subtract the tangential component from the total acceleration vector.
aN = a(t) - aT
Since a(t) = 0 i + 2 j + 0 k, and we already found aT to be (4t) / sqrt(4t^2 + 17), we can substitute these values into the equation:
aN = 0 i + 2 j + 0 k - ((4t) / sqrt(4t^2 + 17))