Find the normal and tangential components for acceleration for an object with a position vector r(t)= <cosπt + πtsinπt, sinπt - πtcosπt, t> at t=1,2, and 3.
To find the normal and tangential components of acceleration, we first need to find the velocity and acceleration vectors of the object by differentiating the position vector with respect to time.
r(t) = <cosπt + πtsinπt, sinπt - πtcosπt, t>
v(t) = dr/dt = <-πtsinπt + πcosπt, πtcosπt + πsinπt, 1>
a(t) = dv/dt = <-π^2tcosπt - π^2sinπt, -π^2tsinπt + π^2cosπt, 0>
Now, we can find the normal component of acceleration by projecting the acceleration vector onto the unit normal vector, which is the unit vector in the direction of the acceleration vector's change in direction:
n(t) = a(t) - proj_(a(t))_T
where proj_(a(t))_T is the projection of a(t) onto the tangent vector T(t), which is the unit vector in the direction of the velocity vector:
T(t) = v(t)/||v(t)||
||v(t)|| = √(π^4t^2 + π^2) [magnitude of velocity vector]
T(1) = <-πsinπ + cosπ, πcosπ + sinπ, 1>/√(π^4 + π^2) = <-1/√(2), (π+1)/√(2), 1/√(π^2 + 1)>
T(2) = <-2πsin2π + cos2π, 2πcos2π + sin2π, 1>/√(16π^2 + π^2) = <-1/√(17), (2π+1)/√(17), 1/√(17π^2 + 1)>
T(3) = <-3πsin3π + cos3π, 3πcos3π + sin3π, 1>/√(81π^2 + π^2) = <-1/√(82), (3π+1)/√(82), 1/√(81π^2 + 1)>
To find the projection of a(t) onto T(t), we can use the dot product:
proj_(a(t))_T = a(t) · T(t)
proj_(a(1))_T = <-π^2cosπ - πsinπ, -π^2sinπ + πcosπ, 0> · <-1/√(2), (π+1)/√(2), 1/√(π^2 + 1)> = -π/√(2)
proj_(a(2))_T = <-4π^2cos2π - 2πsin2π, -4π^2sin2π + 2πcos2π, 0> · <-1/√(17), (2π+1)/√(17), 1/√(17π^2 + 1)> = -4π/√(17)
proj_(a(3))_T = <-9π^2cos3π - 3πsin3π, -9π^2sin3π + 3πcos3π, 0> · <-1/√(82), (3π+1)/√(82), 1/√(81π^2 + 1)> = -9π/√(82)
Therefore:
n(1) = <-π^2cosπ - πsinπ, -π^2sinπ + πcosπ, 0> - (-π/√(2)) <-1/√(2), (π+1)/√(2), 1/√(π^2 + 1)> = <-π/2, -π/2, 0>
n(2) = <-4π^2cos2π - 2πsin2π, -4π^2sin2π + 2πcos2π, 0> - (-4π/√(17)) <-1/√(17), (2π+1)/√(17), 1/√(17π^2 + 1)> = <-8π/17, -8π/17, 0>
n(3) = <-9π^2cos3π - 3πsin3π, -9π^2sin3π + 3πcos3π, 0> - (-9π/√(82)) <-1/√(82), (3π+1)/√(82), 1/√(81π^2 + 1)> = <-9π/41, -9π/41, 0>
These are the normal components of acceleration at t=1,2, and 3. The tangential components of acceleration can be found by taking the projection of the acceleration vector onto the tangent vector:
t(t) = proj_(a(t))_T
t(1) = -π/√(2) <-1/√(2), (π+1)/√(2), 1/√(π^2 + 1)> = <-π/(2√2), π/(2√2) + π/2, π/(2√(π^2 + 1))>
t(2) = -4π/√(17) <-1/√(17), (2π+1)/√(17), 1/√(17π^2 + 1)> = <-4π/(17√17), 4π/(17√17) + (4π+2)/17, 4π/(17√(17π^2 + 1))>
t(3) = -9π/√(82) <-1/√(82), (3π+1)/√(82), 1/√(81π^2 + 1)> = <-9π/(41√82), 9π/(41√82) + (3π+1)/41, 9π/(41√(81π^2 + 1))>
These are the tangential components of acceleration at t=1,2, and 3.