Find T, N, B, k, aT, aN for the motion
R(t) = 3cost i + 4t j + 3sint k
A Slinky?
a circle radius 3 in the xz plane moving at constant velocity in the y direction?
surely your differential geometry book has formulas for all of these vectors.
I know google does.
To find the values of T, N, B, k, aT, and aN for the given motion, we need to use the formulas for the unit tangent vector, unit normal vector, unit binormal vector, curvature, tangential acceleration, and normal acceleration.
Given the position vector R(t) = 3cos(t) i + 4t j + 3sin(t) k, we can calculate the components of velocity and acceleration.
1. Velocity vector V(t):
To find the velocity vector, we differentiate the position vector with respect to time (t):
V(t) = dR(t)/dt = -3sin(t) i + 4 j + 3cos(t) k
2. Acceleration vector a(t):
Taking the derivative of the velocity vector with respect to time gives us the acceleration vector:
a(t) = dV(t)/dt = -3cos(t) i + 0 j - 3sin(t) k
3. Magnitude of velocity vector V:
To find the magnitude of the velocity vector (V), we use the formula:
|V| = sqrt(Vx^2 + Vy^2 + Vz^2)
where Vx, Vy, and Vz are the x, y, and z components of the velocity vector.
Using V(t) = -3sin(t) i + 4 j + 3cos(t) k, we can calculate the magnitude of V.
|V(t)| = sqrt((-3sin(t))^2 + (4)^2 + (3cos(t))^2)
= sqrt(9sin^2(t) + 16 + 9cos^2(t))
= sqrt(9(sin^2(t) + cos^2(t)) + 16)
= sqrt(9 + 16)
= sqrt(25)
= 5
4. Unit tangent vector T(t):
The unit tangent vector (T) is the normalized velocity vector divided by its magnitude:
T(t) = V(t) / |V(t)|
= (-3sin(t) i + 4 j + 3cos(t) k) / 5
= (-3/5)sin(t) i + (4/5) j + (3/5)cos(t) k
5. Curvature (k):
The curvature (k) is the magnitude of the derivative of the unit tangent vector with respect to arc length (s):
k = |dT(s)/ds|
In our case, since we don't have the arc length parameter, we can use:
k = |dT(t)/dt| / |V(t)|
To calculate k, we need to derive T(t) with respect to t and find its magnitude:
dT(t)/dt = d(-3/5)sin(t) i + d(4/5) j + d(3/5)cos(t) k
= (-3/5)cos(t) i + 0 j + (-3/5)sin(t) k
|dT(t)/dt| = sqrt[ [(-3/5)cos(t)]^2+ [(-3/5)sin(t)]^2 ]
= sqrt[ (9/25)(cos^2(t)+sin^2(t)) ]
= sqrt(9/25)
= 3/5
Finally, we calculate k:
k = (3/5) / 5
= 3/25
6. Unit normal vector N(t):
The unit normal vector (N) is the normalized derivative of the unit tangent vector with respect to time (t):
N(t) = dT(t)/dt / |dT(t)/dt|
Using the previously calculated values, we can find N:
N(t) = [(-3/5)cos(t) i + 0 j + (-3/5)sin(t) k] / (3/5)
= -cos(t) i + 0 j - sin(t) k
7. Unit binormal vector B(t):
The unit binormal vector (B) is found using the cross product of the unit tangent vector T and unit normal vector N:
B(t) = T(t) × N(t)
Taking the cross product:
B(t) = (-3/5)sin(t) i + (4/5) j + (3/5)cos(t) k × -cos(t) i + 0 j - sin(t) k
= -4/5 cos(t) i - 3/5 sin(t) j + (-3/5 cos(t) sin(t)) k
8. Tangential acceleration aT(t):
To find the tangential acceleration, we need to calculate the derivative of the velocity vector V(t) with respect to time (t):
aT(t) = dV(t)/dt = -3cos(t) i + 0 j - 3sin(t) k
9. Normal acceleration aN(t):
The normal acceleration is given by the formula:
aN(t) = ||a(t)|| - |aT(t)|
First, we calculate the magnitude of the acceleration vector:
||a(t)|| = sqrt[ (-3cos(t))^2 + 0^2 + (-3sin(t))^2 ]
= sqrt[ 9(cos^2(t) + sin^2(t)) ]
= 3
Next, we calculate the magnitude of the tangential acceleration:
|aT(t)| = sqrt((-3cos(t))^2 + 0^2 + (-3sin(t))^2)
= sqrt(9(cos^2(t) + sin^2(t)))
= 3
Finally, we find the normal acceleration:
aN(t) = ||a(t)|| - |aT(t)|
= 3 - 3
= 0
In summary, for the given motion:
T(t) = (-3/5)sin(t) i + (4/5) j + (3/5)cos(t) k
N(t) = -cos(t) i + 0 j - sin(t) k
B(t) = -4/5 cos(t) i - 3/5 sin(t) j + (-3/5 cos(t) sin(t)) k
k = 3/25
aT(t) = -3cos(t) i + 0 j - 3sin(t) k
aN(t) = 0