How do I find the unit tangent vector with what I have x =2cos(t) , y=2sin(t), z =2t
The formula is T(t)=r'(t)/lr'(t)l. Show steps so I know.
r = <2cost,2sint,2t>
r' = <-2sint,2cost,2>
|r'| = √(4sin^2t + 4cos^2t + 4) = √8
ThankYou
To find the unit tangent vector T(t) of a curve given by the position vector r(t), you will need to follow these steps:
Step 1: Find the derivative of the position vector r(t) with respect to t, denoted as r'(t). This will give you the velocity vector.
Given:
x = 2cos(t)
y = 2sin(t)
z = 2t
Differentiating each component with respect to t:
dx/dt = -2sin(t)
dy/dt = 2cos(t)
dz/dt = 2
Therefore, the velocity vector r'(t) is:
r'(t) = (-2sin(t), 2cos(t), 2)
Step 2: Calculate the magnitude (length) of the velocity vector. The magnitude is given by the formula:
|v| = √(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2
Substituting the values of dx/dt, dy/dt, and dz/dt:
|v| = √((-2sin(t))^2 + (2cos(t))^2 + 2^2)
= √(4sin^2(t) + 4cos^2(t) + 4)
= √(4(sin^2(t) + cos^2(t)) + 4)
= √(4 + 4)
= √8
= 2√2
Step 3: Divide the velocity vector r'(t) by its magnitude |v| to obtain the unit tangent vector T(t):
T(t) = r'(t) / |v|
Substituting the values:
T(t) = (-2sin(t), 2cos(t), 2) / (2√2)
= (-sin(t)/√2, cos(t)/√2, 1/√2)
Therefore, the unit tangent vector T(t) is (-sin(t)/√2, cos(t)/√2, 1/√2).