Answer:
The curvature of the plan curve at t=2 is 0.
Step-by-step explanation:
First, we find take a derivative of r(t) = ti + t^2j
r'(t) = i + 2tj
we can also write it like r'(t) = < 1, 2t>
Now, we take a magnitude of it
|r'(t)| = √(1)^2 + (2t)^2
|r'(t)| = √17
Now, we assume a variable T(t)
T(t) = r'(t)/|r'(t)|
T(t) = (i + 2tj) / √17
T(t) = 1/√17 (< 1, 2t>)
Take derivative of T(t)
T'(t) = 1/√17 (< 0, 2>)
|T'(t)| = √(1/√17)^2 + (0)^2 + (2)^2
|T'(t)| = √69/17
Therefore,
Curvature = |T'(t)| / |r'(t)|
Curvature = (√69/17) / √17
Curvature = √69 .
