- #1
chetzread
- 801
- 1
chetzread said:
My ans is [1/ sqrt (2) ] [ (cos t -sin t ) i + (sin t + cos t ) j +k)PeroK said:
No, your answer is wrong. I get the same answer that you showed in post #1.chetzread said:
The unit tangent vector is a vector that is tangent to a curve at a specific point and has a magnitude of 1. It represents the direction and rate of change of the curve at that point.
To find the unit tangent vector, you first need to find the derivative of the curve. Then, divide the derivative by its magnitude to get a vector with a magnitude of 1. This vector is the unit tangent vector.
Yes, the unit tangent vector can change at different points on a curve. This is because the direction and rate of change of the curve can vary at different points, resulting in a different unit tangent vector.
The unit tangent vector and the velocity vector are closely related. The unit tangent vector is the normalized version of the velocity vector, which means it has a magnitude of 1. They both represent the direction and rate of change of the curve at a specific point.
Yes, the unit tangent vector can be used to find the curvature of a curve. The curvature is equal to the magnitude of the derivative of the unit tangent vector with respect to the parameter t.
