To find the unit normal vector, first find the unit tangent vector. Finding the unit tangent vector requires differentiating each component of a vector function, as well as the length of the vector function. Finally, use the derivative of the unit tangent vector over the length of the unit tangent vector to determine the unit normal vector.
Continue ReadingFirst, differentiate the given vector function: r(t) = cos ti + sin tj + tk. The derivative of this function is r'(t) = -sin ti + cos tj + k. The vector function is the point where t = 0 is: r'(0) = j + k.
The length of the vector function is found by determining the square root of the square of each directional coefficient in the derivative: |r'(t)| = sqrt(1^2 + 1^2) = sqrt(2). Putting all of these elements together, the unit tangent vector is: T(t) = r'(t) / |r'(t)| = (1/sqrt(2)) * (-sin ti + cos tj + k).
Find the derivative of the unit tangent vector: T'(t) = (1/sqrt(2)) * (-cos ti - sin tj). Setting this to zero gives the result: T'(0) = -1/sqrt(2).
For the unit tangent vector, the length is: |T'(t)| = (1/sqrt(2)).
The unit normal vector is the unit tangent vector over the length of the unit tangent vector: N(t) = T'(t)/|T'(t)| = -cos ti - sin tj.