Final Answer:
The unit normal vector n^ is given by 1/√(1+(h(x))²) (-h'(x) i^ + j^), and the unit tangent vector τ^ is given by 1/√(1+(h(x))²) (i^ + h'(x) j^). To express i^ and j^ as linear combinations of n^ and τ^, we find i^ = -h'(x)/√(1+(h(x))²) n^ + 1/√(1+(h(x))²) τ^ and j^ = 1/√(1+(h(x))²) n^ + h'(x)/√(1+(h(x))²) τ^.
Step-by-step explanation:
To derive the expressions for n^ and τ^, we begin with the definition of the unit normal and unit tangent vectors for a curve defined by a vector function r(x) = ⟨x, h(x)⟩. The unit normal n^ is obtained by normalizing the derivative of r with respect to x, and τ^ is obtained by normalizing r'(x).
The computations involve finding the derivatives and magnitudes, leading to the given expressions for n^ and τ^. To express i^ and j^ in terms of n^ and τ^, we solve the resulting equations. The coefficients obtained provide the linear combinations.
These expressions offer a geometric interpretation of the unit normal and unit tangent vectors in terms of the parameterized curve. The linear combinations demonstrate how the standard basis vectors i^ and j^ can be represented using the vectors n^ and τ^ associated with the curve