To find a unit speed curve γ~0 : (−1, 1) with signed curvature κs equal to k and passing through the point γ(0) = 2, we can follow these steps:
1. Let's first consider a unit speed curve γ~ : (−1, 1) → R², parameterized by arc length s. This means that the velocity vector ||γ~'(s)|| = 1 for all s.
2. We can express the velocity vector as γ~'(s) = (x'(s), y'(s)). Since the curve is unit speed, we have x'(s)² + y'(s)² = 1.
3. The signed curvature κs is related to the curvature κ by the formula κs = κ/||γ~'(s)||. In this case, we want κs = k, which implies κ = k/||γ~'(s)||.
4. To find a curve with the desired signed curvature, we need to find functions x(s) and y(s) that satisfy the equations x'(s)² + y'(s)² = 1 and κ = k/||γ~'(s)||.
5. Let's assume k(s) is a given smooth function on the interval (−1, 1). We can write κ = k(s)/sqrt(x'(s)² + y'(s)²).
6. To simplify the calculations, let's assume x'(s) = cos(θ(s)) and y'(s) = sin(θ(s)), where θ(s) is a function that determines the angle of the tangent vector.
7. Plugging the assumed expressions for x'(s) and y'(s) into the equation for κ, we get κ = k(s)/sqrt(cos²(θ(s)) + sin²(θ(s))) = k(s).
8. This simplification implies that θ(s) is constant, meaning the tangent vector of the curve is constant, which implies a straight line. Since we want a curve passing through γ(0) = 2, this approach won't work.
Therefore, it seems that we cannot find a unit speed curve with the desired properties in this case.
Regarding your second request, my favorite nonzero vector ~v is (1, 0).