Given the equation y + Ij = 22 + y^2, we can rewrite it as:
F(x, y) = (0)i + (22 + y^2)j
Now, let's find the position vector r(t) for the circle C:
r(t) = (cos(t))i + (sin(t))j
A. To compute OT, we need to find the tangent vector T at a point on the circle. We
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