Find the curve whose curvature is 2, passes through the point \(\displaystyle (1,0)\) and whose tangent vector at \(\displaystyle (1,0)\) is \(\displaystyle [1/2 , \sqrt{3}/2 ].\)
I know I can apply the fundamental theorem of plane curves here. How will I be able to do that?
Also, I know there is a basic way other than using the fundamental theorem of plane curves since it gave us the curvature and tangent vector. Since the curvature is a constant I can conclude it is a circle thus the equation of a circle with radius r is \(\displaystyle (x-a)^2 + (y-b)^2 = r^2.\) But how is knowing the curvature, the value and tangent vector at (1,0) sufficient to construct the curve and how will I apply it?
I know I can apply the fundamental theorem of plane curves here. How will I be able to do that?
Also, I know there is a basic way other than using the fundamental theorem of plane curves since it gave us the curvature and tangent vector. Since the curvature is a constant I can conclude it is a circle thus the equation of a circle with radius r is \(\displaystyle (x-a)^2 + (y-b)^2 = r^2.\) But how is knowing the curvature, the value and tangent vector at (1,0) sufficient to construct the curve and how will I apply it?
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