Find the curve whose signed curvature is $2$, pass through the point $(1,0)$ and whose tangent vector at $(1,0)$ is $(\frac{1}{2}, \frac{\sqrt{3}}{2})$. I know that I have to use the proof of the fundamental theorem of plane curves.
By this moment, I have an expression for the arc length parameterization using the formula of the proof and using that the signed curvature is 2 in order to find the function used inside de cosine and the sine. How can I use the information about the tangent vector?
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From the given curvature, we have $$y'' = 2$$, which implies that curve has the form
$$y=x^2+bx+c$$
whose first derivative is
$$y'= 2x+b$$
At the point (1,0), $y'(1)$ should match the slope which is known from the tangent vector as $\sqrt{3}$, i.e.
$$y'(1) = 2+b = \sqrt{3}$$
So, the curve becomes,
$$y=x^2+(\sqrt{3}-2)x+c$$
Since the curve passes through (1,0), we have
$$0=1+(\sqrt{3}-2)+c$$
$$c= 1-\sqrt{3}$$
Therefore, the curve is
$$y=x^2+(\sqrt{3}-2)x+1-\sqrt{3}$$
Following the proof, define $$\theta(s) = \int \kappa\,ds = 2s+\theta_0.$$ An arclength parameterization of the curve is then $$\gamma(s) = \left(\int\cos{\theta(s)},\int\sin{\theta(s)}\right) = \left(\frac12\sin(2s+\theta_0)+h,-\frac12\cos(2s+\theta_0)+k\right).$$ From your question I gather that you’re able to get this far on your own. It’s not terribly hard to derive the implicit Cartesian equation $(x-h)^2+(y-k)^2=\frac14$, the equation of a circle centered at $(h,k)$, from this parameterization.
Now use the given conditions to determine the unknown constants of integration: From the tangent condition, we have $$\gamma'(s) = \left(\cos(2s+\theta_0),\sin(2s+\theta_0)\right) = \left(\frac{\sqrt3}2,\frac12\right),$$ so $2s+\theta_0=\frac\pi6+2k\pi$. Substitute into $\gamma(s)$ and set equal to $(1,0)$ to find $h$ and $k$.