0
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ Do you already have the result that a plane curve with constant positive curvature is (part of) a circle? $\endgroup$
    – amd
    Sep 11, 2019 at 4:58
  • $\begingroup$ Presumably this has yet to be proved, since it would make the problem into a coordinate geometry problem, and not a differential geometry problem. $\endgroup$ Sep 11, 2019 at 5:06
  • $\begingroup$ Yes, I have it. But I am required to use the proof of the theorem. So using the equation of a circle is not good for me $\endgroup$ Sep 11, 2019 at 5:07
  • $\begingroup$ Do you mean the existence proof in which you define $\theta(s)=\int\kappa(s)\,ds+\theta_0$ so that the curve is $\left(\int\cos\theta(s)\,ds+c,\int\sin\theta(s)\,ds+d\right)$? Just integrate and use the given conditions to determine $c$ and $d$. $\endgroup$
    – amd
    Sep 11, 2019 at 6:44

2 Answers 2

1
$\begingroup$

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 $k$.

$\endgroup$
3
  • $\begingroup$ Thank you. Just one more question, what happens with $\theta_{0}$? How can I find it? $\endgroup$ Sep 11, 2019 at 14:56
  • $\begingroup$ @vicase98 It’s arbitrary—the parameterization can have its $s=0$ point anywhere on the circle. $\endgroup$
    – amd
    Sep 11, 2019 at 16:40
  • $\begingroup$ Perfect. I really appreciate it. Thank you so much $\endgroup$ Sep 11, 2019 at 16:48
0
$\begingroup$

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

$$y=x^2+(\sqrt{3}-2)x+1-\sqrt{3}$$

$\endgroup$
4
  • $\begingroup$ Thank you for your response. But I want to solve it using the hint, maybe I should have remarked it $\endgroup$ Sep 11, 2019 at 4:49
  • 1
    $\begingroup$ Constant curvature doesn’t imply constant second derivative. A curve with constant positive curvature lies on a circle, not on a parabola. The curvature of a parabola decreases as you move away from its vertex. $\endgroup$
    – amd
    Sep 11, 2019 at 4:58
  • $\begingroup$ +1. The second derivative is a good proxy for curvature when the first derivative is small, but it is definitely not the same thing. $\endgroup$ Sep 11, 2019 at 5:03
  • $\begingroup$ It's not clear from the question that the curvature is constantly 2; it might be saying that the curvature is 2 at the point in question. (I'm sure it's not, but it could be). In that case, this solution's not so far out of line. $\endgroup$ Sep 12, 2019 at 0:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .