I had this exam question a few days ago and I still cannot figure out how to solve it:

Do note that I was taught differential geometry in a different language. If something doesn't make sense, point it out to me and I'll try to translate it better

Let the coefficients of the first fundamental form of a regular surface $X(u,v)$ be:

$$E=1+4u^2\;,\;F=\frac43uv\;,\; G=1+\frac43v^2$$

Find under which constant angle $\theta$ these two following curves intersect:

$c_1(t)=X(\cos(t),\sin(t),\; t\in(0,2\pi)$

$c_2(s)=X(s,\sqrt3s),\; s>0$

I know that $\cos(\phi)=\frac{F}{\sqrt{EG}}$, where $\phi$ is the angle between the two parametric lines(this is a possible mistranslation) of the surface, but that doesn't give me a lot of stuff to go on. I found the Gaussian Curvature to be $K_G=\frac{108}{\left(4 u^2 \left(8 v^2+9\right)+12 v^2+9\right)^2}$ using Brioschi's Formula but that didn't help either.

  • $\begingroup$ I am not sure what you mean by "intersect under constant angle". The two curves intersect in a single point (in the Parameter space this ist the intersection of a ray with the unit circle). In this intersection point, they both have a tangent vector and you can compute the angle between these two vectors in the inner given product, as described in the answer. $\endgroup$ Commented Jun 29, 2020 at 12:41

1 Answer 1


The curvature of the surface is not relevant. It's a strange question; ordinarily we ask if two surfaces intersect at a constant angle along a curve.

Here, we ask when the curves $c_1$ and $c_2$ intersect. So you need values of $t$ for which $\sin t = \sqrt3\cos t$. At such points, you want to find the angle(s) between the tangent vectors $c_1'(t)$ and $c_2'(s)$.

In terms of the parametrization, you have the tangent vectors $c_1'(t)=(-\sin t,\cos t)$ and $c_2'(s) = (1,\sqrt3)$. Use the first fundamental form and the fact that $\sin t = \sqrt3\cos t$ to find the angle between these two vectors.

  • $\begingroup$ I'm not sure I follow. Wouldn't it be $c_1'(t)=X(-\sin t, \cos t),\; c_2'(t)=X(1,\sqrt3)$? Even if it is not, I don't understand where to use the first fundamental form $\endgroup$
    – UserX
    Commented Jun 28, 2020 at 21:48
  • 1
    $\begingroup$ $E$, $F$, and $G$ allow to compute the dot products back in the parametrization space. You should review definitions! (And your formula for $\cos\phi$ is based on a calculation at these particular points, if in fact it is correct. I haven't checked.) $\endgroup$ Commented Jun 28, 2020 at 21:49
  • $\begingroup$ I've calculated the inner product of the tangent vectors to be $\cos{\theta}=E\frac{\mathrm{d}u_1}{\mathrm{d}t}\frac{\mathrm{d}u_2}{\mathrm{d}t}+F\left( \frac{\mathrm{d}u_1}{\mathrm{d}t}\frac{\mathrm{d}v_2}{\mathrm{d}t}+\frac{\mathrm{d}v_1}{\mathrm{d}t}\frac{\mathrm{d}u_2}{\mathrm{d}t}\right )+G\frac{\mathrm{d}v_1}{\mathrm{d}t}\frac{\mathrm{d}v2}{\mathrm{d}t}$ but that yields a non constant angle, dependant on u,v. Did I do this wrong? $\endgroup$
    – UserX
    Commented Jun 29, 2020 at 10:14
  • $\begingroup$ Yes, you have to divide by the product of the lengths. Recall the formula $u\cdot v = \|u\|\|v\|\cos\theta$. $\endgroup$ Commented Jun 29, 2020 at 15:54

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