Interpreting Ambiguous HW Question on Curve Length and Surface Patch Domains

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I got this HW question made up by the professor that I find ambiguous. It says

Consider the curve \theta (t)=\pi/2-t, \phi(t)=\log \cot(\pi/4-t/2) on the sphere r(\theta,\phi)=(\sin\theta \cos\phi,\sin\theta\sin\phi,cos\theta)

Find the length of the curve btw the points t=pi/6 and pi/4


He did not specify domains for either the curve nor the "surface" r. On one hand, if we take r to be a surface patch, this requires that the (maximum) domain be 0 < \theta < \pi, 0 < \phi < 2\pi. Anything bigger and the domain is not open or r is not injective. But this surface patch does not cover the whole sphere.

I could also consider two other surface patches of the form r_{2,3}(\theta,\phi)=(\sin\theta \cos\phi,\sin\theta\sin\phi,cos\theta) with appropriate domains, that together with r above form an atlas for the unit sphere.

Any thoughts? How would you interpret this question?
 
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I don't know what you mean. The domain is pi/6<t<pi/4. From the above relations you can get r as a function of t, and this is just some curve.
 
There is another question after that:

Find the angles of intersection btw this curve and the parallels \theta = const.

would you still say that the curve's domain is (pi/6,pi/4)? Or would you study it more carefully to find what is the maximum domain where the curve is defined and thus find all the possible intersectino points?
 
I would stick to (pi/6,pi/4), at least for this question. If you're curious, keep going, but then you're doing more than what's asked.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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