Parametric Surface Grid Curves: Solving for Tangent Vectors and Angle

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Homework Statement



Consider the parametric surface r(u, v) = <vsin(u), vcos(u), v^2>

The point (1, 1, 2) is on this surface. Find the grid curve with v constant that contains this point.

And the grid curve with u constant that contains the point.

Then find tangent vector to both grid curves at (1, 1, 2).

Find the angle between both grid curves at (1, 1, 2).

Some other stuff I can't even think about right now follows...

The Attempt at a Solution



I figured the grid curve with v constant is <sqrt(2)sin(u), sqrt(2)cos(u), sqrt(2)^2>

But then I couldn't get the one for U being constant, and this question still doesn't make too much sense to me in the first place...help!
 
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Yes, in order that (v sin(u), vcos(u), v2) pass through (1, 1, 2), v must be \sqrt{2}. Therefore the curve through that point so that v is constant is, of course, (\sqrt{2} sin(u), \sqrt{2} cos(u), 2).

Now that we have established that v must be \sqrt{2} at the point (1, 1, 2), for what value of u is (\sqrt{2} sin(u), \sqrt{2} cos(u), 2)= (1, 1, 2)?
 
pi/4. Thanks.
 
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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