Find the length of the curve given by the parametric representation

tamintl
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Find the length of the curve given by the parametric representation...

Homework Statement


Calculate the length of the curve given by the parametric representation
r(t) = t2(cos t; sin t; cos 2t; sin 2t) for 1≤ t ≤+1:


Homework Equations





The Attempt at a Solution



I know that you need to assume: dx/dt ≥ 0 for α≤t≤β

Then you use the formula for 'L'

Imstruggling with the layout of the question.. Why are there semi-colons between the sin and cos terms?

If someone could explain this that would be great.

Regards as always.
 
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There are probably semi-colons to distinguish between the four different dimensions. So x(t) = t^2 cos(t), y(t) = t^2 sint, etc.
 


I presume you know that the length of the curve given by r(t), from t= a to t= b, is
\int_a^b ||r'(t)|| dt
where r'(t) is the tangent vector and the || || is the length of that vector.
 
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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