Determine the unit tangent vector

chwala
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Homework Statement
see attached
Relevant Equations
vector differentiation
1644910436015.png


I need a justification that ##|\dfrac {dr}{dt}|##=##\dfrac {ds}{dt}## cheers guys... all the other steps are easy and clear to me...
 
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chwala said:
Homework Statement:: see attached
Relevant Equations:: vector differentiation

View attachment 297138

I need a justification that [Modulus ##\big | \dfrac {dr}{dt} \big |##=##\dfrac {ds}{dt}##] cheers guys... all the other steps are easy and clear to me...
We have ##ds \equiv |d\vec r|##.
 
Aaaaah Perok cheers!
PeroK said:
We have ##ds \equiv |d\vec r|##.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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