Equation of oscillating circle

sakkid95
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http://gyazo.com/bc7f6da4d4c4aca300bb5efb2410fc8d.png


The problem, and my solution are in the image. I need help on finding the equation of the osculating circle, I found radius but I don't know where to go from there. Also if you could just check if my math seems correct that would help me too, but my main problem is how I find the eqn for the circle... thanks in advance
 
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You made a mistake differentiating t3.
You have the radius of curvature; from the tangent you can find the normal vector.
 
Yea I noticed the error, I'm working on fixing it now. Can you elaborate more on how to find the radius? I'm not very good at this stuff..
 
sakkid95 said:
Yea I noticed the error, I'm working on fixing it now. Can you elaborate more on how to find the radius? I'm not very good at this stuff..

You said you had the radius. If so, all you need is a line the circle's centre must lie on. The normal to the curve is that line.
 

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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