[Solved] Unwanted minus sign in this derivative of a circular curve

nomadreid
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Homework Statement
Find the derivative dy/dx of the curve (circle) expressed in parametric form: (x,y)=(cos t, sin t).
Relevant Equations
dy/dx of (x(t), y(t))= y'(t)/x'(t)
y'(t)/x'(t) = cos t/-sin t = -cot t.
But as t is the angle, and the derivative is the slope, then isn't the slope supposed to be tan t?
 
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Ah, oops. Thanks, phinds. I was mistakenly thinking of the slope of the line from the origin to the point, but now I see that this was a silly mistake. Problem solved, thread closed.
 
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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