Integrating a line tangent to the unit circle We all know the derivative of the unit circle is $-\frac{x}{y}$. When I apply this derivative to $60^\circ$ $(1/2, \sqrt{\frac{3}{2}})$, you get the equation: $y=-1/3*\sqrt{3}x+1/3(2*\sqrt{3})$ which is all great and fine, until it is integrated, which gives us a completely different result than the original $x^2+y^2=1$. So why is that?
 A: I assume by 'differentiating the unit circle' you mean differentiating the equation
$$x^2+y^2=1$$ with respect to the independent variable $x$.
If we do so, we get (by implicit differentiation) 
$$\frac{dy}{dx} = -\frac{x}{y},$$
a formula for the derivative which depends on both $x$ and $y$.
When you say 'apply the derivative to 60 degrees', it means 'find the slope of the line tangent to the circle at the point $(1/2, \sqrt {3/2})$.
Then the equation of the tangent line is 
$$y=-\frac{\sqrt3}{3}x+\frac{2\sqrt3}{3}. $$
Now you shouldn't expect to get the equation of the circle by integrating the equation of the tangent line. The tangent line is NOT the derivative of the circle; it is just a line tangent to the circle.
If you really want to recover the equation of the circle from the derivative, you should solve the differential equation
$$xdx=-ydy.$$
This is a separable equation, so integrating with respect to the corresponding variables yields
$$\frac{x^2}{2}=-\frac{y^2}{2}+C$$ for some constant $C$.
Since the point $(0,1)$ is on the circle, we get $C=1/2$.
And so finally, we get back
$$x^2+y^2=1.$$
