Determine the equation of the tangent line to the unit circle at the point $ left( frac{1}{2}, frac{sqrt{3}}{2}
ight) $

The equation of the unit circle is given by:

$ x^2 + y^2 = 1 $

To find the tangent line at the point $\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$, we need to determine the slope. Differentiating implicitly:

$ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 $

At $\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$, the slope $m$ is:

$ m = – \frac{x}{y} = – \frac{ \frac{1}{2} }{ \frac{\sqrt{3}}{2} } = – \frac{1}{\sqrt{3}} = – \frac{\sqrt{3}}{3} $

The equation of the tangent line is:

$ y – \frac{\sqrt{3}}{2} = – \frac{\sqrt{3}}{3} \left( x – \frac{1}{2} \right) $

Simplifying the equation:

$ y = – \frac{\sqrt{3}}{3} x + \frac{\sqrt{3}}{6} + \frac{\sqrt{3}}{2} $

$ y = – \frac{\sqrt{3}}{3} x + \frac{2\sqrt{3}}{3} $

The unit circle is defined by:

$ x^2 + y^2 = 1 $

To find the slope of the tangent line at $left( frac{1}{2}, frac{sqrt{3}}{2}
ight)$, we differentiate implicitly:

$ 2x + 2y frac{dy}{dx} = 0 $

Solving for $frac{dy}{dx}$:

$ frac{dy}{dx} = – frac{x}{y} = – frac{ frac{1}{2} }{ frac{sqrt{3}}{2} } = – frac{1}{sqrt{3}} = – frac{sqrt{3}}{3} $

The equation of the tangent line is:

$ y – frac{sqrt{3}}{2} = – frac{sqrt{3}}{3} left( x – frac{1}{2}
ight) $

The slope of the tangent line at $left( frac{1}{2}, frac{sqrt{3}}{2}
ight)$ on the unit circle is:

$ – frac{x}{y} = – frac{1}{sqrt{3}} = – frac{sqrt{3}}{3} $

The tangent line equation is:

$ y – frac{sqrt{3}}{2} = – frac{sqrt{3}}{3} left( x – frac{1}{2}
ight) $

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