Find the slope of the tangent line to the unit circle at the point where the angle with the positive $x$-axis is $ pi / 3 $.

The unit circle is defined by the equation $ x^2 + y^2 = 1 $.

At the point where the angle with the positive x-axis is $ \pi / 3 $, the coordinates are $ ( \cos (\pi / 3), \sin (\pi / 3 )) $, which simplifies to $ ( \frac{1}{2}, \frac{\sqrt{3}}{2} ) $.

The derivative of the equation $ x^2 + y^2 = 1 $ implicitly gives the slope of the tangent line. Differentiating implicitly with respect to $ x $ gives:

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

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

$ \frac{dy}{dx} = -\frac{x}{y} $

Substituting $ x = \frac{1}{2} $ and $ y = \frac{\sqrt{3}}{2} $:

$ \frac{dy}{dx} = -\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $

Therefore, the slope of the tangent line is $ -\frac{\sqrt{3}}{3} $.

The unit circle’s equation is $ x^2 + y^2 = 1 $.

The coordinates corresponding to the angle $ pi / 3 $ are $ ( cos (pi / 3), sin (pi / 3)) $, giving $ ( frac{1}{2}, frac{sqrt{3}}{2} ) $.

Using implicit differentiation on $ x^2 + y^2 = 1 $:

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

This simplifies to:

$ frac{dy}{dx} = -frac{x}{y} $

Plugging in $ x = frac{1}{2} $ and $ y = frac{sqrt{3}}{2} $ gives:

$ frac{dy}{dx} = -frac{1/2}{sqrt{3}/2} = -frac{1}{sqrt{3}} = -frac{sqrt{3}}{3} $

Therefore, the slope is $ -frac{sqrt{3}}{3} $.

For the unit circle $ x^2 + y^2 = 1 $, the point at $ pi / 3 $ is $ ( frac{1}{2}, frac{sqrt{3}}{2} ). $

Implicit differentiation gives $ frac{dy}{dx} = -frac{x}{y} $.

Thus, $ frac{dy}{dx} = -frac{1/2}{sqrt{3}/2} = -frac{sqrt{3}}{3} $.

The slope is $ -frac{sqrt{3}}{3} $.

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